Regex handling zero

 

I have string with *(asterisk) symbols as an input. String is considered as invalid if it has two consecutive asterisks. But, there is an escape symbol (backslash).

For example:

  • "**" (invalid)
  • "**" (valid)
  • "case**" (invalid)
  • "case**" (valid)
  • "**" (valid)

    • I'm on stuck on such regex's which produce incorrect result:

  • /[^]**/ - java.util.regex.Pattern.compile("/[^\]**/")
  • /([^]*?**)|(**)/ - java.util.regex.Pattern.compile("/([^\]*?**)|(**)/") .

    • Also, I've read about greedy, reluctant and possessive quantifies from here http://docs.oracle.com/javase/tutorial/essential/regex/quant.html

    I know that problem is about zero-length matches, but could not produce correct regex.

    Are you looking for a regex, that will only match invalid strings? This should do: 

    "(?<!\)**+"

    It will match two or more asterisks in a row, not preceded by a backslash.

    EDIT: (?<!foo) thingy is called "negative look-behind". It matches any zero-length place in the string that is not immediately preceded by a region matching the regex inside parentheses ("foo" in this case, or a backslash in yours). I had this as [^\] at first, which is almost the same thing (in this case), except that it matches any character, other than a backslash, but not an absense of a character, like at the beginning of a string in "**".

    There is a good detailed description of lookarounds (look-behind and look-ahead) as well as a lot of other regex "magic" here

    Use string.matches method. This returns true for valid strings. 

    String s1 = "case**";
String s2 = "case**";
System.out.println(s1.matches("(?=.*(\**|*\*)).*"));
System.out.println(s2.matches("(?=.*(\**|*\*)).*"));

    Output: 

    false
true

    DEMO

    链接地址: http://www.djcxy.com/p/76970.html

    上一篇: 正则表达式贪婪匹配以字符或字符串结尾结束

    下一篇: 正则表达式处理零