how can pass an image location to a php function using jquery's call back function?

Question

<div class="hideAfterSuccess">
                    <label for="avatar">Upload an Avatar :</label>
                    <input type="file" name="avatar" id="avatar" />
                </div>

and below is the jquery that has to pass image location to a php function in another file (parsePortal.php). am able to pass other kinds of data o the file but the image failed. i thought since it posts the location string of the image, i would be able to use $_FILES[$_POST['image']]['name'] and then upload the image. i have tried the web, surely, didn't find what i needed.

        $(function(){
           var imageLoc = $('avatar').val();
           var url = "parsePortal.php";
                $.post(url, {
                    status : "getimage",
                    image : "imageLoc"
                }, function(result){

                 });
            });

yes, i did. but it still failed. it would just pass an empty string. — Busker McGreen Brian, Jan 17 '13 at 08:17

score 0 · Answer 1 · answered Jan 17 '13 at 07:51

0

Unfortunately it's not that simple to pass a file to php via ajax. However, there is a nice library that lets you do this easily:

http://malsup.com/jquery/form/

answered Jan 17 '13 at 07:51

Vic

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score 0 · Accepted Answer · answered Jan 18 '13 at 18:01

i successfully did it using the jquery forms plugin. this means it sent the whole path from which i got the the file name and printed it out. THANKS TO YOU VIC, there is no way i c'd have done this without this kind of help.

here is the javascript.

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.7/jquery.js"></script> 
<script src="http://malsup.github.com/jquery.form.js"></script> 

<script> 
        // prepare the form when the DOM is ready 
        $(document).ready(function() { 
            // bind form using ajaxForm 
            $('#htmlForm').ajaxForm({ 
                // target identifies the element(s) to update with the server response 
                target: '#htmlExampleTarget', 

                // success identifies the function to invoke when the server response 
                // has been received; here we apply a fade-in effect to the new content 
                success: function() { 
                    $('#htmlExampleTarget').fadeIn('slow'); 
                } 
            }); 
        });
</script>

and here is the html

<form id="htmlForm" action="file.php" method="post"> 
Message: <input type="text" name="message" value="Hello HTML" /> <p></p>
Image : <input type="file" name="image" id="image" /><p></p>
<input type="submit" value="Echo as HTML" />

and here is the php

$image =  $_FILES['image']['tmp_name'];
$name = $_FILES['image']['name'];
move_uploaded_file($image, "uploads/".$name);
echo '<div style="background-color:#ffa; padding:20px">' . $_POST['message'] .  '</div>'; 
echo '<img src="uploads/'.$name.'"/>';

how can pass an image location to a php function using jquery's call back function?

2 Answers2