I want to parse strings to get fields from them. The format of the string (which come from a dataset) is as so (the -> represents a tab, and the * represents a space):
Date(yyyymmdd)->Date(yyyymmdd)->*City,*State*-->Description
I am only interested in the 1st date and the State. I tried regex like this:
String txt="19951010 19951011 Red City, WI Description";
String re1="(\\d+)"; // Integer Number 1
String re2=".*?"; // Non-greedy match on filler
String re3="(?:[a-z][a-z]+)"; // Uninteresting: word
String re4=".*?"; // Non-greedy match on filler
String re5="(?:[a-z][a-z]+)"; // Uninteresting: word
String re6=".*?"; // Non-greedy match on filler
String re7="((?:[a-z][a-z]+))"; // Word 1
Pattern p = Pattern.compile(re1+re2+re3+re4+re5+re6+re7,Pattern.CASE_INSENSITIVE | Pattern.DOTALL);
Matcher m = p.matcher(txt);
if (m.find())
{
String int1=m.group(1);
String word1=m.group(2);
System.out.print("("+int1.toString()+")"+"("+word1.toString()+")"+"\n");
}
It works fine id the city has two words (Red City) then the State is extracted properly, but if the City only has one word it does not work. I can't figure it out, I don't need to use regex and am open to any other suggestions. Thanks.
Problem:
Your problem is that each component of your current regex essentially matches a number or [a-z] word, separated by anything that isn't [a-z], which includes commas. So your parts for a two word city are:
Input:
19951010 19951011 Red City, WI Description
Your components:
String re1="(\\d+)"; // Integer Number 1
String re2=".*?"; // Non-greedy match on filler
String re3="(?:[a-z][a-z]+)"; // Uninteresting: word
String re4=".*?"; // Non-greedy match on filler
String re5="(?:[a-z][a-z]+)"; // Uninteresting: word
String re6=".*?"; // Non-greedy match on filler
String re7="((?:[a-z][a-z]+))"; // Word 1
What they match:
re1: "19951010"
re2: " 19951011 "
re3: "Red" (stops at non-letter, e.g. whitespace)
re4: " "
re5: "City" (stops at non-letter, e.g. the comma)
re6: ", " (stops at word character)
re7: "WI"
But with a one-word city:
Input:
19951010 19951011 Pittsburgh, PA Description
What they match:
re1: "19951010"
re2: " 19951011 "
re3: "Pittsburgh" (stops at non-letter, e.g. the comma)
re4: ","
re5: "PA" (stops at non-letter, e.g. whitespace)
re6: " " (stops at word character)
re7: "Description" (but you want this to be the state)
Solution:
You should do two things. First, simplify your regex a bit; you are going kind of crazy specifying greedy vs. reluctant, etc. Just use greedy patterns. Second, think about the simplest way to express your rules.
Your rules really are:
Date
A bunch of characters that aren't a comma (including second date and city name).
A comma.
State (one word).
So build a regex that sticks to that. You can, as you are doing now, take a shortcut by skipping the second number, but note that you do lose support for cities that start with numbers (which probably won't happen). Also you don't care about the state. So, e.g.:
String re1 = "(\\d+)"; // match first number
String re2 = "[^,]*"; // skip everything thats not a comma
String re3 = ","; // skip the comma
String re4 = "[\\s]*"; // skip whitespace
String re5 = "([a-z]+)"; // match letters (state)
String regex = re1 + re2 + re3 + re4 + re5;
There are other options as well, but I personally find regular expressions to be very straightforward for things like this. You could use various combinations of split(), as other posters have detailed. You could directly look for commas and whitespace with indexOf() and pull out substrings. You could even convince a Scanner or perhaps a StringTokenizer or StreamTokenizer to work for you. However, regular expressions exist to solve problems like this and are a good tool for the job.
Here is an example with StringTokenizer:
StringTokenizer t = new StringTokenizer(txt, " \t");
String date = t.nextToken();
t.nextToken(); // skip second date
t.nextToken(","); // change delimiter to comma and skip city
t.nextToken(" \t"); // back to whitespace and skip comma
String state = t.nextToken();
Still, I feel a regex expresses the rules more cleanly.
By the way, for future debugging, sometimes it helps to just print out all of the capture groups, this can give you insight into what is matching what. A good technique is to put every component of your regex in a capture group temporarily, then print them all out.
no need to be so complex with this. you can split on whitespace!
//s is your string
String[] first = s.split("\\s*,\\s*")
String[] firstHalf = first[0].split("\\s+")
String[] secondHalf = first[1].split("\\s+")
String date = firstHalf[0]
String state = secondHalf[0]
now you have youre date and your state! do with them what you want.
Related
(Disclaimer: the title of this question is probably too generic and not helpful to future readers having the same issue. Probably, it's just because I can't phrase it properly that I've not been able to find anything yet to solve my issue... I engage in modifying the title, or just close the question once someone will have helped me to figure out what the real problem is :) ).
High level description
I receive a string in input that contains two information of my interest:
A version name, which is 3.1.build and something else later
A build id, which is somenumbers-somenumbers-eitherwordsornumbers-somenumbers
I need to extract them separately.
More details about the inputs
I have an input which may come in 4 different ways:
Sample 1: v3.1.build.dev.12345.team 12345-12345-cici-12345 (the spaces in between are some \t first, and some whitespaces then).
Sample 2: v3.1.build.dev.12345.team 12345-12345-12345-12345 (this is very similar than the first example, except that in the second part, we only have numbers and -, no alphabetic characters).
Sample 3:
v3.1.build.dev.12345.team
12345-12345-cici-12345
(the above is very similar to sample 1, except that instead of \t and whitespaces, there's just a new line.
Sample 4:
v3.1.build.dev.12345.team
12345-12345-12345-12345
(same than above, with only digits and dashes in the second line).
Please note that in sample 3 and sample 4, there are some trailing spaces after both strings (not visible here).
To sum up, these are the 4 possible inputs:
String str1 = "v3.1.build.dev.12345.team\t\t\t\t\t 12345-12345-cici-12345";
String str2 = "v3.1.build.dev.12345.team\t\t\t\t\t 12345-12345-12345-12345";
String str3 = "v3.1.build.dev.12345.team \n12345-12345-cici-12345 ";
String str4 = "v3.1.build.dev.12345.team \n12345-12345-12345-12345 ";
My code currently
I have written the following code to extract the information I need (here reporting only relevant, please visit the fiddle link to have a complete and runnable example):
String versionPattern = "^.+[\\s]";
String buildIdPattern = "[\\s].+";
Pattern pVersion = Pattern.compile(versionPattern);
Pattern pBuildId = Pattern.compile(buildIdPattern);
for (String str : possibilities) {
Matcher mVersion = pVersion.matcher(str);
Matcher mBuildId = pBuildId.matcher(str);
while(mVersion.find()) {
System.out.println("Version found: \"" + mVersion.group(0).replaceAll("\\s", "") + "\"");
}
while (mBuildId.find()) {
System.out.println("Build-id found: \"" + mBuildId.group(0).replaceAll("\\s", "") + "\"");
}
}
The issue I'm facing
The above code works, pretty much. However, in the Sample 3 and Sample 4 (those where the build-id is separated by the version with a \n), I'm getting two matches: the first, is just a "", the second is the one I wish.
I don't feel this code is stable, and I think I'm doing something wrong with the regex pattern to match the build-id:
String buildIdPattern = "[\\s].+";
Does anyone have some ideas in order to exclude the first empty match on the build-id for sample 3 and 4, while keeping all the other matches?
Or some better way to write the regexs themselves (I'm open to improvements, not a big expert of regex)?
Based on your description it looks like your data is in form
NonWhiteSpaces whiteSpaces NonWhiteSpaces (optionalWhiteSpaces)
and you want to get only NonWhiteSpaces parts.
This can be achieved in numerous ways. One of them would be to trim() your string to get rid of potential trailing whitespaces and then split on the whitespaces (there should now only be in the middle of string). Something like
String[] arr = data.trim().split("\\s+");// \s also represents line separators like \n \r
String version = arr[0];
String buildID = arr[1];
(^v\w.+)\s+(\d+-\d+-\w+-\d+)\s*
It will capture 2 groups. One will capture the first section (v3.1.build.dev.12345.team), the second gets the last section (12345-12345-cici-12345)
It breaks down like: (^v\w.+) ensures that the string starts with a v, then captures all characters that are a number or letter (stopping on white space tabs etc.) \s+ matches any white space or tabs/newlines etc. as many times as it can. (\d+-\d+-\w+-\d+) this reads it in, ensuring that it conforms to your specified formatting. Note that this will still read in the dashes, making it easier for you to split the string after to get the information you need. If you want you could even make these their own capture groups making it even easier to get your info.
Then it ends with \s* just to make sure it doesn't get messed up by trailing white space. It uses * instead of + because we don't want it to break if there's no trailing white space.
I think this would be strong for production (aside from the fact that the strings cannot begin with any white-space - which is fixable, but I wasn't sure if it's what you're going for).
public class Other {
static String patternStr = "^([\\S]{1,})([\\s]{1,})(.*)";
static String str1 = "v3.1.build.dev.12345.team\t\t\t\t\t 12345-12345-cici-12345";
static String str2 = "v3.1.build.dev.12345.team\t\t\t\t\t 12345-12345-12345-12345";
static String str3 = "v3.1.build.dev.12345.team \n12345-12345-cici-12345 ";
static String str4 = "v3.1.build.dev.12345.team \n12345-12345-12345-12345 ";
static Pattern pattern = Pattern.compile(patternStr);
public static void main(String[] args) {
List<String> possibilities = Arrays.asList(str1, str2, str3, str4);
for (String str : possibilities) {
Matcher matcher = pattern.matcher(str);
if (matcher.find()) {
System.out.println("Version found: \"" + matcher.group(1).replaceAll("\\s", "") + "\"");
System.out.println("Some whitespace found: \"" + matcher.group(2).replaceAll("\\s", "") + "\"");
System.out.println("Build-id found: \"" + matcher.group(3).replaceAll("\\s", "") + "\"");
} else {
System.out.println("Pattern NOT found");
}
System.out.println();
}
}
}
Imo, it looks very similar to your original code. In case the regex doesn't look familiar to you, I'll explain what's going on.
Capital S in [\\S] basically means match everything except for [\\s]. .+ worked well in your case, but all it is really saying is match anything that isn't empty - even a whitespace. This is not necessarily bad, but would be troublesome if you ever had to modify the regex.
{1,} simple means one or more occurrences. {1,2}, to give another example, would be 1 or 2 occurrences. FYI, + usually means 0 or 1 occurrences (maybe not in Java) and * means one or more occurrences.
The parentheses denote groups. The entire match is group 0. When you add parentheses, the order from left to right represent group 1 .. group N. So what I did was combine your patterns using groups, separated by one or more occurrences of whitespace. (.*) is used for group 2, since that group can have both whitespace and non-whitespace, as long as it doesn't begin with whitespace.
If you have any questions feel free to ask. For the record, your current code is fine if you just add '+' to the buildId pattern: [\\s]+.+.
Without that, your regex is saying: match the whitespace that is followed by no characters or a single character. Since all of your whitespace is followed by more whitespace, you matching just a single whitespace.
TLDR;
Use the pattern ^(v\\S+)\\s+(\\S+), where the capture-groups capture the version and build respectively, here's the complete snippet:
String unitPattern ="^(v\\S+)\\s+(\\S+)";
Pattern pattern = Pattern.compile(unitPattern);
for (String str : possibilities) {
System.out.println("Analyzing \"" + str + "\"");
Matcher matcher = pattern.matcher(str);
while(matcher.find()) {
System.out.println("Version found: \"" + matcher.group(1) + "\"");
System.out.println("Build-id found: \"" + matcher.group(2) + "\"");
}
}
Fiddle to try it.
Nitty Gritties
Reason for the empty lines in the output
It's because of how the Matcher class interprets the .; The . DOES NOT match newlines, it stops matching just before the \n. For that you need to add the flag Pattern.DOTALL using Pattern.compile(String pattern, int flags).
An attempt
But even with Pattern.DOTALL, you'll still not be able to match, because of the way you have defined the pattern. A better approach is to match the full build and version as a unit and then extract the necessary parts.
^(v\\S+)\\s+(\\S+)
This does trick where :
^(v\\S+) defines the starting of the unit and also captures version information
\\s+ matches the tabs, new line, spaces etc
(\\S+) captures the final contiguous build id
I'm new to regular expressions and would appreciate your help. I'm trying to put together an expression that will split the example string using all spaces that are not surrounded by single or double quotes. My last attempt looks like this: (?!") and isn't quite working. It's splitting on the space before the quote.
Example input:
This is a string that "will be" highlighted when your 'regular expression' matches something.
Desired output:
This
is
a
string
that
will be
highlighted
when
your
regular expression
matches
something.
Note that "will be" and 'regular expression' retain the space between the words.
I don't understand why all the others are proposing such complex regular expressions or such long code. Essentially, you want to grab two kinds of things from your string: sequences of characters that aren't spaces or quotes, and sequences of characters that begin and end with a quote, with no quotes in between, for two kinds of quotes. You can easily match those things with this regular expression:
[^\s"']+|"([^"]*)"|'([^']*)'
I added the capturing groups because you don't want the quotes in the list.
This Java code builds the list, adding the capturing group if it matched to exclude the quotes, and adding the overall regex match if the capturing group didn't match (an unquoted word was matched).
List<String> matchList = new ArrayList<String>();
Pattern regex = Pattern.compile("[^\\s\"']+|\"([^\"]*)\"|'([^']*)'");
Matcher regexMatcher = regex.matcher(subjectString);
while (regexMatcher.find()) {
if (regexMatcher.group(1) != null) {
// Add double-quoted string without the quotes
matchList.add(regexMatcher.group(1));
} else if (regexMatcher.group(2) != null) {
// Add single-quoted string without the quotes
matchList.add(regexMatcher.group(2));
} else {
// Add unquoted word
matchList.add(regexMatcher.group());
}
}
If you don't mind having the quotes in the returned list, you can use much simpler code:
List<String> matchList = new ArrayList<String>();
Pattern regex = Pattern.compile("[^\\s\"']+|\"[^\"]*\"|'[^']*'");
Matcher regexMatcher = regex.matcher(subjectString);
while (regexMatcher.find()) {
matchList.add(regexMatcher.group());
}
There are several questions on StackOverflow that cover this same question in various contexts using regular expressions. For instance:
parsings strings: extracting words and phrases
Best way to parse Space Separated Text
UPDATE: Sample regex to handle single and double quoted strings. Ref: How can I split on a string except when inside quotes?
m/('.*?'|".*?"|\S+)/g
Tested this with a quick Perl snippet and the output was as reproduced below. Also works for empty strings or whitespace-only strings if they are between quotes (not sure if that's desired or not).
This
is
a
string
that
"will be"
highlighted
when
your
'regular expression'
matches
something.
Note that this does include the quote characters themselves in the matched values, though you can remove that with a string replace, or modify the regex to not include them. I'll leave that as an exercise for the reader or another poster for now, as 2am is way too late to be messing with regular expressions anymore ;)
If you want to allow escaped quotes inside the string, you can use something like this:
(?:(['"])(.*?)(?<!\\)(?>\\\\)*\1|([^\s]+))
Quoted strings will be group 2, single unquoted words will be group 3.
You can try it on various strings here: http://www.fileformat.info/tool/regex.htm or http://gskinner.com/RegExr/
The regex from Jan Goyvaerts is the best solution I found so far, but creates also empty (null) matches, which he excludes in his program. These empty matches also appear from regex testers (e.g. rubular.com).
If you turn the searches arround (first look for the quoted parts and than the space separed words) then you might do it in once with:
("[^"]*"|'[^']*'|[\S]+)+
(?<!\G".{0,99999})\s|(?<=\G".{0,99999}")\s
This will match the spaces not surrounded by double quotes.
I have to use min,max {0,99999} because Java doesn't support * and + in lookbehind.
It'll probably be easier to search the string, grabbing each part, vs. split it.
Reason being, you can have it split at the spaces before and after "will be". But, I can't think of any way to specify ignoring the space between inside a split.
(not actual Java)
string = "This is a string that \"will be\" highlighted when your 'regular expression' matches something.";
regex = "\"(\\\"|(?!\\\").)+\"|[^ ]+"; // search for a quoted or non-spaced group
final = new Array();
while (string.length > 0) {
string = string.trim();
if (Regex(regex).test(string)) {
final.push(Regex(regex).match(string)[0]);
string = string.replace(regex, ""); // progress to next "word"
}
}
Also, capturing single quotes could lead to issues:
"Foo's Bar 'n Grill"
//=>
"Foo"
"s Bar "
"n"
"Grill"
String.split() is not helpful here because there is no way to distinguish between spaces within quotes (don't split) and those outside (split). Matcher.lookingAt() is probably what you need:
String str = "This is a string that \"will be\" highlighted when your 'regular expression' matches something.";
str = str + " "; // add trailing space
int len = str.length();
Matcher m = Pattern.compile("((\"[^\"]+?\")|('[^']+?')|([^\\s]+?))\\s++").matcher(str);
for (int i = 0; i < len; i++)
{
m.region(i, len);
if (m.lookingAt())
{
String s = m.group(1);
if ((s.startsWith("\"") && s.endsWith("\"")) ||
(s.startsWith("'") && s.endsWith("'")))
{
s = s.substring(1, s.length() - 1);
}
System.out.println(i + ": \"" + s + "\"");
i += (m.group(0).length() - 1);
}
}
which produces the following output:
0: "This"
5: "is"
8: "a"
10: "string"
17: "that"
22: "will be"
32: "highlighted"
44: "when"
49: "your"
54: "regular expression"
75: "matches"
83: "something."
I liked Marcus's approach, however, I modified it so that I could allow text near the quotes, and support both " and ' quote characters. For example, I needed a="some value" to not split it into [a=, "some value"].
(?<!\\G\\S{0,99999}[\"'].{0,99999})\\s|(?<=\\G\\S{0,99999}\".{0,99999}\"\\S{0,99999})\\s|(?<=\\G\\S{0,99999}'.{0,99999}'\\S{0,99999})\\s"
Jan's approach is great but here's another one for the record.
If you actually wanted to split as mentioned in the title, keeping the quotes in "will be" and 'regular expression', then you could use this method which is straight out of Match (or replace) a pattern except in situations s1, s2, s3 etc
The regex:
'[^']*'|\"[^\"]*\"|( )
The two left alternations match complete 'quoted strings' and "double-quoted strings". We will ignore these matches. The right side matches and captures spaces to Group 1, and we know they are the right spaces because they were not matched by the expressions on the left. We replace those with SplitHere then split on SplitHere. Again, this is for a true split case where you want "will be", not will be.
Here is a full working implementation (see the results on the online demo).
import java.util.*;
import java.io.*;
import java.util.regex.*;
import java.util.List;
class Program {
public static void main (String[] args) throws java.lang.Exception {
String subject = "This is a string that \"will be\" highlighted when your 'regular expression' matches something.";
Pattern regex = Pattern.compile("\'[^']*'|\"[^\"]*\"|( )");
Matcher m = regex.matcher(subject);
StringBuffer b= new StringBuffer();
while (m.find()) {
if(m.group(1) != null) m.appendReplacement(b, "SplitHere");
else m.appendReplacement(b, m.group(0));
}
m.appendTail(b);
String replaced = b.toString();
String[] splits = replaced.split("SplitHere");
for (String split : splits) System.out.println(split);
} // end main
} // end Program
If you are using c#, you can use
string input= "This is a string that \"will be\" highlighted when your 'regular expression' matches <something random>";
List<string> list1 =
Regex.Matches(input, #"(?<match>\w+)|\""(?<match>[\w\s]*)""|'(?<match>[\w\s]*)'|<(?<match>[\w\s]*)>").Cast<Match>().Select(m => m.Groups["match"].Value).ToList();
foreach(var v in list1)
Console.WriteLine(v);
I have specifically added "|<(?[\w\s]*)>" to highlight that you can specify any char to group phrases. (In this case I am using < > to group.
Output is :
This
is
a
string
that
will be
highlighted
when
your
regular expression
matches
something random
1st one-liner using String.split()
String s = "This is a string that \"will be\" highlighted when your 'regular expression' matches something.";
String[] split = s.split( "(?<!(\"|').{0,255}) | (?!.*\\1.*)" );
[This, is, a, string, that, "will be", highlighted, when, your, 'regular expression', matches, something.]
don't split at the blank, if the blank is surrounded by single or double quotes
split at the blank when the 255 characters to the left and all characters to the right of the blank are neither single nor double quotes
adapted from original post (handles only double quotes)
I'm reasonably certain this is not possible using regular expressions alone. Checking whether something is contained inside some other tag is a parsing operation. This seems like the same problem as trying to parse XML with a regex -- it can't be done correctly. You may be able to get your desired outcome by repeatedly applying a non-greedy, non-global regex that matches the quoted strings, then once you can't find anything else, split it at the spaces... that has a number of problems, including keeping track of the original order of all the substrings. Your best bet is to just write a really simple function that iterates over the string and pulls out the tokens you want.
A couple hopefully helpful tweaks on Jan's accepted answer:
(['"])((?:\\\1|.)+?)\1|([^\s"']+)
Allows escaped quotes within quoted strings
Avoids repeating the pattern for the single and double quote; this also simplifies adding more quoting symbols if needed (at the expense of one more capturing group)
You can also try this:
String str = "This is a string that \"will be\" highlighted when your 'regular expression' matches something";
String ss[] = str.split("\"|\'");
for (int i = 0; i < ss.length; i++) {
if ((i % 2) == 0) {//even
String[] part1 = ss[i].split(" ");
for (String pp1 : part1) {
System.out.println("" + pp1);
}
} else {//odd
System.out.println("" + ss[i]);
}
}
The following returns an array of arguments. Arguments are the variable 'command' split on spaces, unless included in single or double quotes. The matches are then modified to remove the single and double quotes.
using System.Text.RegularExpressions;
var args = Regex.Matches(command, "[^\\s\"']+|\"([^\"]*)\"|'([^']*)'").Cast<Match>
().Select(iMatch => iMatch.Value.Replace("\"", "").Replace("'", "")).ToArray();
When you come across this pattern like this :
String str = "2022-11-10 08:35:00,470 RAV=REQ YIP=02.8.5.1 CMID=caonaustr CMN=\"Some Value Pyt Ltd\"";
//this helped
String[] str1= str.split("\\s(?=(([^\"]*\"){2})*[^\"]*$)\\s*");
System.out.println("Value of split string is "+ Arrays.toString(str1));
This results in :[2022-11-10, 08:35:00,470, PLV=REQ, YIP=02.8.5.1, CMID=caonaustr, CMN="Some Value Pyt Ltd"]
This regex matches spaces ONLY if it is followed by even number of double quotes.
Does anyone have an idea how I can split a string returned by a WS to different strings?
String wsResult = "SONACOM RC, RUE DES ETOILES N. 20, 75250 PARIS (MI)";
I'm trying to split it into:
String name = "SONACOM RC";
String adress = "RUE DES ETOILES N. 20";
String postalCode = "75250";
String city = "PARIS";
N.B: the return of the WS changes only what is inside of my parameters
Thank you in advance for your help
You could capture your data in 4 capturing groups. Your provided example uses uppercase characters, which you can match with [A-Z].
If you want to match also lowercase characters, digits and an underscore, you could replace [A-Z] or [A-Z\d] with \w.
You can go about this in multiple ways. An approach could be:
([A-Z ]+), +([A-Z\d .]+), +(\d+) +([A-Z\d() ]+)
Explanation
Group 1: match one or more uppercase characters or a whitespace ([A-Z ]+)
Match a comma and one or more whitespaces , +
Group 2: match one or more uppercase characters or digit or whitespace or dot ([A-Z\d .]+)
Match a comma and one or more whitespaces , +
Group 3: match one or more digits (\d+)
Match one or more whitespaces +
Group 4: match one or more uppercase characters or digit or open/close parenthesis or whitespace ([A-Z\d() ]+)
Output in Java
One easy way to split it as you'd like is using wsResult.split(","). However, you'll have to add a comma between 75250 and Paris:
String wsResult = "SONACOM RC, RUE DES ETOILES N. 20, 75250, PARIS (MI)";
String[] temp = wsResult.split(",");
String name = temp[0];
String adress = temp[1];
String postalCode = temp[2];
String city = temp[3];
Using that you will get the output you're looking for.
EDIT
Another way to get your output without adding a comma would be to do this (using the code above too):
for(int i = 1; i<postalCode.length(); i++){
if(postalCode.charAt(i) == ' ') {
city = postalCode.substring(i,postalCode.length());
postalCode = postalCode.substring(0,i);
break;
}
}
For more information check the String class in the API Java and this Stack Overflow question.
I have string something like :
SKU: XP321654
Quantity: 1
Order date: 01/08/2016
The SKU length is not fixed , so my function sometime returns me the first or two characters of Quantity also which I do not want to get. I want to get only SKU value.
My Code :
int index = Content.indexOf("SKU:");
String SKU = Content.substring(index, index+15);
If SKU has one or two more digits then also it is not able to get because I have specified limit till 15. If I do index + 16 to get long SKU data then for Short SKU it returns me some character of Quantity also.
How can I solve it. Is there any way to use instead of a static string character length as limit.
My SKU last digit will always number so any other thing which I can use to get only SKU till it's last digit?
Using .substring is simply not the way to process such things. What you need is a regex (or regular expression):
Pattern pat = Pattern.compile("SKU\\s*:\\s*(\\S+)");
String sku = null;
Matcher matcher = pattern.matcher(Content);
if(matcher.find()) { //we've found a match
sku = matcher.group(1);
}
//do something with sku
Unescaped the regex is something like:
SKU\s*:\s*(\S+)
you are thus looking for a pattern that starts with SKU then followed by zero or more \s (spacing characters like space and tab), followed by a colon (:) then potentially zero or more spacing characters (\s) and finally the part in which you are interested: one or more (that's the meaning of +) non-spacing characters (\S). By putting these in brackets, these are a matching group. If the regex succeeds in finding the pattern (matcher.find()), you can extract the content of the matching group matcher.group(1) and store it into a string.
Potentially you can improve the regex further if you for instance know more about how a SKU looks like. For instance if it consists only out of uppercase letters and digits, you can replace \S by [0-9A-Z], so then the pattern becomes:
Pattern pat = Pattern.compile("SKU\\s*:\\s*([0-9A-Z]+)");
EDIT: for the quantity data, you could use:
Pattern pat2 = Pattern.compile("Quantity\\s*:\\s*(\\d+)");
int qt = -1;
Matcher matcher = pat2.matcher(Content);
if(matcher.find()) { //we've found a match
qt = Integer.parseInt(matcher.group(1));
}
or see this jdoodle.
You know you can just refer to the length of the string right ?
String s = "SKU: XP321654";
String sku = s.substring(4, s.length()).trim();
I think using a regex is clearly overkill in this case, it is way way simpler than this. You can even split the expression although it's a bit less efficient than the solution above, but please don't use a regex for this !
String sku = "SKU: XP321654".split(':')[1].trim();
1: you have to split your input by lines (or split by \n)
2: when you have your line: you search for : and then you take the remaining of the line (with the String size as mentionned in Dici answer).
Depending on how exactly the string contains new lines, you could do this:
public static void main(String[] args) {
String s = "SKU: XP321654\r\n" +
"Quantity: 1\r\n" +
"Order date: 01/08/2016";
System.out.println(s.substring(s.indexOf(": ") + 2, s.indexOf("\r\n")));
}
Just note that this 1-liner has several restrictions:
The SKU property has to be first. If not, then modify the start index appropriately to search for "SKU: ".
The new lines might be separated otherwise, \R is a regex for all the valid new line escape characters combinations.
How to remove duplicate white spaces (including tabs, newlines, spaces, etc...) in a string using Java?
Like this:
yourString = yourString.replaceAll("\\s+", " ");
For example
System.out.println("lorem ipsum dolor \n sit.".replaceAll("\\s+", " "));
outputs
lorem ipsum dolor sit.
What does that \s+ mean?
\s+ is a regular expression. \s matches a space, tab, new line, carriage return, form feed or vertical tab, and + says "one or more of those". Thus the above code will collapse all "whitespace substrings" longer than one character, with a single space character.
Source: Java: Removing duplicate white spaces in strings
You can use the regex
(\s)\1
and
replace it with $1.
Java code:
str = str.replaceAll("(\\s)\\1","$1");
If the input is "foo\t\tbar " you'll get "foo\tbar " as outputBut if the input is "foo\t bar" it will remain unchanged because it does not have any consecutive whitespace characters.
If you treat all the whitespace characters(space, vertical tab, horizontal tab, carriage return, form feed, new line) as space then you can use the following regex to replace any number of consecutive white space with a single space:
str = str.replaceAll("\\s+"," ");
But if you want to replace two consecutive white space with a single space you should do:
str = str.replaceAll("\\s{2}"," ");
String str = " Text with multiple spaces ";
str = org.apache.commons.lang3.StringUtils.normalizeSpace(str);
// str = "Text with multiple spaces"
Try this - You have to import java.util.regex.*;
Pattern pattern = Pattern.compile("\\s+");
Matcher matcher = pattern.matcher(string);
boolean check = matcher.find();
String str = matcher.replaceAll(" ");
Where string is your string on which you need to remove duplicate white spaces
hi the fastest (but not prettiest way) i found is
while (cleantext.indexOf(" ") != -1)
cleantext = StringUtils.replace(cleantext, " ", " ");
this is running pretty fast on android in opposite to an regex
Though it is too late, I have found a better solution (that works for me) that will replace all consecutive same type white spaces with one white space of its type. That is:
Hello!\n\n\nMy World
will be
Hello!\nMy World
Notice there are still leading and trailing white spaces. So my complete solution is:
str = str.trim().replaceAll("(\\s)+", "$1"));
Here, trim() replaces all leading and trailing white space strings with "". (\\s) is for capturing \\s (that is white spaces such as ' ', '\n', '\t') in group #1. + sign is for matching 1 or more preceding token. So (\\s)+ can be consecutive characters (1 or more) among any single white space characters (' ', '\n' or '\t'). $1 is for replacing the matching strings with the group #1 string (which only contains 1 white space character) of the matching type (that is the single white space character which has matched). The above solution will change like this:
Hello!\n\n\nMy World
will be
Hello!\nMy World
I have not found my above solution here so I have posted it.
If you want to get rid of all leading and trailing extraneous whitespace then you want to do something like this:
// \\A = Start of input boundary
// \\z = End of input boundary
string = string.replaceAll("\\A\\s+(.*?)\\s+\\z", "$1");
Then you can remove the duplicates using the other strategies listed here:
string = string.replaceAll("\\s+"," ");
You can also try using String Tokeniser, for any space, tab, newline, and all. A simple way is,
String s = "Your Text Here";
StringTokenizer st = new StringTokenizer( s, " " );
while(st.hasMoreTokens())
{
System.out.print(st.nextToken());
}
This can be possible in three steps:
Convert the string in to character array (ToCharArray)
Apply for loop on charater array
Then apply string replace function (Replace ("sting you want to replace"," original string"));