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.
I am a newbie in regex, I want to extract values between commas but I don't know how.
I have values like this :
[1000, Value_to_extract, 1150370.5]
and I used this Technic to simplify it:
String val = "[1000, Value_to_extract, 1150370.5]";
String designation=val.replace("[", "").replace("]", "").trim();
It give's me this result :
1000, Value_to_extract, 1150370.5
I don't know how to extract only Value_to_extract
I tried : String designation=val.replace("[", "").replace("]", "").replaceAll(".*, ,.*", "").trim();
but i doesn't work .
Thank you for your help.
String input = "[1000, Value_to_extract, 1150370.5]";
String[] parts = input.replaceAll("\\[\\] ", "") // strip brackets and whitespace
.split(","); // split on comma into an array
String valueToExtract = parts[1]; // grab the second entry
Notes:
You might also be able to use a regex here, q.v. the answer by #Thomas, but a regex will become unwieldy for extracting values from a CSV string of arbitrary length. So in general, I would prefer splitting here to using a regex.
someting like this:
,[ ]?([0-9]+[.]?[0-9]+),
breakdown
, // literal ,
[ ]? // 0 or 1 spaces
([0-9]+[.]?[0-9]+) // capture a number with or without a dot
, // another litteral ,
https://regex101.com/r/oR7nI8/1
Here are some options:
String val = "[1000, Value_to_extract, 1150370.5]";
//you can remove white space by
String noSpaces = val.trim();
System.out.println(noSpaces);
//you can split the string into string[] settting
//the delimiting regular expression to ", "
String[] strings = noSpaces.split(", ");
//the strings[1] will hold the desired string
System.out.println(strings[1]);
//in the private case of val, only Value_to_extract contains letters and "_" ,
//so you can also extract it using
System.out.println(val.replaceAll("[^a-zA-Z_]", ""));
If val does not well represent the more general need, you need to define the need more precisely.
I have a string of SVG markup that contains multiples of these:
url(#586-xr___83_193_101__rgba_243_156_18_1__0-rgba_243_156_18_1__100)
and I need them to be like this:
url('#586-xr___83_193_101__rgba_243_156_18_1__0-rgba_243_156_18_1__100')
with quotes inside the parenthesis.
These will be mixed inside a long string containing lots of different markup, so needs to be very accurate.
You can use a regex like this:
\((.*?)\)
With the replacement string ('$1')
The idea is capture everything within parentheses and concatenates the '
So, you can use a code like this:
String str = "url(#586-xr___83_193_101__rgba_243_156_18_1__0-rgba_243_156_18_1__100)";
str = str.replaceAll("\\((.*?)\\)", "('$1')");
//Outuput: url('#586-xr___83_193_101__rgba_243_156_18_1__0-rgba_243_156_18_1__100')
IdeOne example
In case you want a better performance regex you can use:
str = str.replaceAll("\\(([^)]*)\\)", "('$1')");
ReplaceAll remove a part of the string and put an unrelated and invariant new stuff instead.
Because the replacement string can't be the same at both side, the only solution I imagine (with the constraint of using RegEx and ReplaceAll) is to do it in two time:
String Str = "url(#586-xr___83_193_101__rgba_243_156_18_1__0-rgba_243_156_18_1__100)";
Str = Str.replaceAll("\\(", "('"); // replace left parenthesis
Str = Str.replaceAll("\\)", "')"); // replace right parenthesis
System.out.print("Return Value: " + Str);
// Return Value: url('#586-xr___83_193_101__rgba_243_156_18_1__0-rgba_243_156_18_1__100')
You can test it here.
Using Java, I want to go through the lines of a text and replace all "" symbols with a null entity reference.
Here is a sample string
String str = "asdsadas:\"\":asdasdASD:\"\":aSdasdcsC";
Result wanted is
String resStr = "asdsadas:null:asdasdASD:null:aSdasdcsC"
First, your string is not correct. In java you need to scape QUOTES:
String str = "asdsadas:\"\":asdasdASD:\"\":aSdasdcsC";
After, you need to escape quotes also to replace them:
String resStr= str.replaceAll("\"\"","null");
System.out.println(resStr);
OUTPUT
asdsadas:null:asdasdASD:null:aSdasdcsC
OR if you need QUOTES in the output:
String str = "\"asdsadas:\"\":asdasdASD:\"\":aSdasdcsC\"";
String resStr= str.replaceAll("\"\"","null");
System.out.println(resStr);
OUTPUT
"asdsadas:null:asdasdASD:null:aSdasdcsC"
i have a link http://localhost:8080/reporting/pvsUsageAction.do?form_action=inline_audit_view&days=7&projectStatus=scheduled&justificationId=5&justificationName= No Technicians in Area in my struts based web application.
The variable in URL justificationName have some spaces before its vales as shown. when i get value of justificationName using request.getParameter("justificationName") it gives me that value with spaces as given in the URL. i want to remove those spaces. i tried trim() i tries str = str.replace(" ", ""); but any of them did not removed those spaces. can any one tell some other way to remove the space.
Noted one more thing that i did right click on the link and opened the link into new tab there i noticed that link looks like.
http://localhost:8080/reporting/pvsUsageAction.do?form_action=inline_audit_view&days=7&projectStatus=scheduled&justificationId=5&justificationName=%A0%A0%A0%A0%A0%A0%A0%A0No%20Technicians%20in%20Area
Notable point is that in the address bar it shows %A0 for white spaces and also show %20 for space as well see the link and tell the difference please if any one have idea about it.
EDIT
Here is my code
String justificationCode = "";
if (request.getParameter("justificationName") != null) {
justificationCode = request.getParameter("justificationName");
}
justificationCode = justificationCode.replace(" ", "");
Note: replace function remove the space from inside the string but not removing starting spaces.
e-g if my string is " This is string" after using replace it becomes " Thisisstring"
Thanks in advance
Strings are immutable in Java, so the method doesn't change the string you pass but returns a new one. You must use the returned value :
str = str.replace(" ", "");
Manual trim
You need to remove the spaces the string. This will remove any number of consecutive spaces.
String trimmed = str.replaceAll(" +", "");
If you want to replace all whitespace characters:
String trimmed = str.replaceAll("\\s+", "");
URL Encoding
You could also use an URLEncoder, which sounds like a more appropriate way to go:
import java.net.UrlEncoder;
String url = "http://localhost:8080/reporting/" + URLEncoder.encode("pvsUsageAction.do?form_action=inline_audit_view&days=7&projectStatus=scheduled&justificationId=5&justificationName= No Technicians in Area", "ISO-8859-1");
You have to assign the result of the replace(String regex, String replacement) operation to another variable. See the Javadoc for the replace(String regex, String replacement) method. It returns a brand new String object and this is because the String(s) in Java are immutable. In your case, you can simply do the following
String noSpacesString = str.replace("\\s+", "");
You can use replaceAll("\\s","") It will remove all white space.
If you are trying to remove the trailing and ending white spaces, then
s = s.trim();
Or if you want to remove all the spaces the use :
s = s.replace(" ","");
There are two ways of doing one is regular expression based or your own way of implementing the logic
replaceAll("\\s","")
or
if (text.contains(" ") || text.contains("\t") || text.contains("\r")
|| text.contains("\n"))
{
//code goes here
}