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pass url variable to javascript


By : Rajveer Singh Chauhan
Date : May 07 2020, 10:07 AM

I have next script that is not giving me the result I want. JSFiddle link at the bottom.

Note: This is not my image. I grabbed it randomly from imgur.

I expect:

<img src="https://i.imgur.com/MiOeWrk.jpg" style="max-width:800px;">

But I get this output , I believe, but I'm unable to write this in a code block while debugging, for some reason.

<img src="http://thisismydomain.com/https://i.imgur.com/MiOeWrk.jpg" style="max-width:800px;">

Here's my script code. Please advise:

<div><a href="javascript:viewimage('https://i.imgur.com/MiOeWrk.jpg','video_aba');">Show Image</a></div>

<div id="jsdebug">jsdebug</div>
<div id="video_aba">video_aba</div>

<script>    
function viewimage(ytlink, uid) {
  var imagelink = ytlink;
  document.getElementById("jsdebug").innerHTML = imagelink;
  var uid = uid;
  if (imagelink.length == 0) {
    imageembed = "<strong>Sorry, unable to load.</strong>";
  } else {
    imageembed = "<img src=\"/" + imagelink + "\" style=\"max-width:800px\">";
  }
  document.getElementById("video_aba").innerHTML = imageembed;
}
</script>

https://jsfiddle.net/gde5630h/

Answer :

The issue is what your image link coded . It's like this

imageembed = "<img src=\"/" + imagelink + "\" style=\"max-width:800px\">";
 => this will prepend to your current domain url eg.if your domain is localhost it will goes localhost/https://i.imgur.com/MiOeWrk.jpg

So remove / from your image link



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