Replace images source for all images

三世轮回 提交于 2019-11-28 12:59:53

Your regular expression is the problem, you need to wrap it in slashes (otherwise the interpreter has no clue that it should treat that as a regexp):

var allImg=document.getElementsByTagName("img"), i=0, img;

var pattern = /^http\:\/\/aff\.kooora\.com\/(.*)$/;

while (img = allImg[i++])
{
    if (img.src.match(pattern)) {
        img.src = img.src.replace(pattern, 'http://example.com/aff.kooora.com/$1');
    }
}

Just an alternative.

In this Greasemonkey script, ProxyLoadIMG, this is accomplished without using regex.

It was created to load images through a proxy.
Only the images. Links and other content remains untouched.

var proxy = 'https://proxytea.appspot.com/'

for (var x=0; x<document.images.length; x++) {
    imagem = document.images[x];
    imagem.src = imagem.src.replace(location.protocol + '//', proxy);
}

http://aff.kooora.com/i/any.jpg becomes https://proxytea.appspot.com/aff.kooora.com/i/any.jpg
http://aff.kooora.com/i/z3.gif becomes https://proxytea.appspot.com/aff.kooora.com/i/z3.gif
and so on...

No RegEx needed :)

Use this:

function replaceImages(oldUrl, newUrl) {
    var imgs = document.getElementsByTagName('img')
    for (i = 0; i<imgs.length; i++) {
        imgs[i].src = imgs[i].src.replace(oldUrl, remoteUrl)
    }
}

replaceImages('http://aff.kooora.com/i/z3.gif', 'http://example.com/aff.kooora.com/i/z3.gif')

This is also great if you're working on a local version of a site and you can't be bothered to download all their image content

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