using javascript regex to capture text in a url

Question

I need to use a regular expression in javascript to match certain text in a url.

Here is an example:

"qty1=0&qty2=0&qty3=0&"

I want to match the text(number) that comes after the '=' and before the '&' whenever the word 'qty' appears in the string. The text(number) will be subject to change every time the url is generated, so I want to match whatever the new text(number) may be.

What regular expression could I use to solve this?

Any help would be appreciated!

I need to clarify a bit. Match was the wrong word to use, my apologies. I simply need to use a regular expression to find whatever text(number) comes between "=" and "&" whenever "qty" shows up in the text string. I then need to store this text(number) in a variable so I can run some conditional statements against it. — gijoe426
– gijoe426, Commented Feb 2, 2012 at 3:40

driangle · Accepted Answer · 2012-02-02 17:24:11Z

1

var queryStr = "qty1=0&qty2=0&qty3=0";
var regex = /qty(\d+)=(\d+)/g;
var match = regex.exec(queryStr);
while(match != null){
  var numBeforeEquals = match[1];
  var numAfterEquals = match[2];
  // do your comparison, for example, if it's > 0:
  if(numAfterEquals > 0){
     // do something
  }
  match = regex.exec(queryStr);
}

Note the /g at the end of the regex, in some javascript implementations you will end up with an infinite loop without it.

edited Feb 2, 2012 at 17:24

answered Feb 2, 2012 at 3:18

driangle

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Rezigned · Accepted Answer · 2012-02-02 03:50:21Z

0

try this

var regex = /qty[\d]+=(\d+)/g,
    text  = "qty1=5&qty2=7&qty3=8&";

while(match = regex.exec(text)) {
    console.log(match[1]);
}

@ggreiner missing + in \d that make the script above does not work with number greather than 9 e.g. 10, 11

answered Feb 2, 2012 at 3:50

Rezigned

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2 Comments

gijoe426 Over a year ago

This definitely gives me number beyond 0-9 between "qty" and "=" as well as "=" and "&" (as far as I can tell). But the regular expression matches "qty1=0","qty2=0","qty3=0" instead of just the number after "=" as in "0","0","0".

Rezigned Over a year ago

You missed my point match[1] this is what you want (match only that number).

Matt Lo · Accepted Answer · 2012-02-02 04:36:10Z

0

Here is an alternative solution (inside a working URL). I call it alternative because I'm using the parts of the expression twice, I think it could be optimized.

var breakdown = "http://local/?wtf=omg&qty12=a&qty2=b&qty3=4&test=5".match(/qty[0-9]{0,9999}(.*?)&/ig);
for(var i in breakdown) {
   var v = breakdown[i].match(/qty[0-9]{0,9999}\=(.*?)&/i)[1];
   if(v == 4 || v == "a") {
      alert("Found '"+v+"'!")
   }
}

Working Example: http://jsfiddle.net/MattLo/FxDyg/

edited Feb 2, 2012 at 4:36

answered Feb 2, 2012 at 3:48

Matt Lo

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2 Comments

gijoe426 Over a year ago

This is almost exactly what I need. Is there a way to change the regular expression to match any number after "qty" and the "="? Right now it only checks numbers 0-9. Also, is there a way I can capture each value individually so I can run a conditional if statement against each value in order to run separate blocks of code?

Matt Lo Over a year ago

Yes and yes, I updated the snippet to accept up to 9999 and an area for your conditional statement

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