I want to get temp name of image upload using php,ajax

1
votes

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.Any body

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
?>

Can you please give me a suggestion on how to get $_FILES['file']['tmp_name']?

2
phpjqueryajax
i need the value $_FILES["file"]["tmp_name"] in ajax page .Now i got the file name.but not temp name in ajax page – sh kk
I dont give form,i used onclick event not submit – sh kk
I have posted a full tutorial for image upload it may help you stackoverflow.com/questions/26861807/… – Veerendra

2 Answers

0
votes

For Ajax based file upload i used the http://jquery.malsup.com/form/. Even more functionality like upload progress etc.

Take a look on DEMO Page

0
votes
<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 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
                    });
                }
</script>

Use $_FILES to get file properties. For example:

$file_temp_name = $_FILES['file']['tmp_name'];

Bcause $_POST does not hold any file information you have posted.