javascript regular expression to not match a word

Question

How do I use a javascript regular expression to check a string that does not match certain words?

For example, I want a function that, when passed a string that contains either abc or def, returns false.

'abcd' -> false

'cdef' -> false

'bcd' -> true

EDIT

Preferably, I want a regular expression as simple as something like, [^abc], but it does not deliver the result expected as I need consecutive letters.

eg. I want myregex

if ( myregex.test('bcd') ) alert('the string does not contain abc or def');

The statement myregex.test('bcd') is evaluated to true.

score 165 · Accepted Answer · edited Mar 24 '21 at 22:55

165

This is what you are looking for:

^((?!(abc|def)).)*$

The ?! part is called a negative lookahead assertion. It means "not followed by".

The explanation is here: Regular expression to match a line that doesn't contain a word

edited Mar 24 '21 at 22:55

Matthias

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answered Mar 19 '13 at 19:27

ssgao

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1

This is the answer I expect! Thanks. I need a regular expression rather than a function. My question was edited and answers came up to answer the new version of my question. This is why I used an "EDIT" part to avoid confusion. – bxx Mar 21 '13 at 03:53
2

Is there an answer that doesn't match a whole word? Your example "abc", "babc" and "abcd" all fail and where as "xyz" passes. I need "abc" to fail but "abcd" to pass. Removing the `.` and `*` don't seem to work – gman Jan 02 '17 at 17:15

Petar Ivanov · Answer 2 · 2011-06-23T04:05:44.093

23

if (!s.match(/abc|def/g)) {
    alert("match");
}
else {
    alert("no match");
}

edited Jun 23 '11 at 04:05

answered Jun 23 '11 at 03:58

Petar Ivanov

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Why are you capturing in your regex? – Flimzy Jun 23 '11 at 04:02
Good point! I think I thought if I don't capture it might mean ab(c|d)ef. – Petar Ivanov Jun 23 '11 at 04:05
1

It is maybe better to use `/abc|def/g.test(s)` that return a boolean in this case ;) – TOPKAT Jan 05 '19 at 21:38

score 7 · Answer 3 · answered Oct 10 '12 at 14:23

7

Here's a clean solution:

function test(str){
    //Note: should be /(abc)|(def)/i if you want it case insensitive
    var pattern = /(abc)|(def)/;
    return !str.match(pattern);
}

answered Oct 10 '12 at 14:23

NoBrainer

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Flimzy · Answer 4 · 2011-06-23T04:18:58.383

2

function test(string) {
    return ! string.match(/abc|def/);
}

edited Jun 23 '11 at 04:18

answered Jun 23 '11 at 03:57

Flimzy

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score 1 · Answer 5 · answered Jun 17 '15 at 10:43

1

This can be done in 2 ways:

if (str.match(/abc|def/)) {
                       ...
                    }


if (/abc|def/.test(str)) {
                        ....
                    }

answered Jun 17 '15 at 10:43

Girish Gupta

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score 1 · Answer 6 · answered Jun 23 '11 at 04:00

1

function doesNotContainAbcOrDef(x) {
    return (x.match('abc') || x.match('def')) === null;
}

answered Jun 23 '11 at 04:00

Bemmu

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javascript regular expression to not match a word

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