Basically, I want to do "zipzam&&&?&&&?&&&" -> "zipzam%26%26%26?&&&?&&&". I can do that without regex many different ways, but it'd cleanup things a tad bit if I could do it with regex.

Thanks

Edit: "zip=zam&&&=?&&&?&&&" -> "zip=zam%26%26%26=?&&&?&&&" should make things a little clearer.

Edit: "zip=zam=&=&=&=?&&&?&&&" -> "zip=zam=%26=%26=%26=?&&&?&&&" should make things clearer.

However, theses are just examples. I still want to replace all '&' before the first '?' no matter where the '&' are before the first '?' and no matter if the '&' are consecutive or not.

Basically, I want to do "zipzam&&&?&&&?&&&" -> "zipzam%26%26%26?&&&?&&&". I can do that without regex many different ways, but it'd cleanup things a tad bit if I could do it with regex.

Thanks

Edit: "zip=zam&&&=?&&&?&&&" -> "zip=zam%26%26%26=?&&&?&&&" should make things a little clearer.

Edit: "zip=zam=&=&=&=?&&&?&&&" -> "zip=zam=%26=%26=%26=?&&&?&&&" should make things clearer.

However, theses are just examples. I still want to replace all '&' before the first '?' no matter where the '&' are before the first '?' and no matter if the '&' are consecutive or not.

• javascript
• regex

Share Improve this question Follow edited Apr 3, 2009 at 22:19 Chad Birch 74.7k2323 gold badges155155 silver badges150150 bronze badges asked Jan 26, 2009 at 13:03 Shadow2531Shadow2531 12.2k55 gold badges3838 silver badges5151 bronze badges

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7 Answers 7

Sorted by: Reset to default 4

This should do it:

"zip=zam=&=&=&=?&&&?&&&".replace(/^[^?]+/, function(match) { return match.replace(/&/g, "%26"); });

you need negative lookbehinds which are tricky to replicate in JS, but fortunately there are ways and means:

var x = "zipzam&&&?&&&?&&&";

x.replace(/(&+)(?=.*?\?)/,function ($1) {for(var i=$1.length, s='';i;i--){s+='%26';} return s;})

mentary: this works because it's not global. The first match is therefore a given, and the trick of replacing all of the matching "&" chars 1:1 with "%26" is achieved with the function loop

---

edit: a solution for unknown groupings of "&" can be achieved simply (if perhaps a little clunkily) with a little modification. The basic pattern for replacer methods is infinitely flexible.

var x = "zipzam&foo&bar&baz?&&&?&&&";

var f = function ($1,$2)
{
  return $2 + ($2=='' || $2.indexOf('?')>-1 ? '&' : '%26')
}

x.replace(/(.*?)&(?=.*?\?)/g,f)

This should do it:

^[^?]*&[^?]*\?

Or this one, I think:

^[^?]*(&+?)\?

In this case regexes are really not the most appropiate things to use. A simple search for the first index of '?' and then replacing each '&' character would be best. However, if you really want a regex then this should do the job.

(?:.*?(&))*?\?

This close enough to what you are after:-

alert("zipzam&&&?&&&?&&&".replace(/^([^&\?]*)(&*)\?/, function(s, p, m)
{
for (var i = 0; i < m.length; i++) p += '%26';
return  p +'?';
}));

Since the OP only wants to match ampersands before the first question mark, slightly modifying Michael Borgwardt's answer gives me this Regex which appears to be appropriate :

^[^?&]*(\&+)\?

Replace all matches with "%26"

This will not match zipzam&&abc?&&&?&&& because the first "?" does not have an ampersand immediately before it.