Ealier I tried using same functionality with colobox but could not success, now I'm trying to use nyroModal and trying to send ajax data using post method with iframe. But its not going through. Here is the code which I have tried:
var src = "this is test data';
$.nmManual('try_python.php',{
callbacks: {
initFilters: function(nm) {
nm.filters.push('link');
nm.filters.push('iframe');
},
ajax:{data:'code='+src, type:"post"}
}
});
I'm able to open iframe and content from try_python.php is properly displayed in the frame but passed data is not accessible to try_python.php.
I will really appreciate you for any input in this regard.
Kind Regards
Mohtashim
I could not success in posting my content using POST method but I found a workaround where did the following:
(a) Saved my content in a hidden <div id="source" style="display:none;">...</div>
(b) Access this content from the iframe window upon successful load using the following:
var code = window.parent.$("#source");
and this is what I wanted to have....May be it will help many others. If you find a solution to pass direct content using post method kindly share.
Related
Ok, this idea might seem quite a bit crazy and it kindo' is (at least for me at my level).
I have a fairly standarad rails app (some content pages, a blog, a news block, some authentication). And I want to make it into a single page app.
What I want to accomplish is:
All the pages are fetched through AJAX like when using turbolinks, except that the AJAX returns only the view part (the yield part in the layout) withought the layout itself, which stays the same (less data in the responces, quicker render and load time).
The pages are mostly just static html with AngularJS markup so not much to process.
All the actual data is loaded separately through JSON and populated in the view.
Also the url and the page title get changed accordingly.
I've been thinking about this concept for quite a while and I just can't seem to come up with a solution. At this point I've got to some ideas on how this actualy might be done along with some problems I can't pass. Any ideas or solutions are greatly appreciated. Or might be I've just gone crazy and 3 small requests to load a page are worse then I big that needs all the rendering done on server side.
So, here's my idea and known problems.
When user first visits the app, the view template with angular markup is rendered regularly and the second request comes from the Angular Resource.
Then on ngClick on any link that adress is sent to ngInclude of the content wrapper.
How do I bind that onClick on any link and how can I exclude certain links from that bind (e.g. links to external authentication services)?
How do I tell the server not to render the layout if the request is comming from Angular? I though about adding a parameter to the request, but there might be a better idea.
When ngInclude gets the requested template, it fires the ngInit functions of the controllers (usually a single one) in that template and gets the data from the server as JSON (along with the proper page title).
Angular populates the template with the received data, sets the browser url to the url of the link and sets the page title to what it just got.
How do I change the page title and the page url? The title can be changed using jQuery, but is there a way through Angular itself?
Again, I keep thinking about some kind of animation to make this change more fancy.
Profit!
So. What do you guys think?
OK, in case enyone ever finds this idea worth thinking about.
The key can be solved as follows.
Server-side decision of whether to render the view or not.
Use a param in the ngInclude and set the layout: false in the controller if that param is present.
Have not found an easier way.
Client-side binding all links except those that have a particular class no-ajax
Here's a directive that does it.
App.directive('allClicks', function($parse) {
return {
restrict: 'A',
transclude: true,
replace: true,
link: function(scope, element, attrs) {
var $a = element.find('a').not($('a.no-ajax')),
fn = $parse(attrs['allLinks']);
$a.on('click', function(event) {
event.preventDefault();
scope.$apply(function() {
var $this = angular.element(event.target);
fn(scope, {
$event: event,
$href: $this.attr('href'),
$link: $this
});
});
});
}
};
})
And then use it on some wrapper div or body tag like <body ng-controller="WrapperCtrl" all-links="ajaxLink($href)"> and then in your content div do <div id="content" ng-include="current_page_template">
In your angular controller set the current_page template to the document.URL and implement that ajaxLink function.
$scope.ajaxLink = function(path) {
$scope.current_page_template = path+"?nolayout=true";
}
And then when you get your JSON with your data from the server don't forget to use history.pushState to set the url line and document.title = to setr the title.
I put a Ajax link using the following code:
echo chtml::ajaxLink('GO', 'http://localhost/index.php?r=user/delete', array('method'=>'POST'));
But, regardless of giving the second parameter as URL i,e 'http://localhost/index.php?r=user/delete'. It generates link with the current URL in the browser not the URL I just specified.
What is the issue? How could I create AJAX link? Google several hours but can't solve the issue.
Any kind of help is highly appreciated.
First of all, you should always try and create normalized urls.
But i think your doubt lies in the # that is generated/appended. If you go and check the source of yii ajaxLink you'll see this:
public static function ajaxLink($text,$url,$ajaxOptions=array(),$htmlOptions=array())
{
if(!isset($htmlOptions['href']))
$htmlOptions['href']='#';
$ajaxOptions['url']=$url;
$htmlOptions['ajax']=$ajaxOptions;
self::clientChange('click',$htmlOptions);
return self::tag('a',$htmlOptions,$text);
}
so if you don't set the href property of the a tag in the htmloptions array, the # will be appended.
You should also understand that yii uses jquery, so if you check out the source of the page, you'll see at the bottom, how jquery is used to carry out an ajax request, your actual url that is called will also be seen in that script. So the third option/parameter in ajaxLink is for options for jquery's ajax function. You can create better ajax links using this option.
Regardless of where(which controller) your url points to in your project, the action associated with that url will be called.
So anyway, you can modify your code like this if you want the url to be shown and not a # :
echo CHtml::ajaxLink('GO', 'http://localhost/index.php?r=user/delete',
array('type'=>POST), //there are various other options for jquery ajax
array('href'=>'http://localhost/index.php?r=user/delete'));
To make better ajax links i would suggest going through jquery's ajax documentation. There is an option for a success function, that you can use to let the user know that the operation was completed.
Hope this helps, don't hesitate to leave comments if i haven't answered your question completely.
Have you tried:
echo CHtml::ajaxLink('GO', array('/user/delete'), array('method'=>'POST'));
as the ajaxLink documentation suggests...? Look also at the normalizeUrl method.
Using these methods, which in turn are using createUrl, is usually better since it will take care to create a valid url for your site.
I had the same issue(or maybe similar).
I've used renderPartial to load view and later in that view i was using ajaxLink and it was not working.
What i have found, that when using renderPartial, there was no jquery script for ajax action.
What you have to do is to add 4th argument(true) in renderPartial function to generate jquery script.
See the documentation: http://www.yiiframework.com/doc/api/1.1/CController/#renderPartial-detail
Hope it helps and saves time to figure it out.
From my view I am sending via $.ajax a JSON object to my controller to save it in the database.
If all succeeded i want to redirect to another action which will show a diferent view.
If i use this code:
return RedirectToAction("CreatePage", "Survey", new {id = question.PageId});
The execution goes to the Survey controller which returns a view but it is not shown.
I have read some post which said that it is not posible to redirect via ajax.
The solution I use so far is to redirect via javascript like this:
success: function (ret) {
window.location.href = "/Survey/CreatePage/" + $("#PageId").val();
}
Although this always works, sometimes i need to refresh the CreatePage view to show the last changes made.
Any idea of how to solve this problem better?
Thanks in advance
As mccow002 suggested, I wasn't really needing to make the call via AJAX for that part. After studying the solutions suggested, i realized that i could simple submit it in a form. My confusion came because I have a save and continue editing and a save. For the save and continue I use the AJAX call, but for the save option with the form being submitted is ok.
Thanks very much for your help.
Instead of redirecting to a new page, you can send a rendered html from .net code back to client and load that html in page, like this $("#main").load(renderedHtml).
But for refreshing the page you can write a simple script that run at specified intervals and refresh the page contens.
You could use [OutputCache] on the CreatePage action so that it doesn't cache the page or only caches for so long.
output caching
hi
sorry for the bad title but I'm not 100% sure what I need for this problem
I created a welcome page and then when you click on links you get more information, for example:
Click Me
And then the php would get the information based on the id.
so the information received is reloaded on the page after the pages refreshes
what I would like to be able to do is when user clicks on the link, use jquery to not allow the link to run but still run the url in the background (without refreshing the page)
I have no idea where to start from so I really hope you could help
thanks
In a nutshell, it's called Ajax: sending an HTTP request to your server through javaScript, and receiving a response which can contain results, data, or other information.
You mention jQuery, here are the docs about that:
http://api.jquery.com/jQuery.get/
http://api.jquery.com/jQuery.post/
are convenience methods, which encapsulate $.ajax with preset options.
http://api.jquery.com/category/ajax/ is an overview of the whole system in jQuery.
The basics go like
//include jquery, etc.
$(document).ready(function(){
$('#some_element').click(function(){
$.get('some_url_on_your_server.php',{'data':'whatever params'},function(data){
do_something();//
},'json');
});
This will bind an element to make an Ajax call on click, and then you use the function ('success' function, in $.ajax) to handle the json data.
Have your server send back the data in JSON by using json_encode in php. Be sure to send the right header back, like
<?php
header('Content-Type: application/json');
echo json_encode($some_array);
exit;
There's a lot of resources on the web and SO for learning about Ajax, it's a big topic. Best of luck.
Make a JavaScript function, like sendData(linkId) and then each tag would have an onclick event called sendData(this). SendData(linkId) can then do an HTTPRequest (also known as an asynchronous or AJAX request) to a php file, let's call it handler.php, which receives GET or POST methods. I prefer using the prototype framework to do this kind of thing (you can get it at prototypejs.org).
Okay, now that I have said all that, let's look into the nitty-gritty of how to do this (way simplified for illustrative purposes).
Download the prototype script, save it on your server (like prototype/prototype.js, for example) and then put somewhere in your html <script type='text/javascript' language='Javascript' src='prototype/prototype.js'></script>
Your tags would look like this:<a id='exampleLink' onclick = 'sendData(this)'>Click me!</a>
You need JavaScript to do this: function sendData(tagId){
var url = 'handler.php?' + 'id=' + tagId;
var request = new AJAX.Request(url, {method = 'get'});
}
Finally, you need a php file (let's call it handler.php) that has the following: <?php
$tag_to_get = $_GET['tagId'];
do_a_php_function($tag_to_get);
?>
That's it in a nutshell, but it's worth mentioning that you should give your user some sort of feedback that clicking link did something. Otherwise he will click the link furiously waiting for something to happen, when it is actually doing just what its supposed to but in secret. You do that by making your php script echo something at the end, like 'Success!', and then add an onSuccess parameter to your JavaScript's new Ajax.Request. I'll let you read how to do that on your own because the prototype website explains how to receive a response from the handler and put the feedback somewhere in your HTML without making the user refresh.
you can achieve that behavior with a jquery function called $.get ... you can get more information on how to use here http://api.jquery.com/jQuery.get/
If you really want to (and I don't think you really do), you can use XMLHTTPRequest (wrapped in jQuery.get) to facilitate loading content into the page without page refreshing. You want an id or class on that tag, i.e. Click Me, and then:
<script>
$(".fetch").bind("click", function(evt)
{
$.get(this.attr("href"), function(data)
{
$("#whereIWantMyContent").html(data);
});
evt.preventDefault();
});
</script>
I would recommend you use AJAX to start with. A good place to being is http://www.w3schools.com/Ajax/Default.Asp
The link comes with a handy AJAX ASP/PHP Example too =))
Good Luck.
After processing a jQuery Ajax Post from an HTML form successfully within a Go program, how do I load a new form? I first tried sending the form as the response and the Javascript displayed it, but it did not clear the old (existing) form. I then tried within the HTML Javascript to set the URL using "window.location = 'localhost:8088/MaintForm/'". That resulted in a warning from the browser but did not load the form and did not change the URL. I would like to ideally know both methods - via the Go program acting as a server, and via Javascript. If I manually change the URL, the form loads OK. What I am trying to do is receive a response in Javascript (jQuery Ajax), and then request the new form if the response is positive. I would prefer to do this without changing the URL. As I said above, this partially worked.
You would have to put your original form inside a tag, for example a div, and use your JQuery code to replace the contents of that tag with the new form. This way you are not changing the URL.
This is more of a javascript/JQuery question than a go-specific one.
In javascript:
location.href = '/MaintForm/';
In golang, you can use the http.Redirect function, like this:
http.Redirect(w, r, "/MaintForm/", http.StatusFound)
Please note: this appears to be solved by : I just need to do an "document.write(data);" in Javascript. "Data" contains the new HTML.