CODEWITHSUNDEEP
javascriptregex
user1634845
user1634845

Reputation:

Regex matching in Javascript returning the original string

I'm attempting to use Javascript Regex to select something in this fashion: !something: somethingelse; The exclamation point, the colon, and the semicolon have to be there, but the other parts can be any value. I attempted to do this:

"!test: value; random text here !another: valueagain;".match(/!(.*):(.*);/g);

This returns the original string, !test: value; random text here !another: valueagain;. Is there something I'm doing incorrectly, or is the Regex incorrect?

Upvotes: 0

Views: 191

Answers (4)

StupidDev
StupidDev

Reputation: 342

try this expression :

"!test: value; random text here !another: valueagain;".match(/![^!]*:[^!]*;/g);

return : 

["!test: value;", "!another: valueagain;"]

Upvotes: 0

Cerbrus
Cerbrus

Reputation: 72947

Try this:

.match(/!(.*?):(.*?);/g);

The added questionmarks make the * non-greedy, meaning it matches a string that's as short as possible.

Upvotes: 0

thefourtheye
thefourtheye

Reputation: 239643

You are using greedy matchers, use non-greedy matchers like this

"!test: value; random text here !another: valueagain;".match(/!(.*?):(.*?);/g);
# [ '!test: value;', '!another: valueagain;' ]

(.*): consumes everything till the last : in the string, as it is greedy. That is why you are getting only a single string. Simply make it non-greedy by using ?, like this .*?. This makes sure that it matches till the first :.

Regular expression visualization

Debuggex Demo

Upvotes: 0

merlin2011
merlin2011

Reputation: 75579

As devnull suggested, use non-greedy quantifiers.

"!test: value; random text here !another: valueagain;".match(/!(.*?):(.*?);/g);

Output:

 [ '!test: value;', '!another: valueagain;' ]

In general, regular expression quantifiers like * are greedy, in the sense that they will try to match as large of an expression as possible.

The non-greedy versions *? will match only as much as they need to, to match.

Upvotes: 1

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