Regex pattern SyntaxException

Question

In my app, I'd like to remove all text in between "example" and the first occurance after this of } from a string. And I want to do this for all occurences. So I use this code:

myString.replaceAll("\"example\"(.+?)}", "");

However, this gives me a PatternSyntaxException. Why? And: how do I solve it?

stack trace:

05-10 23:32:16.129: W/System.err(724): java.util.regex.PatternSyntaxException: Syntax error in regexp pattern near index 16:
05-10 23:32:16.129: W/System.err(724): "example"(.+?)}
05-10 23:32:16.129: W/System.err(724):                 ^
05-10 23:32:16.159: W/System.err(724):  at java.util.regex.Pattern.compileImpl(Native Method)
05-10 23:32:16.190: W/System.err(724):  at java.util.regex.Pattern.compile(Pattern.java:400)
05-10 23:32:16.190: W/System.err(724):  at java.util.regex.Pattern.<init>(Pattern.java:383)
05-10 23:32:16.219: W/System.err(724):  at java.util.regex.Pattern.compile(Pattern.java:374)
05-10 23:32:16.219: W/System.err(724):  at java.lang.String.replaceAll(String.java:1784)
...

Well, never mind... I just tried System.out.println("\"example\" hello world } hello world".replaceAll("\"example\"(.+?)}", "")); and it worked fine for me. Do you have a stack trace to show? — awksp
– awksp, Commented May 10, 2014 at 21:43
Hmmmm, how strange. What happens if you put a double backslash (`\` x2) before the last curly brace? (edit: turns out I needed to escape the backslashes here!) — awksp
– awksp, Commented May 10, 2014 at 22:07
The ^ is supposed to point to the character that's causing the PatternSyntaxException. That's what makes this exception so strange -- the character in question appears to be pointing at nothing in particular. — awksp
– awksp, Commented May 10, 2014 at 22:08

awksp · Accepted Answer · 2014-05-10 22:31:43Z

2

OK, so I don't understand why this gave an exception, but it seems that what you need to do is escape the last curly brace. So instead of

myString.replaceAll("\"example\"(.+?)}", "");

you do

myString.replaceAll("\"example\"(.+?)\\}", "");
                                     ^^

The first string worked for me in Java 1.7.0_51 and 1.8.0_05, so I'm not sure how this came about... But it works?

answered May 10, 2014 at 22:31

awksp

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Onur · Accepted Answer · 2014-05-10 21:48:46Z

0

"\\\"example\\\"(.+?)}"

should work. Your string will be converted to a pattern, which means it will process the escape chars twice. so you need to type \\ for \ and \" for " .

Here is the example http://www.myregextester.com/?r=ab9d1f06

answered May 10, 2014 at 21:48

Onur

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3 Comments

awksp Over a year ago

The thing is the example as provided worked just fine for me, and the issue seems to be that the regex compiler appears to be reading past the end of the regex string...

awksp Over a year ago

And I think the escape chars are actually for the Java compiler, not for the pattern compiler

Xander Over a year ago

This solution didn't help, but thank you for your help anyway!

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