I'm trying to upload multiple images using AJAX and PHP but there is an error

Question

In my form.php

<form method="POST" enctype="multipart/form-data">
  <input type="file" name="file_upload[]" id="file_upload" style="display:none;" required multiple accept="images/*" onchange="preview_image()">
</form>

And there is a submit button below.

In my jquery part

$('#submit').on('click', function(e) {
  var files = $('#file_upload')[0].files;
  var form_data = new FormData();
  form_data.append("file_upload",files);

  $.ajax({
    url:"execute/Script.php?request=submit",
    type: "POST",
    data: form_data,
    cache: false,
    contentType: false,
    processData: false,
    success: function(dataResult){
      var dataResult = JSON.parse(dataResult);
    
      if(dataResult.statusCode == 1){
        alert("Success!");
      }
    }
  });
});

And in my script.php

if($_GET['request']=="submit"){
  $gallery_G = "";

  foreach($_FILES['file_upload']['error'] as $key => $error){

    if($error==UPLOAD_ERR_OK){
      $tmp_name = $_FILES['file_upload']['tmp_name'][$key];
      $PhotoName = $_FILES['file_upload']['name'][$key];
                    
      move_uploaded_file($tmp_name,"../img/properties/$name");
      $gallery_G = $gallery_G.';'.$PhotoName;
    }
  }

  $sql = mysqli_query($conn,"INSERT INTO property(photos) VALUES('$gallery_G')");

  if($sql) {
    echo json_encode(array("statusCode"=>1));
  }
  else{
    echo json_encode(array("statusCode"=>2));
  }
}

But it returns me this:

Notice: Undefined index: file_upload in G:\Xampp\htdocs\execute\Script.php on line 31

Warning: Invalid argument supplied for foreach() in G:\Xampp\htdocs\execute\Script.php on line 31
{"statusCode":2}

Is there any solution? I need to send those photos using 'append' method. And need to get those photos in 'Script.php' so that I can process or upload them from there.

@mplungjan <button class="btn-4 btn-round-3" id="submit">Submit</button> — AmarGallery - an OpenGallery
– AmarGallery - an OpenGallery, Commented Feb 20, 2022 at 14:03
Warning: You are wide open to SQL Injections and should use parameterized prepared statements instead of manually building your queries. They are provided by PDO or by MySQLi. Never trust any kind of input! Even when your queries are executed only by trusted users, you are still in risk of corrupting your data. Escaping is not enough! — Dharman
– Dharman ♦, Commented Feb 20, 2022 at 14:37

Travis Neel · Accepted Answer · 2022-02-20 17:26:44Z

1

Try passing the form element to FormData. No need to call the append method afterwards.

  var files = $('form')[0];
  var form_data = new FormData(files);

By using the append() method, you are putting a [object FileList] in the $_POST array under the file_upload key

If you just want the files from the form, you would need a loop:

  var files = $('#file_upload')[0].files;
  var form_data = new FormData();
    
  for (var i = 0; i < files.length; i++) {
    form_data.append("file_upload[]",files[i]);
  }

edited Feb 20, 2022 at 17:26

answered Feb 20, 2022 at 16:16

Travis Neel

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2 Comments

AmarGallery - an OpenGallery Over a year ago

But I've other fields and multiple checkbox too.

Travis Neel Over a year ago

updated answer to pull only the file input

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