PHP JAVASCRIPT HTML, upload image upon BROWSE competion

Question

I have this little form that has a button loading a little icon onto the form itself. The button says 'browse', and there is another button says 'upload image'. I find it kinda unnecessary to have two buttons to upload the image.

How can I make the upload happen on the end of Browse ?

here's the code by the way. (The upload button just activate the js that calls a php file that does the trick, the js function updates it on the form)

<form class="jQ-form" action="includes/ajaxupload.php" method="post" name="standard_use" id="standard_use" enctype="multipart/form-data">
    <fieldset>  
        <button id="image_upload_button" style="float:left;"  onclick= "$('#upload_area').css('display','none');
        $('#upload_area').fadeIn('slow');ajaxUpload(this.form,'includes/ajaxupload.php?filename=filename&amp;maxSize=200000&amp
        ... more func data.'); return false;" disabled>upload icon</button>

        <input type="file" id="upload_file" name="filename" style="float:left;width:70%;" size="42"/>

        <p style="float:right;color:#a2983c;margin:10px;width:373px;">
        Pick a nice icon that is max 300x300 pixels please
        </p>

        <?php 
            echo '<div id="upload_area" style="'.((($theiconname!='') && (file_exists($thumb_path.$theiconname))) ? '' : 'display:none;').'float:left;
            width:50px;height:50px;border:3px solid #000;margin-top:12px;margin-left:3px;">';
            if ($theiconname){

                if (file_exists($thumb_path.$theiconname))
                {
                    echo'<img id="the_logo" src="'.$thumb_path.$theiconname.'"/>';
                }

            }
        ?>
        </div>
    </fieldset>
</form>

Thanks in advance!

Answer 1

Sure! Just detect when the file input field has been changed, then submit the parent form. I notice you're using jQuery so I'll provide an answer using that:

$("#upload_file").change( function(){
     $("#standard_use").submit();
});

Alternatively, if you're submitting this via AJAX and not actually intending to submit the form, then replace $("#standard_use").submit(); with whatever function would normally fire when someone pressed your "upload image" button.