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 to get String which is outside of brackets

i am searching a way(regex expression) to get anything outside the brackets in java
like
String sample = "Hi all(igi)";
Output like
Hi all
Searched a lot but not able to find it on stack or google
What i am trying
Matcher m = Pattern.compile("(?:[^<>(^)]++)").matcher(abc);
while (m.find()) {
String newName = m.group(0);
System.out.println(newName);
}
It is giving me both text
also i want a regex expression not a workaround (i guess it can be done in regex only)
also explain the regex expression if you got the answer i want to learn how to achieve it

You can easily use replaceAll with some regex \(.*?\) which mean replace every thing between brackets so in the end you will get only the result that is not between the brackets :
sample = sample.replaceAll("\\(.*?\\)", "");
Example:
Input Output
Hi all(igi) Hi all
Hi all(igi) Some string Hi all Some string

How to skip a specific word when replace the String using regexp in java

Consider the string
String s = "H_ello pe_rfec_t wor_ld"
I want to replace all '_' symbols on... no matter what, let`s say on '1', except those which are placed inside the 'pe_rfec_t'.
I could not find any solution even to just skip the word 'pe_rfec_t':
s = s.replaceAll("(?<=pe_rfec_t).*|.*(?=pe_rfec_t)", "1");
looks nice at a glance but results to:
11pe_rfec_t1 //instead of 1111111pe_rfec_t1111111
Ideally I need the following result:
Hello pe_rfec_t world
Could anyone help me please?

You can use alternation and captured group:
String str = "H_ello pe_rfec_t wor_ld";
String repl = str.replaceAll("(pe_rfec_t)|_", "$1");
//=> Hello pe_rfec_t world
RegEx Demo
Here in alternation we match first pe_rfec_t and capture it in group #1. In repalcement we put $1 (back-reference to group #1) back.

most elegant way of escaping $ in java

I have a base String "abc def", I am trying to replace my base string with "abc$ def$" using replaceFirst(), which is running into errors as $ is not escaped.
I tried doing it with Pattern and Matcher APIs, as given below,
newValue = "abc$ def$";
if(newValue.contains("$")){
Pattern specialCharacters = Pattern.compile("$");
Matcher newMatcherValue = specialCharacters.matcher(newValue) ;
newValue = newMatcherValue.replaceAll("\\\\$") ;
}
This runs into an error. Is there any elegant way of replacing my second string "abc$ def$" with "abc\\\\$ def\\\\$" so as to use the replacefirst() API successfully?

Look at Pattern.quote() to quote a regex and Matcher.quoteReplacement() to quote a replacement string.
That said, does this do what you want it to?
System.out.println("abc def".replaceAll("([\\w]+)\\b", "$1\\$"));
This prints out abc$ def$

You can use replaceAll just in one step:
String newValueScaped = newValue.replaceAll("\\$", "\\\\$")
$ has a special mining in regex, so you need to scape it. It's used to match the end of the data.

Change group using regex java

I need help in regular expression using in regex java.
I need change group in string:
Example:
Input:
=sum($var1;2) or =if($result<10;"little";"big") ...
Need Output:
=sum(teste;2) or =if(teste<10;"little";"big") ...
Code I have:
Pattern p = Pattern.compile("(\\.*)(\\$\\w)(\\.*)");
Matcher m = p.matcher(total);
if (m.find()) {
System.out.println(m.replaceAll("$2teste"));
}
Output I have:
=sum($vtestear1;2)
=if($r testeesultado<5;"maior";"menor")

Why match everything when all you need is to match variable tokens?
Pattern p = Pattern.compile("\\b\\$[a-z0-9]+\\b");
p.matcher(total).replaceAll("teste");
Change the [a-z0-9] part if you can have more than lowercase ASCII letters and digits.
Also, you don't need to test for .find() or anything if you .replace(): no match means nothing will be replaced.