Java regex error - Look-behind group does not have an obvious maximum length

穿精又带淫゛_ 提交于 2019-11-27 22:11:59

Java doesn't support variable length in look behind.
In this case, it seems you can easily ignore it (assuming your entire input is one word):

([a-z])(?!.*\1)([a-z])(?!.*\2)(.)(\3)(.)(\5)

Both lookbehinds do not add anything: the first asserts at least two characters where you only had one, and the second checks the second character is different from the first, which was already covered by (?!.*\1).

Working example: http://regexr.com?2up96

luobo25

To avoid this error, you should replace + with a region like {0,10}:

([a-z])(?!.*\1)(?<!\1.{0,10})([a-z])(?!.*\2)(?<!\2.{0,10})(.)(\3)(.)(\5)
boiledwater

Java takes things a step further by allowing finite repetition. You still cannot use the star or plus, but you can use the question mark and the curly braces with the max parameter specified. Java determines the minimum and maximum possible lengths of the lookbehind.
The lookbehind in the regex (?<!ab{2,4}c{3,5}d)test has 6 possible lengths. It can be between 7 to 11 characters long. When Java (version 6 or later) tries to match the lookbehind, it first steps back the minimum number of characters (7 in this example) in the string and then evaluates the regex inside the lookbehind as usual, from left to right. If it fails, Java steps back one more character and tries again. If the lookbehind continues to fail, Java continues to step back until the lookbehind either matches or it has stepped back the maximum number of characters (11 in this example). This repeated stepping back through the subject string kills performance when the number of possible lengths of the lookbehind grows. Keep this in mind. Don't choose an arbitrarily large maximum number of repetitions to work around the lack of infinite quantifiers inside lookbehind. Java 4 and 5 have bugs that cause lookbehind with alternation or variable quantifiers to fail when it should succeed in some situations. These bugs were fixed in Java 6.

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