Question

After looking around for weeks and being totally confounded by the tutorials that were already on the books, I have decided to write my own jQuery upload script. I thought I could get it to work but I am having issues. I will enumerate them:

On the page that I am writing this for, I have placed a jQuery script that looks a little like this:

$(document).ready(function() {
  var thumb = $('#thumb');  

  $('#profilepicinput').live('change', function(){
    $("#preview").html('');
    $("#preview").html('<img src="loader.gif" alt="Uploading...."/>');
    $("#registerpt3").ajaxForm({
      target: $('#registerpt3').attr('action')
    }).submit();
  });

  $('#profilepicbutton').change(function(){
    alert("Boy I hate PHP");
    $.ajax({
      type: "POST",
      url: "register3.php",
      success: function(msg) {
        alert( "Data Saved: " + msg );
        $.ajax({
          type: "POST",
          url: "retrievePic.php",
          success: function(msg) {
            var image = new Image();
            $(image).load(function(){
              console.log("we have uploaded this image");
            }).attr('src', 'images/');
            alert("AJAX Success!");
          },
          error: function(msg) {
            alert(" didn't work");
          }
        });
      },
      error:function(msq){
        alert("didn't work!: " + msg);
      }
    });
  });
});

This calls the following script which will load a picture into a database:

<?php 

session_start();

$email = '';

if (isset($_SESSION['user_email'])) {
  $email = $_SESSION['user_email'];
} else {
  $email = '[e-mail]';
}

$link = mysql_connect('censcoredFool.com', 'greetmeet', 'Maverick$41'); 

mysql_select_db(first_1) or die("Opps, You are pretty Got-Damned Stupid!  Did you         realize that?!?!?");

$target = './Uploads/'; 
$target = $target . basename( $_FILES['uploaded']['name']); 

$ok = 1; 

$path = "uploads/";
$Email = $_SESSION['user_email'];
$valid_formats = array("jpg", "png", "gif", "bmp","jpeg");

if (isset($_POST) and $_SERVER['REQUEST_METHOD'] == "POST") {
  $name = $_FILES['photoimg']['name'];
  $size = $_FILES['photoimg']['size'];

  if (strlen($name)) {
    list($txt, $ext) = explode(".", $name);
    if (in_array($ext,$valid_formats)) {
      if ($size<(1024*1024)) { // Image size max 1 MB
        $actual_image_name = time().$session_id.".".$ext;
        $tmp = $_FILES['photoimg']['tmp_name'];
        if (move_uploaded_file($tmp, $path.$actual_image_name)) {
          mysql_query("insert into personal_photos      (Email, Pics) values('$email', '$tmp')");
          echo "<img   src='uploads/".$actual_image_name."' class='preview'>";
        } else {
          echo "failed";
        }
      } else {
        echo "Image file size max 1 MB";
      }
    } else {
      echo "Invalid file format..";
    }
  } else {
    echo "Please select image..!";
  }

  exit;
}

What I have doesn’t even seem to work. I get the success alert, but the picture doesn’t even upload. I am guessing the PHP script ran the whole way through but didn’t throw an exception or anything. How do I get this thing to work?

UPDATE: After running Firebug My PHP script isn’t getting the Picture
Not sure why that is. Need to figure out how the $_File datastructure works.

Answer 1

Most browsers won’t send files via ajax for security purposes. To have a pretend-ajax effect on uploads, you might try the excellent jQuery form plugin, which uses an hidden iframe to do the uploading.

Since this is the only answer so far – and it’s also right – allow me to give you some OT advice:

• Most of the people using PHP like PHP (more things in Heaven and Earth, Horatio). Most of them will take your “PHP Sucks” thing as a pretty pretentious thing to say, since your problem has more to do with a lack of basic knowledge of Ajax and http implementations than it has to do with the language’s faults. Thus the lack of answers and the unwelcoming comments.

• Your witty is my distracting: please make the debugging output in the code you post as descriptive as possible – actually, it would be a useful thing to do to simply write descriptive output anyway – it’s easier to find problems that way, trust me.

• If a form is made of one hundred fields ond only one is relevant, please post the relevant field only.