up vote 2 down vote favorite
2

I've been looking for an efficient way to do this but haven't been able to find it, basically what I need is that given this url for example:

http://localhost/mysite/includes/phpThumb.php?src=http://media2.jupix.co.uk/v3/clients/4/properties/795/IMG_795_1_large.jpg&w=592&aoe=1&q=100

I'd like to be able to change the URL in the src parameter with another value using javascript or jquery, is this possible?

Thanks in advance.

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3 Answers

up vote 6 down vote accepted

Wouldn't this be a better solution?

var text = 'http://localhost/mysite/includes/phpThumb.php?src=http://media2.jupix.co.uk/v3/clients/4/properties/795/IMG_795_1_large.jpg&w=592&aoe=1&q=100';
var newSrc = 'www.google.com';
var newText = text.replace(/(src=).*?(&)/,'$1' + newSrc + '$2');

EDIT:

added some clarity in code and kept 'src' in the resulting link

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great this worked perfectly! – javiervd Aug 24 '11 at 6:59
Doesn't handle the general case of replacing the last parameter. I changed the regex to /(src=).*?(&)?/ for a similar problem. – Bearddo Aug 14 '12 at 18:54
@ZenMaster Perfect :), can you explain meaning of text.replace(/(src=).*?(&)/,'$1' + newSrc + '$2'); – VaibhaV Feb 17 at 9:48
2 capturing groups... See more here: stackoverflow.com/questions/3512471/non-capturing-group – ZenMaster Feb 17 at 13:07
up vote 0 down vote

How about something like this:

<script>
function changeQueryVariable(keyString, replaceString) {
    var query = window.location.search.substring(1);
    var vars = query.split("&");
    for (var i = 0; i < vars.length; i++) {
        var pair = vars[i].split("=");
        if (pair[0] == keyString) {
            vars[i] = pair[0] + "=" replaceString
        }
    }
    return vars.join("&");
}
</script>
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The only problem is that the URL doesn't come from the browser but from a variable, so I can't use the search method :( – javiervd Aug 24 '11 at 6:14
That is okay, you can do the same thing with regex. Here is an example: stackoverflow.com/questions/901115/… – Robert Martin Aug 24 '11 at 6:21
up vote 0 down vote

UpdatE: Make it into a nice function for you: http://jsfiddle.net/wesbos/KH25r/1/

function swapOutSource(url, newSource) {
    params = url.split('&');
    var src = params[0].split('=');
    params.shift();
    src[1] = newSource;
    var newUrl = ( src.join('=') + params.join('&')); 
    return newUrl; 
}

Then go at it!

var newUrl = swapOutSource("http://localhost/mysite/includes/phpThumb.php?src=http://media2.jupix.co.uk/v3/clients/4/properties/795/IMG_795_1_large.jpg&w=592&aoe=1&q=100","http://link/to/new.jpg");


console.log(newUrl);
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The problem here is that you are always replacing the 1st parameter. So if I rearrange the string to be "localhost/mysite/includes/phpThumb.php?q=100&src=http://…; All the sudden it will break... – Robert Martin Aug 24 '11 at 6:25

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