Javascript regex always returns true

Question

var re = /^\[*:*\]*/;
alert(re.test("aa"));

Always alerts true (on any string). What am i doing wrong? I need to check is there something like [445:123] in the beginning of the string.

Your regex means "Start of string, then zero of more [ then zero or more :, then zero of more ] . aa does contains 0 or more of the above. What are you trying to test here? — Benjamin Gruenbaum
– Benjamin Gruenbaum, Commented Oct 22, 2013 at 0:02

Pointy · Accepted Answer · 2013-10-22 00:10:01Z

5

Your regex means

at the start of the string ...
there can be zero or more [ characters ...
then zero or more : characters ...
then zero or more ] characters.

The string "aa" matches that. You probably want something like:

var re = /^\[\d+:\d+\]/;

The + quantifier means "one or more", while * means "zero or more". The \d escape means "any digit".

*edit — if the regex needs to match something like

[12:2:17:419]

as well, then it would be

var re = /^\[\d+(:\d+)+\]/;

edited Oct 22, 2013 at 0:10

answered Oct 22, 2013 at 0:02

Pointy

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5 Comments

Xandrmoro Over a year ago

Thank you! Can you give an advice how to build a correct regex?

Ruan Mendes Over a year ago

What the OP could use /^\[\d*:\d*\]/

Xandrmoro Over a year ago

@Pointy unfortunatelly, there can be any number (but at least two) of \d: groups. Is there any way to solve this?

Pointy Over a year ago

@user32876 sure; you can group the ":" and the second number. I'll add stuff. Here's a decent website about regular expressions.

Xandrmoro Over a year ago

@Pointy last regex worked fine. Thank you very much!

Josh · Accepted Answer · 2013-10-22 00:03:00Z

1

You're checking for any number of [, followed by any number of :, followed by any number of ]. Note that that's any number - 0 occurrences of any of those is a valid result.

What it sounds like you mean to do is something more like var re= /^\[\d+:\d+\]/;

answered Oct 22, 2013 at 0:03

Josh

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Comments

Billy Moon · Accepted Answer · 2013-10-22 00:05:24Z

0

in regular expressions, * does not mean wildcard match, it means zero or more of the previous token. To match any character, use . instead. A regex for what you want to match is more like this...

/^\[.*:.*\].*/

But better is to be more specific, using \d to match decimals, and use + to match one or more of the previous token, and remove the erroneous characters after the match...

/^\[\d+:\d+\]/

answered Oct 22, 2013 at 0:05

Billy Moon

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