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.

Escape special characters using Regex in java [duplicate]

Does Java have a built-in way to escape arbitrary text so that it can be included in a regular expression? For example, if my users enter "$5", I'd like to match that exactly rather than a "5" after the end of input.

Since Java 1.5, yes:
Pattern.quote("$5");

Difference between Pattern.quote and Matcher.quoteReplacement was not clear to me before I saw following example
s.replaceFirst(Pattern.quote("text to replace"),
Matcher.quoteReplacement("replacement text"));

It may be too late to respond, but you can also use Pattern.LITERAL, which would ignore all special characters while formatting:
Pattern.compile(textToFormat, Pattern.LITERAL);

I think what you're after is \Q$5\E. Also see Pattern.quote(s) introduced in Java5.
See Pattern javadoc for details.

First off, if
you use replaceAll()
you DON'T use Matcher.quoteReplacement()
the text to be substituted in includes a $1
it won't put a 1 at the end. It will look at the search regex for the first matching group and sub THAT in. That's what $1, $2 or $3 means in the replacement text: matching groups from the search pattern.
I frequently plug long strings of text into .properties files, then generate email subjects and bodies from those. Indeed, this appears to be the default way to do i18n in Spring Framework. I put XML tags, as placeholders, into the strings and I use replaceAll() to replace the XML tags with the values at runtime.
I ran into an issue where a user input a dollars-and-cents figure, with a dollar sign. replaceAll() choked on it, with the following showing up in a stracktrace:
java.lang.IndexOutOfBoundsException: No group 3
at java.util.regex.Matcher.start(Matcher.java:374)
at java.util.regex.Matcher.appendReplacement(Matcher.java:748)
at java.util.regex.Matcher.replaceAll(Matcher.java:823)
at java.lang.String.replaceAll(String.java:2201)
In this case, the user had entered "$3" somewhere in their input and replaceAll() went looking in the search regex for the third matching group, didn't find one, and puked.
Given:
// "msg" is a string from a .properties file, containing "<userInput />" among other tags
// "userInput" is a String containing the user's input
replacing
msg = msg.replaceAll("<userInput \\/>", userInput);
with
msg = msg.replaceAll("<userInput \\/>", Matcher.quoteReplacement(userInput));
solved the problem. The user could put in any kind of characters, including dollar signs, without issue. It behaved exactly the way you would expect.

To have protected pattern you may replace all symbols with "\\\\", except digits and letters. And after that you can put in that protected pattern your special symbols to make this pattern working not like stupid quoted text, but really like a patten, but your own. Without user special symbols.
public class Test {
public static void main(String[] args) {
String str = "y z (111)";
String p1 = "x x (111)";
String p2 = ".* .* \\(111\\)";
p1 = escapeRE(p1);
p1 = p1.replace("x", ".*");
System.out.println( p1 + "-->" + str.matches(p1) );
//.*\ .*\ \(111\)-->true
System.out.println( p2 + "-->" + str.matches(p2) );
//.* .* \(111\)-->true
}
public static String escapeRE(String str) {
//Pattern escaper = Pattern.compile("([^a-zA-z0-9])");
//return escaper.matcher(str).replaceAll("\\\\$1");
return str.replaceAll("([^a-zA-Z0-9])", "\\\\$1");
}
}

Pattern.quote("blabla") works nicely.
The Pattern.quote() works nicely. It encloses the sentence with the characters "\Q" and "\E", and if it does escape "\Q" and "\E".
However, if you need to do a real regular expression escaping(or custom escaping), you can use this code:
String someText = "Some/s/wText*/,**";
System.out.println(someText.replaceAll("[-\\[\\]{}()*+?.,\\\\\\\\^$|#\\\\s]", "\\\\$0"));
This method returns: Some/\s/wText*/\,**
Code for example and tests:
String someText = "Some\\E/s/wText*/,**";
System.out.println("Pattern.quote: "+ Pattern.quote(someText));
System.out.println("Full escape: "+someText.replaceAll("[-\\[\\]{}()*+?.,\\\\\\\\^$|#\\\\s]", "\\\\$0"));

^(Negation) symbol is used to match something that is not in the character group.
This is the link to Regular Expressions
Here is the image info about negation:

How to split by every space preceded by a dot or colon?

I've got a string in Java: Acidum acetylsalic. Acid.ascorb, Calcium which I want to split. The string has to be cut after every space preceded by a dot or colon: ,[space] or .[space]
In result I need three strings: Acidum acetylsalic, Acid.ascorb, Calcium
I know I need some regex and according to this and this I tried "\, |\. " but I doubt that's not how regex work.

Split by
"[,.] "
[,.] - character set with one comma or dot
The problem with your original regex is that you need to escape the dot once to make it a literal dot and a second time to escape the slash escaping it. It will also work if you change it to:
", |\\. "

Try this:
str.split("[\\.,]\\s")
....

Use split("(\.\s)|(\,\s)")
You need to encode special characters see https://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html

Try this
String str="Acidum acetylsalic. Acid.ascorb, Calcium";
String[] resStr= str.split("[.,][\\s]");
for (String res : resStr) {
System.out.println(res);
}
Output :
Acidum acetylsalic
Acid.ascorb
Calcium

How to add regular expression to match alphanumeric character with some special characters

I have the words with special character like
Ex: ABC12-xy
ABCD
ABC12_12
12-AB_xy
I have tried the following but not working
'(-\\w+)' , '[-A-Za-z_0-9]'
But not working.

Try this regex
[\w-]+
Which matches all below
ABC12-xy
ABCD
ABC12_12
12-AB_xy

use [\w-]+ to match the entire string. You can use ^ and $ to specify the start and the end of the line. For example ^[\w-]+$ would match the entire line only if the line has all word or - characters.

String regex = "[A-Za-z0-9_\\-]+";
System.out.println(java.util.regex.Pattern.matches(regex, "ABC12-xy"));
System.out.println(java.util.regex.Pattern.matches(regex, "ABCD"));
System.out.println(java.util.regex.Pattern.matches(regex, "ABC12_12"));
System.out.println(java.util.regex.Pattern.matches(regex, "12-AB_xy"));

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.