So I'm trying to pull two strings via a matcher object from one string that is stored in my online databases.
Each string appears after s:64: and is in quotations
Example s:64:"stringhere"
I'm currently trying to get them as so but any regex that I've tried has failed,
Pattern p = Pattern.compile("I don't know what to put as the regex");
Matcher m = p.matcher(data);
So with that said, all I need is the regex that will return the two strings in the matcher so that m.group(1) is my first string and m.group(2) is my second string.
Try this regex:-
s:64:\"(.*?)\"
Code:
Pattern pattern = Pattern.compile("s:64:\"(.*?)\"");
Matcher matcher = pattern.matcher(YourStringVar);
// Check all occurance
int count = 0;
while (matcher.find() && count++ < 2) {
System.out.println("Group : " + matcher.group(1));
}
Here group(1) returns the each match.
OUTPUT:
Group : First Match
Group : Second Match
Refer LIVE DEMO
String data = "s:64:\"first string\" random stuff here s:64:\"second string\"";
Pattern p = Pattern.compile("s:64:\"([^\"]*)\".*s:64:\"([^\"]*)\"");
Matcher m = p.matcher(data);
if (m.find()) {
System.out.println("First string: '" + m.group(1) + "'");
System.out.println("Second string: '" + m.group(2) + "'");
}
prints:
First string: 'first string'
Second string: 'second string'
Regex you need should be compile("s:64:\"(.*?)\".*s:64:\"(.*?)\"")
Related
How to extract the strings between the delimiters '<' and '>' from the string
“Rahul<is>an<entrepreneur>”
I tried using substring() method, but I could only extract one string out of the primary string. How to loop this and get all the strings between the delimiters from the primary string
You could use Pattern and Matcher for pattern lookup. For example, see code below:
String STR = "Rahul<is>an<entrepreneur>";
Pattern pattern = Pattern.compile("<(.*?)>", Pattern.CASE_INSENSITIVE);
Matcher matcher = pattern.matcher(STR);
while (matcher.find()) {
System.out.println(matcher.start() + " " + matcher.end() + " " + matcher.group());
}
Output of above will give you start and end indexes and group substring:
5 9 <is>
11 25 <entrepreneur>
More specifically if you just want the strings, you can get string between the group start and end indexes.
STR.substring(matcher.start() + 1, matcher.end() - 1);
This gives you only the matching strings.
This worked for me:
String str = "Rahul<is>an<entrepreneur>";
String[] tempStr = str.split("<");
for (String st : tempStr) {
if (st.contains(">")) {
int index = st.indexOf('>');
System.out.println(st.substring(0, index));
}
}
Output:
is
entrepreneur
I have this Strings :
String test1=":test:block1:%a1%a2%a3%a4:block2:BL";
and
String test2=":test:block2:BL:block1:%a1%a2%a3%a4";
I've created an regex pattern in order to isolate this piece of String
block1:%a1%a2%a3%a4:
from the rest of the String letting those Strings like this :
in the case of test1="block1:%a1%a2%a3%a4:"; (with ':' at the end)
in the case of test2=":block1:%a1%a2%a3%a4"; (with ':' at the beggining)
The regex i've created is :
"(block1:(.*?):|:block1:(.*))";
With test1 is working , but with test2 is retrieving me this :
block1:%a1%a2%a3%a4:block2:BL";
Can someone give me a hand with this ?
Cheers!
You may use
block1:([^:]*)
It matches block1: text and then captures into Group 1 any 0 or more chars other than :.
See Java demo:
String patternString = "block1:([^:]*)";
String[] tests = {":test:block1:%a1%a2%a3%a4:block2:BL",
":test:block2:BL:block1:%a1%a2%a3%a4"};
for (int i=0; i<tests.length; i++)
{
Pattern p = Pattern.compile(patternString, Pattern.DOTALL);
Matcher m = p.matcher(tests[i]);
if(m.find())
{
System.out.println(tests[i] + " matched. Match: " +
m.group(0) + ", Group 1: " + m.group(1));
}
}
Output:
:test:block1:%a1%a2%a3%a4:block2:BL matched. Match: block1:%a1%a2%a3%a4, Group 1: %a1%a2%a3%a4
:test:block2:BL:block1:%a1%a2%a3%a4 matched. Match: block1:%a1%a2%a3%a4, Group 1: %a1%a2%a3%a4
I am very new in regex and need your help. I wanna take numbers and letters between two span.
<span>454.000 $</span>
I wanna take 454.000 $. There are 12 space before . Please help me.
This Should Work.
Regexp:
\s+<.+>(.+)<.+>
Input:
<span>454.000 $</span>
Output:
454.000 $
JAVA CODE:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
final String regex = "\\s+<.+>(.+)<.+>";
final String string = " <span>454.000 $</span>";
final Pattern pattern = Pattern.compile(regex);
final Matcher matcher = pattern.matcher(string);
while (matcher.find()) {
System.out.println("Full match: " + matcher.group(0));
for (int i = 1; i <= matcher.groupCount(); i++) {
System.out.println("Group " + i + ": " + matcher.group(i));
}
}
See: https://regex101.com/r/2zg5Ws/1
Capturing group using pattern matching is something like below
String x = " <span>454.000 $</span> ";
Pattern p = Pattern.compile("<span>(.*?)</span>");
Matcher m = p.matcher(x);
if (m.find()) {
System.out.println(">> "+ m.group(1)); // output 454.000 $
}
But for such cases I always prefer to use the replaceAll() as it is shorter version of code:
String num = x.replaceAll(".*<span>(.*?)</span>.*", "$1");
// num has 454.000 $
For the replace it is actually capturing the group from the text and replacing the whole text with that group ($1). This solution depends upon how your input string is.
I have a below string which comes from an excel column
"\"USE CODE \"\"Gef, sdf\"\" FROM 1/7/07\""
I would like to set regex pattern to retrieve the entire string,so that my result would be exactly like
"USE CODE ""Gef, sdf"" FROM 1/7/07"
Below is what I tried
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class RegexMatches
{
public static void main( String args[] ){
// String to be scanned to find the pattern.
String line = "\"USE CODE \"\"Gef, sdf\"\" FROM 1/7/07\", Delete , Hello , How are you ? , ";
String line2 = "Test asda ds asd, tesat2 . test3";
String dpattern = "(\"[^\"]*\")(?:,(\"[^\"]*\"))*,|([^,]+),";
// Create a Pattern object
Pattern d = Pattern.compile(dpattern);
Matcher md = d.matcher(line2);
Pattern r = Pattern.compile(dpattern);
// Now create matcher object.
Matcher m = r.matcher(line);
if (m.find( )) {
System.out.println("Found value: 0 " + m.group(0) );
// System.out.println("Found value: 1 " + m.group(1) );
//System.out.println("Found value: 2 " + m.group(2) );
} else {
System.out.println("NO MATCH");
}
}
}
and the result out of it breaks after ,(comma) and hence the output is
Found value: 0 "USE CODE ""Gef,
It should be
Found value: 0 "USE CODE ""Gef sdf"" FROM 1/7/07",
and for the second line Matcher m = r.matcher(line2); the output should be
Found value: 0 "Test asda ds asd",
You may use
(?:"[^"]*(?:""[^"]*)*"|[^,])+
See the regex demo
Explanation:
" - leading quote
[^"]* - 0+ chars other than a double quote
(?:""[^"]*)* - 0+ sequences of a "" text followed with 0+ chars other than a double quote
" - trailing quote
OR:
[^,] - any char but a comma
And the whole pattern is matched 1 or more times as it is enclosed with (?:...)+ and + matches 1 or more occurrences.
IDEONE demo:
String line = "\"USE CODE \"\"Gef, sdf\"\" FROM 1/7/07\", Delete , Hello , How are you ? , ";
String line2 = "Test asda ds asd, tesat2 . test3";
Pattern pattern = Pattern.compile("(?:\"[^\"]*(?:\"\"[^\"]*)*\"|[^,])+");
Matcher matcher = pattern.matcher(line);
if (matcher.find()){ // if is used to get the 1st match only
System.out.println(matcher.group(0));
}
Matcher matcher2 = pattern.matcher(line2);
if (matcher2.find()){
System.out.println(matcher2.group(0));
}
I have a string in format:
<+923451234567>: Hi here is the text.
Now I want to get the mobile number(without any non-alphanumeric characters) ie 923451234567 in the start of the string in-between < > symbols, and also the text ie Hi here is the text.
Now I can place a hardcoded logic, which I am currently doing.
String stringReceivedInSms="<+923451234567>: Hi here is the text.";
String[] splitted = cpaMessage.getText().split(">: ", 2);
String mobileNumber=MyUtils.removeNonDigitCharacters(splitted[0]);
String text=splitted[1];
How can I neatly get the required strings from the string with regular expression? So that I don't have to change the code whenever the format of the string changes.
String stringReceivedInSms="<+923451234567>: Hi here is the text.";
Pattern pattern = Pattern.compile("<\\+?([0-9]+)>: (.*)");
Matcher matcher = pattern.matcher(stringReceivedInSms);
if(matcher.matches()) {
String phoneNumber = matcher.group(1);
String messageText = matcher.group(2);
}
Use a regex that matches the pattern - <\\+?(\\d+)>: (.*)
Use the Pattern and Matcher java classes to match the input string.
Pattern p = Pattern.compile("<\\+?(\\d+)>: (.*)");
Matcher m = p.matcher("<+923451234567>: Hi here is the text.");
if(m.matches())
{
System.out.println(m.group(1));
System.out.println(m.group(2));
}
You need to use regex, the following pattern will work:
^<\\+?(\\d++)>:\\s*+(.++)$
Here is how you would use it -
public static void main(String[] args) throws IOException {
final String s = "<+923451234567>: Hi here is the text.";
final Pattern pattern = Pattern.compile(""
+ "#start of line anchor\n"
+ "^\n"
+ "#literal <\n"
+ "<\n"
+ "#an optional +\n"
+ "\\+?\n"
+ "#match and grab at least one digit\n"
+ "(\\d++)\n"
+ "#literal >:\n"
+ ">:\n"
+ "#any amount of whitespace\n"
+ "\\s*+\n"
+ "#match and grap the rest of the string\n"
+ "(.++)\n"
+ "#end anchor\n"
+ "$", Pattern.COMMENTS);
final Matcher matcher = pattern.matcher(s);
if (matcher.matches()) {
System.out.println(matcher.group(1));
System.out.println(matcher.group(2));
}
}
I have added the Pattern.COMMENTS flag so the code will work with the comments embedded for future reference.
Output:
923451234567
Hi here is the text.
You can get your phone number by just doing :
stringReceivedInSms.substring(stringReceivedInSms.indexOf("<+") + 2, stringReceivedInSms.indexOf(">"))
So try this snippet:
public static void main(String[] args){
String stringReceivedInSms="<+923451234567>: Hi here is the text.";
System.out.println(stringReceivedInSms.substring(stringReceivedInSms.indexOf("<+") + 2, stringReceivedInSms.indexOf(">")));
}
You don't need to split your String.