This is my form code:
<form enctype="multipart/form-data" name="upload-file" method="post" action="http://example.com/upload">
<div class="formi">
<input id="text-link-input" type="text" name="url" class="link_form" value="your link" onfocus="if(this.value==this.defaultValue) this.value='';" onblur="if(this.value=='') this.value=this.defaultValue;" />
<input type="submit" value="OK" class="link_but" />
</div>
<div class="upl" title="Upload"><img src="http://example.com/example.png" alt="" style="vertical-align:middle;"/>Upload
<input type="file" name="file" size="1" class="up" onchange = "document.getElementById('text-link-input').value = String(this.value).replace('C:\\fakepath\\','')"/>
</div>
</form>
Now, I want to redirect the submitter to any page of my choice after the form data has been submitted, but the action resides on another website where I cannot edit. Is it possible to redirect the user to any page after he has submitted the data to that site?
From some Googling, this is what I have found. Adding this to form code:
onSubmit=window.location='http://google.com'
This didnt work. Maybe I didnt implement it correctly? This is what I did:
<form enctype="multipart/form-data" name="upload-file" method="post" onSubmit=window.location='http://google.com' action="http://example.com/upload">
Another person says adding a hidden field should work:
<input type="hidden" name="redirect" value="http://your.host/to/file.html">
How should I implement this is my code?
Suggestions and help awaited...
Try this Javascript (jquery) code. Its an ajax request to an external URL. Use the callback function to fire any code:
<script type="text/javascript">
$(function() {
$('form').submit(function(){
$.post('http://example.com/upload', function() {
window.location = 'http://google.com';
});
return false;
});
});
</script>
Source: Stackoverflow.com