Send Multiple data with ajax - ajax

I want to use ajax for add data in database and i found following code in net and it is working fine.
<script language='javascript'>
reqObj=null;
function saveCust(){
document.getElementById('res').innerHTML='processing';
if(window.XMLHttpRequest){
reqObj=new XMLHttpRequest();
}else {
reqObj=new ActiveXObject('Microsoft.XMLHTTP');
}
reqObj.onreadystatechange=processSave;
reqObj.open('POST','./custSave?reqObj.open('POST','./cName?id='+document.getElementById('CustCode').value,true);,true);
reqObj.send(null);
}
function processSave(){
if(reqObj.readyState==4){
document.getElementById('res').innerHTML=reqObj.responseText;
}
}
</script>
Above code sends only one String but, i have 5 Strings in my form.
Please anybody alter the code for sending multiple data.

The problem is that you're sending a single parameter in the reqObj.open function:
reqObj.open('POST','./custSave?reqObj.open('POST','./cName?id='+document.getElementById('CustCode').value,true);,true);
Note that the only parameter you send is id.
You can add more parameters in the flavor of QueryString:
id=something&otherParameter=else //and more parameters
IMO the easiest way to handle an ajax request would be using jQuery, as shown and heavily explained by BalusC in How to use Servlets and Ajax?.
Based on the samples there and jQuery Ajax POST example with PHP, you can come with the following code:
Assuming the 5 Strings are in the form
function saveCust(){
$('#res').html('processing');
var $form = $(this);
var serializedData = $form.serialize();
$.post('./custSave', serializedData, function(responseText) {
$('#res').html(responseText);
});
}
Assuming there's data outside the form
function saveCust(){
$('#res').html('processing');
var $form = $(this);
var serializedData = $form.serialize() + "&id=" + $('#CustCode').val();
$.post('./custSave', serializedData, function(responseText) {
$('#res').html(responseText);
});
}
And you can even enhance this using more jQuery functions, but that's outside the scope of this answer.

Related

Laravel render for differend controller method

I'm struggling with the render() method in Laravel 5.
When $whatever->render() is runned, it takes the controller method name as the route by default.
Example:
When i run this command in DelasController#updateFilter, the pagination route is set to whatever.com/marketplace/updateFiler?page=2, which does not make a sense to me.
Problem:
I want to keep the route as simple as whatever.com/marketplace?page=2.
Question:
Can anybody gives me a hint on how to solve this?
Thank you for your time and a discussion.
Looking forward for a reply.
I have an application in which various paginated lists are displayed in "windows" on the page and are updated via AJAX calls to the server. Here's how I did it:
Set up a route to render the whole page, something like this:
Route::get('/marketplace', function ($arguments) {
....
});
Set up a route which will return the current page of the list. For example, it might be something like this:
Route::get('/marketplace/updateFiler', function ($arguments) {
....
});
In your Javascript code for the page, you need to change the pagination links so that, instead of loading the new page with the URL for the link, it makes the AJAX request to the second route. The Javascript could look something like this:
$('ul.pagination a').on('click', function (event) {
// stop the default action
event.stopPropagation();
event.preventDefault();
// get the URL from the link
var url = $(event.currentTarget).attr('href');
// get the page number from the URL
var page = getURLParameterByName(url, 'page');
$.get('marketplace/updateFiler', { page: page }, function (data){
// do something with the response from the server
});
});
The getURLParameterByName function is simply a helper that extracts the named parameter from a URL:
var getURLParameterByName = function (url, name, defaultValue) {
// is defaultValue undefined? if so, set it to false
//
if (typeof defaultValue === "undefined") {
defaultValue = false;
}
name = name.replace(/[\[]/, "\\[").replace(/[\]]/, "\\]");
var regex = new RegExp("[\\?&]" + name + "=([^&#]*)"),
results = regex.exec(url);
return results === null ?
defaultValue :
decodeURIComponent(results[1].replace(/\+/g, " "));
};
I adapted this code from an answer I found here on Stack Overflow: https://stackoverflow.com/a/901144/2008384.

MVC3 redirect to action after ajax call

In an ASP.NET MVC3 Application I have a button in the view.
When the button is clicked a function is called and it jquery ajax call is made to save items to the database
function SaveMenuItems() {
var encodeditems = $.toJSON(ids);;
$.ajax({
type: 'POST',
url: '#Url.Action("SaveItems", "Store")',
data: 'items=' + encodeditems + '&storeKey=#Model.StoreID',
complete: function () {
}
}
});
}
What i want is after the items are saved to the database I want to redirect to another view. (Redirect to action)
How can I do that?
I tried to use return RedirectToAction("Stores","Store") in the controller at the end of the SaveItems function. But it is not working
I also tried to add window.location.replace("/Store/Stores"); in the complete function of the ajax call but didn't work either
Any help is greatly appreciated
Thanks a lot
You can use javascript to redirect to the new page. Set the value of window.location.href to the new url in your ajax call's success/complete event.
var saveUrl = '#Url.Action("SaveItems","Store")';
var newUrl= '#Url.Action("Stores","Store")';
$.ajax({
type: 'POST',
url: saveUrl,
// Some params omitted
success: function(res) {
window.location.href = newUrl;
},
error: function() {
alert('The worst error happened!');
}
});
Or in the done event
$.ajax({
url: someVariableWhichStoresTheValidUrl
}).done(function (r) {
window.location.href = '#Url.Action("Stores","Store")';
});
The above code is using the Url.Action helper method to build the correct relative url to the action method. If your javascript code is inside an external javascript file, you should build the url to the app root and pass that to your script/code inside external js files and use that to build the url to the action methods as explained in this post.
Passing parameters ?
If you want to pass some querystring parameters to the new url, you can use this overload of the Url.Action method which accepts routevalues as well to build the url with the querystring.
var newUrl = '#Url.Action("Stores","Store", new { productId=2, categoryId=5 })';
where 2 and 5 can be replaced with some other real values.
Since this is an html helper method, It will work in your razor view only,not in external js files. If your code is inside external js file, you need to manually build the url querystring parameters.
Generating the new url at server side
It is always a good idea to make use of the mvc helper methods to generate the correct urls to the action method. From your action method, you can return a json strucutre which has a property for the new url to be redirected.
You can use the UrlHelper class inside a controller to do this.
[HttpPost]
public ActionResult Step8(CreateUser model)
{
//to do : Save
var urlBuilder = new UrlHelper(Request.RequestContext);
var url = urlBuilder.Action("Stores", "Store");
return Json(new { status = "success", redirectUrl = url });
}
Now in your ajax call's success/done callback, simply check the return value and redirect as needed.
.done(function(result){
if(result.status==="success")
{
window.location.href=result.redirectUrl;
}
else
{
// show the error message to user
}
});
In action you can write this:
if(Request.IsAjaxRequest()) {
return JavaScript("document.location.replace('"+Url.Action("Action", new { ... })+"');"); // (url should be encoded...)
} else {
return RedirectToAction("Action", new { ... });
}
Try
window.location = "/Store/Stores";
Instead.

ASP.Net MVC 3.0 Ajax Action Link OnBegin Function Collect return Value and Append to URL

new AjaxOptions
{
OnBegin = "MyFunction",
Url="/Controller/JSONAction/"+OnbeginRetunrValue,
HttpMethod="GET"
}
In my Ajax properties, I have On Begin function which returns a value.
I wan tot append that return value to the URl.
How can I do that?
Well I realized i can't get this done the way I have asked.
But the Other way to Achieve it is have regular link and call jQuery Ajax Link
function CallActionResult(){
//do your logic
var newparam = //get your new param;
$.ajax({
url:"",
DataType:"JSON",
.....
continue rest of the properties.
});
}
this is how, I go this worked for me.

Cross-Domain Requests with jQuery

For a project I need to get the source code of web page of different other domains.
I have tried following code:
$('#container').load('http://google.com');
$.ajax({
url: 'http://news.bbc.co.uk',
type: 'GET',
success: function(res) {
var headline = $(res.responseText).find('a.tsh').text();
alert(headline);
}
});
Still I am not getting any results but just a blank alert box.
By default all browsers restrict cross-domain requests, you can get around this by using YQL as a proxy. See a guide here: http://ajaxian.com/archives/using-yql-as-a-proxy-for-cross-domain-ajax
For security reasons scripts aren't able to access content from other domains. Mozilla has a long article about HTTP access control, but the bottom line is that without the website themselves adding support for cross-domain requests, you're screwed.
This code is Working Perfectly with the help of JQuery and YQL
$(document).ready(function(){
var container = $('#target');
$('.ajaxtrigger').click(function(){
doAjax($(this).attr('href'));
return false;
});
function doAjax(url){
if(url.match('^http')){
$.getJSON("http://query.yahooapis.com/v1/public/yql?"+
"q=select%20*%20from%20html%20where%20url%3D%22"+
encodeURIComponent("http://www.yahoo.com")+
"%22&format=xml'&callback=?",
function(data){
if(data.results[0]){
var data = filterData(data.results[0]);
container.html(data);
} else {
var errormsg = '<p>Error: could not load the page.</p>';
container.html(errormsg);
}
}
);
} else {
$('#target').load(url);
}
}
function filterData(data){
data = data.replace(/<?\/body[^>]*>/g,'');
data = data.replace(/[\r|\n]+/g,'');
data = data.replace(/<--[\S\s]*?-->/g,'');
data = data.replace(/<noscript[^>]*>[\S\s]*?<\/noscript>/g,'');
data = data.replace(/<script[^>]*>[\S\s]*?<\/script>/g,'');
data = data.replace(/<script.*\/>/,'');
return data;
}
});
The solution for your case is JSON with padding or JSONP.
You will need an HTML element that specified for its src attribute a URL that returns JSON like this:
<script type="text/javascript" src="http://differentDomain.com/RetrieveUser?UserId=1234">
You can search online for a more in-depth explanation, but JSONP is definitely your solution for this.
Do the following steps.
1: Add datatype:jsonp to the script.
2: Add a "callback" parameter to the url
3: Create a javascript function with name same as "callback" param value.
4: The output can be received inside javascript function.
Found one more solution for this :
function getData(url){
if(url.match('^http')){
$.get(url,
function(data){
process(data);
}//end function(data)
);//end get
}
}
This is really a pretty easier way to handle cross-domain requests. As some of the sites like www.imdb.com rejects YQL requests.

get $.post to work with the validate plugin on multiple forms without seperate functions

On a fansite im doing http://yamikowebs.com/ee/
I have a few forms (2 atm). I used $.post to find out what form is being submited. submit the form and display that pages results where the form was originally with .html().
My next step was to use the validator which is working fine but im not sure how to put the 2 together.
submitHandler: function(form){} seems to be the setting for how its submitted. However, I can't get this to work with my $.post function or find out what form is being processed.
If I leave the defaults for validation plug-in if there no errors it will send you to the page. the ajax plug-in that it works with doesn't do what I want. Below is my $.post function
form validation:
//ajax post
$("form").submit(function(event)
{
event.preventDefault();//stop from submiting
//set needed variables
var $form = $(this)
var $div = $form.parent("div")
$url = $form.attr("action");
//submit via post and put results in div
$.post( $url, $form.serialize() , function(data)
{ $div.html(data) })
})
http://docs.jquery.com/Plugins/validation#source is the validation plugin
You're correct in thinking that submitHandler is the right callback to use. However, I ran into some interesting issues while using it with multiple forms (like you're trying to do). For example, in this code:
$("#form1, #form2").validate({
submitHandler: function(form) {
alert(form.action);
alert(form.id);
}
});
The submitHandler callback does not get supplied the correct parameter (it always gets #form1). I believe this is actually a bug in jQuery-validate (so I've filed it here).
Anyway, a decent workaround would be to wrap the validate call in .each():
$("form").each(function() {
$(this).validate({
submitHandler: function(form) {
/* 'form' has the correct value */
var values = $(form).serialize(),
$div = $(form).parent("div");
alert(form.action);
alert(form.id);
/* Perform AJAX call here */
}
});
});
Example: http://jsfiddle.net/andrewwhitaker/MmCXN/

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