regex to allow special characters but not leading/trailing white space

Question

I would like to allow all special characters and white space in between words only for a password input field.

If whitespace entered at the leading, trailing of string, regex should fail

Any useful javascript regex?

I tried \S this does not accept any white space, would that be sufficient?

I tried \A\s|\s*\Z , but not able to negate this.

If the `/\s/` regex matches, the result is fail. If you need to match a string that does not allow whitespace, use `/^\S+$/` — Wiktor Stribiżew, Feb 13 '18 at 22:01
You may benefit from reading [Reference - Password Validation](https://stackoverflow.com/q/48345922) — ctwheels, Feb 13 '18 at 23:10
@wiktor-stribiżew he says white spaces **ARE** allowed. OP wants to reject input that has leading or trailing white spaces. — Stephen Quan, Feb 15 '18 at 02:17
If this is not longer a duplicate then you need to explain why it is different — New Alexandria, Feb 15 '18 at 04:45

score 1 · Answer 1 · answered Feb 13 '18 at 22:00

1

Using something like [^\s] would suffice.

answered Feb 13 '18 at 22:00

Drewness

5,004
4
32
50

1

`\S` is the same as `[^\s]`. – melpomene Feb 13 '18 at 22:16

score 1 · Answer 2 · answered Feb 15 '18 at 07:30

The \A (start of string) and \Z (end of string) anchors are not supported by JS RegExp. If you use /\S/ it will only match any non-whitespace char, anywhere inside a string. If you use /^\s|\s*$/ it will match a whitespace at the start or any 0 or more whitespaces at the end.

You need

/^\S+(?:\s+\S+)*$/

See the regex demo.

It will match:

^ - start of string
\S+ - 1 or more non-whitespace chars
(?:\s+\S+)* - any 0 or more occurrences of
- \s+ - 1+ whitespaces
- \S+ - 1+ non-whitespace chars
$ - end of string.

JS demo:

var strs = ['Abc 123 !@#', 'abc123@', '   abc34', ' a ', 'bvc   '];
var rx = /^\S+(?:\s+\S+)*$/;
for (var s of strs) {
  console.log("'"+s+"'", "=>", rx.test(s));
}

score -1 · Answer 3 · answered Feb 13 '18 at 22:14

-1

I don't know if it's totally fine but in your case, I think this could apply better

^((\w)*){1}$

answered Feb 13 '18 at 22:14

Antonio Rodriguez

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This does not allow any special characters and contains two completely superfluous capturing groups and one redundant quantifier (`{1}` is a no-op). – melpomene Feb 13 '18 at 22:17
You are right, I want to apologize @sri because I read something totally opposite to the question done. Next time I will read it carefully. – Antonio Rodriguez Feb 17 '18 at 23:22

regex to allow special characters but not leading/trailing white space

3 Answers3