At this moment I am using laravel. In this context I am having a form which is successfully submitted by using ajax to a controller. and that controller make it to the database. But the problem is as the ajax is doing its job the whole page remain unmoved / unchanged after the submission even the database is updated.
Now what I want
I want to give feedback to the user that your post is successfully submitted there. or what I want to do in further, I want to refresh the section in which the post is collected from the database as this post can be retrieved from there. But by using ajax only.
So there is no need to collect the whole page or refresh.
here is my form structure
`
{{ Form::open(array('route' => array('questions.store'), 'class' => 'form-horizontal' )) }}
blah blah blaaa .......
<script type="text/javascript">
$(".form-horizontal").submit(function(e){
$(this).unbind("submit")
$("#ask").attr("disabled", "disabled")
var that = $(this),
url = that.attr('action'),
type = that.attr('method'),
data = {};
that.find('[name]').each(function(index, value){
var that = $(this),
name = that.attr('name'),
value = that.val();
data[name] = value;
});
$.ajax({
url: url,
type: type,
data: data,
success: function(response){
console.log(response);
}
});
return false;
});
</script>
{{ Form::close() }}
` As it is very much visible that the post is updated through a route & controller I want to have another dive and a success message at this script to be displayed after the success of posting. I am looking for some professional structure using what there is minimal need to have interaction with the server side and give user a better page viewing experience.
Thanks a lot for helping me in this research.
I am not sure if I understand you well, but if you want to notify the user about the result of an ajax-called db update you need to have
check out the below code for ajax call itself (inside the form submit handler) to see what I mean
var db_ajax_handler = "URL_TO_YOUR_SITE_AND_ROUTE";
var $id = 1; //some id of post to update
var $content = "blablabla" //the cotent to update
$.ajax({
cache: false,
timeout: 10000,
type: 'POST',
tryCount : 0,
retryLimit : 3,
url: db_ajax_handler,
data: { content: $content, id: $id }, /* best to give a CSRF security token here as well */
beforeSend:function(){
},
success:function(data, textStatus, xhr){
if(data == "OK")
{
$('div.result').html('The new Question has been created');
}
else
{
$('div.result').html('Sorry, the new Question has not been created');
}
},
error : function(xhr, textStatus, errorThrown ) {
if (textStatus == 'timeout') {
this.tryCount++;
if (this.tryCount <= this.retryLimit) {
//try again
$.ajax(this);
return;
}
return;
}
if (xhr.status == 500) {
alert("Error 500: "+xhr.status+": "+xhr.statusText);
} else {
alert("Error: "+xhr.status+": "+xhr.statusText);
}
},
complete : function(xhr, textStatus) {
}
});
EDIT: as per comment, in step 2 (the method that is called with AJAX) replace
if($s)
{
return Redirect::route('questions.index') ->with('flash', 'The new Question has been created');
}
with
return ($s) ? Response::make("OK") : Response::make("FAIL");
EDIT 2: To pass validation errors to the ajax-returned-results, you cannot use
return Response::make("FAIL")
->withInput()
->withErrors($s->errors());
as in your GIST. Instead you have to modify the suggested solution to work on JSON response instead of a plain text OK/FAIL. That way you can include the errors in the response and still benefit from the AJAX call (not having to refresh the page to retrieve the $errors from session). Check this post on the Laravel Forum for a working solution - you will get the idea and be able to fix your code.
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