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.

Java(Apex) RegEx not working?

I am having trouble with a regex in salesforce, apex. As I saw that apex is using the same syntax and logic as apex, I aimed this at java developers also.
I debugged the String and it is correct. street equals 'str 3 B'.
When using http://www.regexr.com/, the regex works('\d \w$').
The code:
Matcher hasString = Pattern.compile('\\d \\w$').matcher(street);
if(hasString.matches())
My problem is, that hasString.matches() resolves to false. Can anyone tell me if I did something somewhere wrong? I tried to use it without the $, with difference casing, etc. and I just can't get it to work.
Thanks in advance!

You need to use find instead of matches for partial input match as matches attempts to match complete input text.
Matcher hasString = Pattern.compile("\\d \\w$").matcher(street);
if(hasString.find()) {
// matched
System.out.println("Start position: " + hasString.start());
}

regular expressions for random letters/symbols

I have searched for hours using regular expression generators and checkers but i cant seem to get it to work...
I have this string: hdr("");cr(92);cl(3,"",4,"420720250","random message here");etr();
and so far my code is :
private void strchecker() {
Pattern pattern = Pattern.compile("(\\d{9})");
Matcher matcher = pattern.matcher(strLine);
if (matcher.find()) {
System.out.println(matcher.group(0)); //prints /{item}/
} else {
//System.out.println("Match not found");
}
}
This code is working and it finds the 9 digit number in the string. What im trying to do it find the regex code to search for "cl(3" or "cl(2", if it exists then send the 9 digit number to a variable. i just don't know how to find that cl(3 or 2..
any advice?
Thanks
Matt

/cl\([23].*(\d{9})/
The final parentheses will capture the 9 digits in group 1.

Since you note you're using javascript, I think you could do it like this, as you can't use lookbehind; you just have to grab the capture group rather than use the full match.
cl\\([23].*?(\\d{9})

Pattern syntax error

The following regex works in the find dialog of Eclipse but throws an exception in Java.
I can't find why
(?<=(00|\\+))?[\\d]{1}[\\d]*
The syntax error is at runtime when executing:
Pattern.compile("(?<=(00|\\+))?[\\d]{1}[\\d]*")
In the find I used
(?<=(00|\+))?[\d]{1}[\d]*
I want to match phone numbers with or without the + or 00. But that is not the point because I get a Syntax error at position 13. I don't get the error if I get rid of the second "?"
Pattern.compile("(?<=(00|\\+))[\\d]{1}[\\d]*")
Please consider that instead of 1 sometime I need to use a greater number and anyway the question is about the syntax error

If your data looks like 00ddddd or +ddddd where d is digit you want to get #Bergi's regex (?<=00|\\+)\\d+ will do the trick. But if your data sometimes don't have any part that you want to ignore like ddddd then you probably should use group mechanism like
String[] data={"+123456","00123456","123456"};
Pattern p=Pattern.compile("(?:00|\\+)?(\\d+)");
Matcher m=null;
for (String s:data){
m=p.matcher(s);
if(m.find())
System.out.println(m.group(1));
}
output
123456
123456
123456

Here is an example that works for me:
public static void main(String[] args) {
Pattern pattern = Pattern.compile("(?<=00|\\+)(\\d+)");
Matcher matcher = pattern.matcher("+1123456");
if (matcher.find()) {
System.out.println(matcher.group(1));
}
}

You might shorten your regex a lot. The character classes are not needed when there is only one class inside - just use \d. And {1} is quite useless as well. Also, you can use + for matching "one or more" (it's short for {1,}). Next the additional grouping in your lookbehind should not be needed.
And last, why is that lookbehind optional (with ?)? Just leave it away if you don't need it. This might even be the source of your pattern syntax error - a lookaround must not be optional.
Try this:
/(?<=00|\+)\d+/
Java:
"(?<=00|\\+)\\d+"

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.