$ .ajax和php响应

Question

When i make requests this way i never get the response to manipulate . Well i get as response ***Array ( [name] => Mastering React Native (2).pdf [type] => application/pdf ... but i need to show to user the file info.

  <form id='formid'>
        <input id='fileid' type='file' name='file1' hidden/>
        <input type='submit' value='IMPORT'/>
    </form>
    <script>
        $(document).on("submit", "form", function (e) {
            e.preventDefault();
            var fdat = new FormData(this);
            fdat.append('operation', 'IMPORT');
            function importFile() {
                return $.ajax({
                    url: "data-source/oracle_controller_upload.php",
                    type: 'POST',
                    dataType: "JSON",
                    data: fdat,
                    processData: false,
                    contentType: false,
                    success: function(response){
                                   }
                });
            }
            debugger;
            importFile().done(function (result) {
            }).fail(function () {
            });
        });
    </script>

this is my php code:

if (@$_REQUEST['operation'] === "IMPORT" && $_FILES) {
    echo print_r($_FILES['file1']);
    return;
}

Cant reach either the success callback or done(). Any help?

Answer 1

Solved with

if (@$_REQUEST['operation'] === "IMPORT" && $_FILES) {
  echo json_encode($_FILES['file1']);
  return;
}

I have dataType: "JSON", so of course the response must be JSON.