Java正则表达式-转义所有特殊的正则表达式字符

Question

I do have a lot of strings with special regex characters. An examples:

- Test1 + Test2 -> plus should not be a regex special character but a normal character

Is there a Java regex method to escape all regex special characters?

Answer 1

The comments Advice to better escape it manually is generally right - but if your input String is "unknown", for instance a string the user can enter, you can't do this. So, assuming, your string is a variable, you are most likely looking for Pattern.quote : https://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html#quote(java.lang.String)

String variableFromSomewhere="- Test1 + Test2";
String escapedString = Pattern.quote(variableFromSomewhere);

(This does nothing else than using \\Q and \\E , but requires less effort while typing and avoids problems if \\Q and \\E is a part of the input string itself.)

Answer 2

This is how to escape all regex metacharacters.

Background:

1. If you need to convert some input or dynamic source into a regex, it is assured
   that it is %100 a literal.

2. \\Q .. \\E is used if you have a MIX of regex constructs and literals that use regex constructs via metacharacters.

   Example: (?:\\Q(?:dogs|cats)*\\E)+
   This will match one or more literal (?:dogs|cats)*

There are other issues with \\Q .. \\E like nesting and interpreting the
final resultant escaped regex. It becomes very difficult to debug.

So, the easiest and safest thing is to just use String.replaceAll() .

Java sample:

 String src = "he,<>!!llo \\ + * ? [ ] ( ) { } | . ^ $ wo-r@l#d";
 System.out.println( src );
 src = src.replaceAll("([\\\\+*?\\[\\](){}|.^$])", "\\\\$1");
 System.out.println( src );

Output:

he,<>!!llo \ + * ? [ ] ( ) { } | . ^ $ wo-r@l#d
he,<>!!llo \\ \+ \* \? \[ \] \( \) \{ \} \| \. \^ \$ wo-r@l#d

Answer 3

要转义单个特殊字符，可以使用\\\\

boolean b = Pattern.matches("\\- .* \\+ .*",  "- Test + Test"); // true