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Home/ Questions/Q 7600125
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Editorial Team
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Editorial Team
Asked: May 30, 20262026-05-30T22:50:14+00:00 2026-05-30T22:50:14+00:00

This tutorial http://www.blocsoft.com/blog/imageshack.asp shows a great way to upload to imageshack and get a

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This tutorial http://www.blocsoft.com/blog/imageshack.asp shows a great way to upload to imageshack and get a image url back although its for ASP and I need it in PHP.

Any ideas if and how this could be done?

Thanks heaps.

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  1. Editorial Team
    Editorial Team
    2026-05-30T22:50:15+00:00Added an answer on May 30, 2026 at 10:50 pm

    I just rewrote the asp to php, no clue if it works haven’t tested it, but if the asp does work then the php should also.

    index.php

    <html>
      <head>
        <title>AJAX image upload</title>
      </head>
      <body>
        <form action="http://imageshack.us/redirect_api.php" target="AXframe" method="post" enctype="multipart/form-data">
          <input type="file" name="media"/>
          <input type="hidden" name="key" value="YOUR_DEVELOPER_KEY">
          <input type="hidden" name="error_url" value="http://example.com/error.php">
          <input type="hidden" name="success_url" value="http://example.com/success.php?one=%y&two=%u&three=%s&four=%b&five=%i">
          <input type="submit"/>
        </form>
        <iframe style="visibility:hidden" id="AXframe" name="AXframe"></iframe>
        <div id="link"></div>
        <div id="yfrog"></div>
        <div id="image"></div>
      </body>
    </html>
    

    success.php

    <?php
      $str1 = $_GET["one"];
      $str2 = $_GET["two"];
      $str3 = $_GET["three"];
      $str4 = $_GET["four"];
      $str5 = $_GET["five"];
    ?>
    
    <script type="text/javascript">
      parent.document.getElementById('yfrog').innerHTML =  '<?php echo($str1); ?>';
      parent.document.getElementById('link').innerHTML = '<?php echo($str2); ?>';
      parent.document.getElementById('image').innerHTML = '<img src="http://img<?php echo($str3); ?>.imageshack.us/img<?php echo($str3); ?>/<?php echo($str4); ?>/<?php echo($str5); ?>">';
    </script>
    

    error.php

    <script type="text/javascript">
      alert("There was an error uploading the file.");
    </script>
    
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