Regex help needed, finding repeating patterns

Question

I have some text

I01:00:00:05
I01:00:00:04
I01:00:00:03
I01:00:00:02
I01:00:00:01

Is there a regex that will find each one?

I tried: var locs = txt.match(/(([A-Z]\d\d\:\d\d\:\d\d\:\d\d)+)+/);

And it finds 5 copies of the first pattern that matches.

Is there a way to get an array with each of them in it?

Thanks

Is this text part of a larger document? What exactly are you trying to match? — mellamokb
– mellamokb, Commented Mar 7, 2011 at 18:17
The regex pattern. but I want to end up with all the matches, not just the first one — Tim Brown
– Tim Brown, Commented Mar 7, 2011 at 18:20
Write your regular expression to match a single one, instead of trying to match them all in a single expression. Then use the /g flag, as noted by the answers below. — mellamokb
– mellamokb, Commented Mar 7, 2011 at 18:22

Groovetrain · Accepted Answer · 2011-03-07 18:20:04Z

2

if you include the 'g' flag at the end it should work.

var locs = txt.match(/(([A-Z]\d\d\:\d\d\:\d\d\:\d\d)+)+/g);

answered Mar 7, 2011 at 18:20

Groovetrain

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1 Comment

Tim Brown Over a year ago

That did it! Thanks very much. I should have remembered that from my unix days

Tomas · Accepted Answer · 2011-03-07 18:19:58Z

0

You are simply missing the g flag on your regexp and it will find all the occurences that match the pattern. Like so:

var locs = txt.match(/(([A-Z]\d\d\:\d\d\:\d\d\:\d\d)+)+/g);

of course if that pattern is actually good ;)

answered Mar 7, 2011 at 18:19

Tomas

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mellamokb · Accepted Answer · 2011-03-07 18:20:44Z

0

Write your regular expression to match a single instance, and use the global specifier /g to get a collection of matches (note the modification to regular expression):

var locs = txt.match(/([A-Z]\d\d\:\d\d\:\d\d\:\d\d)/g);

http://rubular.com/r/L2ZNyz2yJy

answered Mar 7, 2011 at 18:20

mellamokb

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Wei Liu · Accepted Answer · 2013-05-24 03:15:04Z

0

Indeed in this case the "+" sign is not necessary:

var locs = txt.match(/[A-Z]\d\d\:\d\d\:\d\d\:\d\d/g);

locs will be ["I01:00:00:05", "I01:00:00:04", "I01:00:00:03", "I01:00:00:02", "I01:00:00:01"].

answered May 24, 2013 at 3:15

Wei Liu

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