Javascript正则表达式匹配以某些字符结尾但不与这些字符的特定组合结束的字符串

Question

Let's say using Javascript, I want to match a string that ends with [abcde]* but not with abc .

So the regex should match xxxa , xxxbc , xxxabd but not xxxabc .

I am utterly confused.

Edit: I have to use regex for some reason, i cannot do something if (str.endsWith("abc"))

Answer 1

The solution is simple: use negative lookahead:

(?!.*abc$)

This asserts that the string doesn't end with abc .

You mentioned that you also need the string to end with [abcde]* , but the * means that it's optional, so xxx matches. I assume you really want [abcde]+ , which also simply means that it needs to end with [abcde] . In that case, the assertions are:

(?=.*[abcde]$)(?!.*abc$)

See regular-expressions.info for tutorials on positive and negative lookarounds.

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I was reluctant to give the actual Javascript regex since I'm not familiar with the language (though I was confident that the assertions, if supported, would work -- according to regular-expressions.info , Javascript supports positive and negative lookahead). Thanks to Pointy and Alan Moore's comments, I think the proper Javascript regex is this:

var regex = /^(?!.*abc$).*[abcde]$/;

Note that this version (with credit to Alan Moore) no longer needs the positive lookahead. It simply matches .*[abcde]$ , but first asserting ^(?!.*abc$) .

Answer 2

Either the question is not properly defined, or everyone is overlooking a simple answer.

var re = /abc$/;

!re.test("xxxa");    // pass
!re.test("xxxbc");   // pass
!re.test("xxxabd");  // pass
!re.test("xxxabc");  // fail

All of these end in /[abcde]* /

Answer 3

Firstly, note every string ends with [abcde]* , as that allows zero width. Thus we're really just looking for a regex that matches strings that don't end in abc . Easy.

([^c]|[^b]c|[^a]bc)$

That's something that's not c , something that's not b followed by c , or something that's not a followed by bc , and whichever option of those, then followed by the end of the string.

Answer 4

Just want to say to @maček that you answer is simple and clear with ! in the beginning But I really wanted the not operator to be inside the regexp so i could use it with eg: Array::some , Array::filter , Array::find , Array::findIndex and without having to create a function or a wile loop

Here is an example:

 var items = ["ja", "ka", "mf", "bg", "vb", "b"] var regex = /^((?!a$|b$).)*$/; // match a string that doesn't end with a or b var matched = items.filter(RegExp.prototype.test, regex); alert(matched); // result: ["mf", "bg"] 

Answer 5

Hmm ...

var regex = /(ab[abde]|[abcde])$/; // wrong

maybe? wait no; hold on:

var regex = /(ab[abde]|([^a].|.[^b])[abcde]|\b.?[abcde])$/;

So that's "ab" followed by "a", "b", "d", or "e"; or any two-character sequence where the first character isn't "a" or the second character isn't "b", followed by "a" through "e"; or any word boundary followed by any character (possibly) followed by "a" through "e". The last clause is to deal with short strings; it's sort-of cheating but in this case it works.