Can't upload file in PHP using $ajax()

Question

I was trying to upload file on server using php with jQuery's ajax() function. Below is the code which I was trying. Things worked fine when I wrote PHP code on same page without jQuery, so there is no doubt that file upload is working.

HTML


<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js"></script>
<script>
    $(function() {
        $("form").submit(function(e){
            e.preventDefault();

            var fd = new FormData();    
            fd.append("files", $("#fileinput").prop("files"));
            $.ajax({
                url: "imgupload_.php", 
                type:"POST",
                processData: false,
                contentType: false,
                data: fd,
                success: function(result){
                    alert(result);
                }
            });
        });
    });
</script>
<form method="POST" action="#" enctype="multipart/form-data">
    <input type='file' name='files' id="fileinput"/>
    <input type='submit' value='Submit' name='submit'/>
</form>

imgupload_.php

if(isset($_POST["submit"])) {
    $target_dir = "images/";
    $target_file = $target_dir . basename($_FILES["files"]["name"]);

    if (move_uploaded_file($_FILES["files"]["tmp_name"], $target_file)) {
        echo "The file ". basename( $_FILES["files"]["name"]). " has been uploaded.";
    } else {
        echo "Sorry, there was an error uploading your file.";
    }
}

If you want any other info please comment below.

miken32 · Accepted Answer · 2019-11-18 17:51:57Z

2

You're checking if(isset($_POST["submit"])) but this is not set.

Instead of just this:

var fd = new FormData();

Use this to pull in any non-file input values (such as your submit button.)

var fd = new FormData(this);

answered Nov 18, 2019 at 17:51

miken32

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1 Comment

MrEyebr0w5 Over a year ago

Yupp. If I replace if(isset($_POST["submit"])) with if($_SERVER["REQUEST_METHOD"] == "POST") and add this in FormData, it works. Thank you sir.

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