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16

I am encoding a string that will be passed in a url (get). But, if I use escape, encodeURI or encodeURIComponent, & will be replaced with %26amp%3B and I want it to be replaced with just %26... What am I doing wrong?

Dodo

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Where does the string come from? Can you post the code you have so far? –  Andy E Aug 22 '10 at 13:56
1  
& is the proper way to escape the ampersand in an HTML context...where is your source coming from? and what's the destination? It may be better to do this server-side for example. –  Nick Craver♦ Aug 22 '10 at 13:59
    
I grap something from the HTML body (and that is HTML encoded (so, there is & I realize now)) and I have to pass it in an URL... So, I need to decode the html (but how?) en then encode the string (with encodeURIComponent)... –  dododedodonl Aug 22 '10 at 14:04
1  
found it... I used in jquery .html(), not .text()... stupid (A) –  dododedodonl Aug 22 '10 at 14:07
1  
jQuery's .html() maps to the innerHTML property, so the issue is as I said in my answer :-) –  Andy E Aug 22 '10 at 14:19

2 Answers 2

up vote 286 down vote accepted

Without seeing your code, it's hard to answer other than a stab in the dark. I would guess that the string you're passing to encodeURIComponent(), which is the correct method to use, is coming from the result of accessing the innerHTML property. The solution is to get the innerText/textContent property value instead:

var str, 
    el = document.getElementById("myUrl");

if ("textContent" in el)
    str = encodeURIComponent(el.textContent);
else
    str = encodeURIComponent(el.innerText);

If that isn't the case, you can use the replace() method to replace the HTML entity:

encodeURIComponent(str.replace(/&/g, "&"));
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up vote 67 down vote

If you did literally this:

encodeURIComponent('&')

Then the result is %26, you can test it here. Make sure the string you are encoding is just & and not & to begin with...otherwise it is encoding correctly, which is likely the case. If you need a different result for some reason, you can do a .replace(/&/g,'&') before the encoding.

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3  
... that is probably his problem. –  Pekka 웃 Aug 22 '10 at 13:58

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