Regex to match against something that is not a specific substring

I am looking for a regex that will match a string that starts with one substring and does not end with a certain substring.

Example:

// Updated to be correct, thanks @Apocalisp
^foo.*(?<!bar)$

Should match anything that starts with "foo" and doesn't end with "bar". I know about the [^...] syntax, but I can't find anything that will do that for a string instead of single characters.

I am specifically trying to do this for Java's regex, but I've run into this before so answers for other regex engines would be great too.

Thanks to @Kibbee for verifying that this works in C# as well.

---

我认为在这种情况下，你想要消极的后顾之忧 ，就像这样：

foo.*(?<!bar)

---

Verified @Apocalisp's answer using:

import java.util.regex.Pattern;
public class Test {
  public static void main(String[] args) {
    Pattern p = Pattern.compile("^foo.*(?<!bar)$");
    System.out.println(p.matcher("foobar").matches());
    System.out.println(p.matcher("fooBLAHbar").matches());
    System.out.println(p.matcher("1foo").matches());
    System.out.println(p.matcher("fooBLAH-ar").matches());
    System.out.println(p.matcher("foo").matches());
    System.out.println(p.matcher("foobaz").matches());
  }
}

This output the the right answers:

false
false
false
true
true
true

---

I'm not familiar with Java regex but documentation for the Pattern Class would suggest you could use (?!X) for a non-capturing zero-width negative lookahead (it looks for something that is not X at that postision, without capturing it as a backreference). So you could do:

foo.*(?!bar) // not correct

Update : Apocalisp's right, you want negative lookbehind. (you're checking that what the .* matches doesn't end with bar)

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