regular expression in javascript? [duplicate]

Question

Possible Duplicate:
Using a string variable as regular expression

I have a var that contains all English words. I want to use regular exp in order to filter the words according to specific letter that are stored in a var.

 var letter1 = "a";
    var letter2 = "b";
var words= "aardvark aardwolf aaron aback abacus abaft abalone abandon abandoned abandonment abandons abase abased abasement abash abashed abate abated abatement abates abattoir";

I tried this in order to find all the words that have the letter "a" and "b", but it is not working:

document.writeln(words.match(/letter1 +/ + letter2 +/g));

Your intent is not quite clear. What exactly do you want to get? All the words containing letter1 and letter2 at the same time? — Alexander Pavlov
– Alexander Pavlov, Commented Mar 24, 2012 at 8:04
<script type="text/javascript"> var abc = ["asdf","asfdf","sdgsdwvc","avdvcccv","acbfd","sdfsd","gvvfvbb","sdfge34"]; var pattern = /.*[ab].*/i; var res; for(a in abc) { res = abc[a].match(pattern); if(res!=null || res!=undefined) alert(res); } </script> — user900360
– user900360, Commented Mar 24, 2012 at 8:26

Joseph · Accepted Answer · 2012-03-24 08:18:37Z

3

this pattern should do it:

/\S*[ab]\S*/g

and so

var pattern = new RegExp('\S*['+letter1+'|'+letter2+']\S*','g');
var matches = words.match(pattern);
document.writeln(matches);

edited Mar 24, 2012 at 8:18

answered Mar 24, 2012 at 8:05

Joseph

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1 Comment

user1200640 Over a year ago

I need to find the whole word not only the two letters

tjb · Accepted Answer · 2012-03-24 08:22:51Z

0

>>> "acadfb".match(/a.*b|b.*a/)
["acadfb"]
>>> "basdfc".match(/a.*b|b.*a/)
["ba"]
>>> "asdfbasdfcvb".match(/a.*b|b.*a/)
["asdfbasdfcvb"]
>>> "cadsfg".match(/a.*b|b.*a/)


var myStringArray = ["Hello","World","asdfsadf","asdbc","cdjfieab"];
for (var i = 0; i < myStringArray.length; i++) {
   if(myStringArray[i].match(/a.*b|b.*a/)){
    console.log(myStringArray[i]);
   };
}

edited Mar 24, 2012 at 8:22

answered Mar 24, 2012 at 8:15

tjb

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