Change href in another PHP page using a variable -

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i'm brand new , trying scab 2 different ideas. each of them work individually, i'm trying integrate them.

  1. i able upload , image (which creates list itme href="javascript:void(0).
  2. i can allow user input assign url existing text link.

what i'm trying target dynamically created list item in upload.php page , change href="javascript:void(0) user specified variable index.php

index.php code

<!doctype html> <html lang="en"> <head> <meta http-equiv="content-type" content="text/html; charset=utf-8" /> <title>image upload</title> <link href="style.css" rel="stylesheet" type="text/css" /> <script type="text/javascript" src="jquery-1.8.3.min.js"></script> <script type="text/javascript" src="jquery.form.js"></script> <script type="text/javascript"> $(document).ready(function(){     $('#images').on('change',function(){         $('#multiple_upload_form').ajaxform({             target:'#images_preview',             beforesubmit:function(e){                 $('.uploading').show();             },             success:function(e){                 $('.uploading').hide();             },             error:function(e){             }         }).submit();     }); });  </script> </head>  <body> <div style="margin-top:50px;">     <div class="upload_div" >     <form name="multiple_upload_form" id="multiple_upload_form" enctype="multipart/form-data" action="upload.php">         <input type="hidden" name="image_form_submit" value="1"/>             <label>choose image</label>             <input type="file" name="images[]" id="images" multiple >         <div class="uploading none">             <label>&nbsp;</label>             <img src="uploading.gif"/>          </div>     </form>     </div>      <div class="gallery" id="images_preview">      </div>  </div> <div class="form_div" id="url_input"> <form id="form1"> url - example http://www.yourdomain.com: <input name="name" type="text" size="20"> </form> <button onclick="outputname()">submit</button> <script>  function outputname() { var x,y,name,a,b,answer; x=document.getelementbyid("form1") ; y=x.elements["name"].value; document.getelementbyid("upload.php#web").setattribute("href",y);  }  </script>  <p id="demo">  <a id="link">this link</a>  </p> </div> </body> </html>

upload.php code

    <?php if($_post['image_form_submit'] == 1) {     $images_arr = array();     foreach($_files['images']['name'] $key=>$val){         $image_name = $_files['images']['name'][$key];         $tmp_name   = $_files['images']['tmp_name'][$key];         $size       = $_files['images']['size'][$key];         $type       = $_files['images']['type'][$key];         $error      = $_files['images']['error'][$key];          ############ remove comments if want upload , stored images "uploads/" folder #############          /*$target_dir = "uploads/";         $target_file = $target_dir.$_files['images']['name'][$key];         if(move_uploaded_file($_files['images']['tmp_name'][$key],$target_file)){             $images_arr[] = $target_file;         }*/          //display images without stored         $extra_info = getimagesize($_files['images']['tmp_name'][$key]);         $images_arr[] = "data:" . $extra_info["mime"] . ";base64," . base64_encode(file_get_contents($_files['images']['tmp_name'][$key]));     }      //generate images view     if(!empty($images_arr)){ $count=0;         foreach($images_arr $image_src){ $count++?>             <ul class="reorder_ul reorder-photos-list">                 <li id="image_li_<?php echo $count; ?>" class="ui-sortable-handle">                     <a href="javascript:void(0);" style="float:none;" class="image_link"><img src="<?php echo $image_src; ?>" alt=""></a>                 </li>             </ul>     <?php }     } } ?>

like said, i'm brand new , there may far more elegant solution, i've gotten far hunting , pecking... in advance help.


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