I dont get temp name of image upload using php,ajax

Question

I want to get the temp name of image upload using php and ajax.I got the file name of upolad image. But dont get the temp name of image upload.My code is given below

main.php

<form action=" " method="POST" enctype="multipart/form-data">
    <div class="form-group">
        <label for="exampleInputFile">File Upload</label>
        <input type="file" name="file" id="file" size="150">
        <p class="help-block">Only Excel/CSV File Import.</p>
    </div>
    <button type="button" class="btn btn-default"
            name="Import" value="Import" onclick="file_up()">
        Upload</button>
</form>
<div id="upload_show"></div>
<script src="//code.jquery.com/jquery-1.11.0.min.js"></script>
<script type="text/javascript">
    function file_up() {
        var file = $('#file').val();
        $.ajax({
            type: "POST",
            url: 'first_ajax.php',
            data: 'file=' + file,
            success: function (msg){
                $("#upload_show").html(msg);
            }
        });
    }
</script>

first_ajax.php

<?php
   echo $file1 = $_POST['file'];   // for get file name
   echo $file1 = $_POST['file']['tmp_name']; //for get temp name
?>

maybe this can help -> stackoverflow.com/questions/17651073/… — Marco Mura
– Marco Mura, Commented Dec 10, 2014 at 8:30
the fle will not be included since you only passed the string value of the input file. use ajaxForm instead. you can search it on google — roullie
– roullie, Commented Dec 10, 2014 at 8:33

Deep Gandhi · Accepted Answer · 2014-12-10 08:40:29Z

1

Instead of,
var file = $('#file').val();

try,
var file = $("#file").prop("files")[0];

answered Dec 10, 2014 at 8:40

Deep Gandhi

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Mira Padalia · Accepted Answer · 2014-12-10 10:36:46Z

1

Use FormData to get $_FILES and $_POST array in AJAX file...

function file_up() {
 var formData = new FormData($('#form')[0]);
                    $.ajax({
                        url: 'first_ajax.php',
                        type: 'POST',
                        data: formData,
                        async: false,
                        success: function(data) {
                            $("#upload_show").html(msg);
                        },
                        cache: false,
                        contentType: false,
                        processData: false
                    });
                }

answered Dec 10, 2014 at 10:36

Mira Padalia

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