Develop Reference

JAVA REGEX: Match until the specific character

I have this Java code
String cookies = TextUtils.join(";", LoginActivity.msCookieManager.getCookieStore().getCookies());
Log.d("TheCookies", cookies);
Pattern csrf_pattern = Pattern.compile("csrf_cookie=(.+)(?=;)");
Matcher csrf_matcher = csrf_pattern.matcher(cookies);
while (csrf_matcher.find()) {
json.put("csrf_key", csrf_matcher.group(1));
Log.d("CSRF KEY", csrf_matcher.group(1));
}
The String contains something like this:
SessionID=sessiontest;csrf_cookie=e18d027da2fb95e888ebede711f1bc39;ci_session=3f4675b5b56bfd0ba4dae46249de0df7994ee21e
Im trying to get the csrf_cookie data by using this Regular Expression:
csrf_cookie=(.+)(?=;)
I expect a result like this in the code:
csrf_matcher.group(1);
e18d027da2fb95e888ebede711f1bc39
instead I get a:
3492f8670f4b09a6b3c3cbdfcc59e512;ci_session=8d823b309a361587fac5d67ad4706359b40d7bd0
What is the possible work around for this problem?

Here is a one-liner using String#replaceAll:
String input = "SessionID=sessiontest;csrf_cookie=e18d027da2fb95e888ebede711f1bc39;ci_session=3f4675b5b56bfd0ba4dae46249de0df7994ee21e";
String cookie = input.replaceAll(".*csrf_cookie=([^;]*).*", "$1");
System.out.println(cookie);
e18d027da2fb95e888ebede711f1bc39
Demo
Note: We could have used a formal regex pattern matcher, and in face you may want to do this if you need to do this search/replacement often in your code.

You are getting more data than expected because you are using an greedy '+' (It will match as long as it can)
For example the pattern a+ could match on aaa the following: a, aa, and aaa. Where the later is 'preferred' if the pattern is greedy.
So you are matching
csrf_cookie=e18d027da2fb95e888ebede711f1bc39;ci_session=3f4675b5b56bfd0ba4dae46249de0df7994ee21e;
as long as it ends with a ';'. The first ';' is skipped with .+ and the last ';' is found with the possitive lookahead
To make a patter ungreedy/lazy use +? instead of + (so a+? would match a (three times) on aaa string)
So try with:
csrf_cookie=(.+?);
or just match anything that is not a ';'
csrf_cookie=([^;]*);
that way you don't need to make it lazy.

Regex match only if text contains something before

Given the following text
KEYWORD This is a test
We want to match the following groups 1:YES 2:YES 3:YES
I want to match with "1:YES", "2:YES" and "3:YES" using
((\d):YES)
If and only if the first word in the complete text is "KEYWORD"
Given this test:
This is a test
We want to match the following groups 1:YES 2:YES 3:YES
No matches should be found

Java (as with most regex engines) doesn't support unbounded length look behinds, however there is a work-around!
String str = "KEYWORD This is a test\n" +
"We want to match the following groups 1:YES 2:YES 3:YES";
Matcher matcher = Pattern.compile("(?s)(?<=\\AKEYWORD\\b.{1,99999})(\\d+:YES)").matcher(str);
while (matcher.find()) {
System.out.println(matcher.group(1));
}
Which outputs:
1:YES
2:YES
3:YES
The trick here is the look behind (?<=\\AKEYWORD.{1,99999}) which has a large (but not unbounded) length. (?s) means DOTALL flag (dot matches newline too) and \A means start of input which is needed because ^ matches start of line when DOTALL flag is used.

Without tricking lookbhinds in Java you can capture \d+:YES\b strings with using \G. \G causes a match to start from where previous match ended or it will match beginning of string the same as \A.
We are in need of its first capability:
(?:\AKEYWORD|\G(?!\A))[\s\S]*?(\d:YES\b)
Breakdown:
(?: Start of non-capturing group
\A Match beginning of subject string
KEYWORD Match keyword
| Or
\G(?!\A) Continue from where previous match ends
) End of NCG
[\s\S]*? Match anything else un-greedily
(\d+:YES\b) Match and capture our desired part
Live demo
Java code:
Pattern p = Pattern.compile("(?:\\AKEYWORD|\\G(?!\\A))[\\s\\S]*?(\\d+:YES\\b)");
Matcher m = p.matcher(string);
while (m.find()) {
System.out.println(m.group(1));
}
Live demo

Java recursive/repeated regex

I am trying to replace all .(periods) with keyword XXX which lie within an alphanumeric word in a large text.
For example: I am trying to match a.b.c.d.e ...
Expected output: I am trying to match aXXXbXXXcXXXdXXXe ...
Pattern I used: (\w+)([\.]+)(\w+)
Actual result: I am trying to match aXXXb.cXXXd.e ...
How can I get expected output via regex without using any code/stubs.

You can use lookarounds:
str = str.replaceAll("(?<=[a-zA-Z0-9])\\.(?=[a-zA-Z0-9])", "XXX");
RegEx Demo
Lookaround Reference

Why don't you do something like if you want to change all . -
str = str.replaceAll("\\.", "XXX");
Or below if you don't want to change . if any first or last index -
str = str.replaceAll("\\.", "XXX").replaceAll("^XXX", ".").replaceAll("XXX$", ".");

How can I get all content between two pipes using regular expression

I have a String say
String s = "|India| vs Aus";
In this case result should be only India.
Second case :
String s = "Aus vs |India|";
In this case result should be only India.
3rd case:
String s = "|India| vs |Aus|"
Result shouls contain only India, Aus. vs should not present in output.
And in these scenarios, there can be any other word in place of vs. e.g. String can be like this also |India| in |Aus|. and the String can be like this also |India| and |Sri Lanka| in |Aus|. I want those words that are present in between two pipes like India, Sri Lanka , Aus.
I want to do it in Java.
Any pointer will be helpful.

You would use a regex like...
\|[^|]+\|
...or...
\|.+?\|
You must escape the pipe because the pipe has special meaning in a regex as or.

You are looking at something similar to this:
String s = "|India| vs |Aus|";
Pattern p = Pattern.compile("\\|(.*?)\\|");
Matcher m = p.matcher(s);
while(m.find()){
System.out.println(m.group(1));
}
You need to use the group to get the contents inside the paranthesis in the regexp.

Regex for removing part of a line if it is preceded by some word in Java

There's a properties language bundle file:
label.username=Username:
label.tooltip_html=Please enter your username.</center></html>
label.password=Password:
label.tooltip_html=Please enter your password.</center></html>
How to match all lines that have both "_html" and "</center></html>" in that order and replace them with the same line except the ending "</center></html>". For example, line:
label.tooltip_html=Please enter your username.</center></html>
should become:
label.tooltip_html=Please enter your username.
Note: I would like to do this replacement using an IDE (IntelliJ IDEA, Eclipse, NetBeans...)

Since you clarified that this regex is to be used in the IDE, I tested this in Eclipse and it works:
FIND:
(_html.*)</center></html>
REPLACE WITH:
$1
Make sure you turn on the Regular expressions switch in the Find/Replace dialog. This will match any string that contains _html.* (where the .* greedily matches any string not containing newlines), followed by </center></html>. It uses (…) brackets to capture what was matched into group 1, and $1 in the replacement substitutes in what group 1 captured.
This effectively removes </center></html> if that string is preceded by _html in that line.
If there can be multiple </center></html> in a line, and they are all to be removed if there's a _html_ to their left, then the regex will be more complicated, but it can be done in one regex with \G continuing anchor if absolutely need be.
Variations
Speaking more generally, you can also match things like this:
(delete)this part only(please)
This now creates 2 capturing groups. You can match strings with this pattern and replace with $1$2, and it will effectively delete this part only, but only if it's preceded by delete and followed by please. These subpatterns can be more complicated, of course.

if (line.contains("_html=")) {
line = line.replace("</center></html>", "");
}
No regExp needed here ;) (edit) as long as all lines of the property file are well formed.

String s = "label.tooltip_html=Please enter your password.</center></html>";
Pattern p = Pattern.compile("(_html.*)</center></html>");
Matcher m = p.matcher(s);
System.out.println(m.replaceAll("$1"));

Try something like this:
Pattern p = Pattern.compile(".*(_html).*</center></html>");
Matcher m = p.matcher(input_line); // get a matcher object
String output = input_line;
if (m.matches()) {
String output = input_line.replace("</center></html>", "");
}

/^(.*)<\/center><\/html>/
finds you the
label.tooltip_html=Please enter your username.
part. then you can just put the string together correctly.