i have a html page, which contains a form and i want when the form is successfully submited, show the below div:
<div class="response" style="display: none;">
<p>you can download ithere</p>
</div>
i also have a jquery function:
<script type="text/javascript">
$(function() {
$('#sendButton').click(function(e) {
e.preventDefault();
var temp = $("#backupSubmit").serialize();
validateForm();
$.ajax({
type: "POST",
data: temp,
url: 'backup/',
success: function(data) {
$(".response").show();
}
});
});
});
</script>
and in my views.py (code behind) i create a link and pass it to html page. i have:
def backup(request):
if request.is_ajax():
if request.method=='POST':
//create a link that user can download a file from it. (link)
variables = RequestContext(request,{'link':link})
return render_to_response('backup.html',variables)
else:
return render_to_response('backup.html')
else:
return render_to_response("show.html", {
'str': "bad Request! :(",
}, context_instance=RequestContext(request))
backup = login_required(backup)
my problem: it seems that my view doesn't execute. it doesn't show me the link that i send to this page. it seems that only jQuery function is executed. i'm confused. how can i make both of them to execute(i mean jQuery function and then the url i set in this function which make my view to be executed.)
i don't know how to use serialize function. whenever i searched, they wrote that:
The .serialize() method creates a text string in standard URL-encoded notation and produces query string like "a=1&b=2&c=3&d=4&e=5.
i don't know when i have to use it, while i can access to my form field in request.Post["field name"]. and i don't know what should be the data which is in success: function(data) in my situation.
thank very much for your help.
You have to get and display the data from your ajax post function, where data is the response you render through your DJango server, for example:
t = Template("{{ link }}")
c = Context({"link": link})
t.render(c):
Your JS / jQuery should become something like this:
<script type="text/javascript">
$(function() {
$('#sendButton').click(function(e) {
e.preventDefault();
var temp = $("#backupSubmit").serialize();
validateForm();
$.ajax({
type: "POST",
data: temp,
url: 'backup/',
success: function(data) {
// 'data' is the response from your server
// (=the link you want to generate from the server)
// Append the resulting link 'data' to your DIV '.response'
$(".response").html('<p>you can download ithere</p>');
$(".response").show();
}
});
});
});
</script>
Hope this helps.
Related
I have a sort of twitter like button function in my app such that, when the button is clicked, it triggers an AJAX call and performs the action specified in the views. However, when i click the button, it does not perform action in views. The code reaches the 'like view' but does not execute anything after 'if request.POST:'. Please help.
Menu.html
<form action="{% url 'like'%}" id="plt_{{menu.id}}" data-id="{{menu.id}}" method="post">
{%csrf_token%}
<input name="menu_id" type="hidden" value="{{ menu.id }}">
<div class="like-button" id="btn_{{menu.id}}"></div>
</form>
<script>
$('.like-button').on('click', function () {
var id = $(this).attr('id');
id = id.replace('btn_','');
$(this).toggleClass('animate').promise().done(function () {
var link = $("#plt_"+id).attr('action')
$.ajax({
type: 'POST',
url: link,
headers: {'X-CSRFToken': '{{ csrf_token }}'},
})
});
});
</script>
Views.py
def like(request):
print('reached') //this prints
if request.POST:
menu = Menu.objects.get(pk=request.POST.get('menu_id'))
//execute code to like
return HTTPResponse('')
Maybe you want to check
if request.is_ajax() and request.method== "POST":
request.POST is a dict .Empty here because body is empty in your request.
Empty dicts are treated like False by python like
if {}:
print("Hello World")
Above won't print anything
But below works
if {"hi" : "there"}:
print("Hello World")
And docs suggests this check is wrong if request.POST:
It’s possible that a request can come in via POST with an empty POST
dictionary – if, say, a form is requested via the POST HTTP method but
does not include form data. Therefore, you shouldn’t use if
request.POST to check for use of the POST method; instead, use if
request.method == "POST" (see HttpRequest.method).
It is fairly simple, use serialize() of jquery. Serialize function will take all the values from the form, even csrftokenmiddleware which is hidden input type. So, doing so you will be able to handle post request successfully. Use sthg like this:
<script>
$('.like-button').on('click', function () {
var id = $(this).attr('id');
id = id.replace('btn_','');
$(this).toggleClass('animate').promise().done(function () {
var link = $("#plt_"+id).attr('action');
var data = $("#plt_"+id).serialize(); // It will serialize all form data
$.ajax({
type: 'POST',
url: link,
data: data
});
});
});
</script>
In views.py do as you do for other request. serialize()
I have a dropdown list in a blade view. I want to send the value of the selected item to the controller immediately onchange. I have 2 routes in web.php:
Route::get('/plots', 'PlotController#index');
Route::get('/plots/{testId}', 'PlotController#getData');
The first one populates the dropdown list. The second one is supposed send the value of the dropdown list to the controller, which pulls stuff from mysql and sends the data back to the view, which draws a chart. I can get the dropdown to populate ok, but I can't figure out how to send the selected value to the controller. I'm trying to use ajax to do it like this:
$(document).ready(function() {
$('#sel_test').change(function() {
var testId = $(this).val();
console.log("testId=" + testId);
$.ajax({
url: 'plots/' + testId,
type: 'get',
dataType: 'json',
success: function(response) {
console.log("success");
}
});
});
});
The testId output to the console is correct but it never makes it to the controller. The error I see in the console is:
GET http://homestead.test/plots/1 500 (Internal Server Error)
I'm pretty new to laravel and find it extremely confusing. Can anyone explain the correct way to do this?
EDIT:
After testing and confirming Rian's answer as correct, I then tried to implement the real code, which of course is much more complicated. Instead of the controller returning the input test_id:
return $request->test_id;
It actually returns a more complex structure:
return view('plot')
->with('measurements',json_encode($result))
->with('events',json_encode($timeline))
->with('limits',json_encode($limits));
When I uncomment the original controller code, including the return section above, it seems to affect the ability of the controller to return anything at all. Here is the first few lines of the PlotController getData method:
public function getData(Request $request) {
Log::debug("made it to PlotController.php#getData");
Log::debug("test_id="+$request->testId);
And here is the log output:
[2020-02-23 16:43:52] laravel.DEBUG: made it to
PlotController.php#getData
The second line does not output anything. Here is what I see in the javascript console after I select an item from the dropdown list:
testId=49 jquery.min.js:2 GET
http://homestead.test/get-data-by-id?test_id=49 500 (Internal Server
Error)
Any ideas?
The easiest way is to get the data in Laravel Request. At least that's how I do it.
So your route shouldn't contain any parameter for that.
Your route will look like this:
Route::get('get-data-by-id', 'PlotController#getData')->name('get.data.by.id');
Your ajax should be like this:
$(document).on('change', '#sel_test',function(){
var testId = $(this).val();
$.ajax({
type:'GET',
url:"{{ route('get.data.by.id') }}",
data:{'test_id':testId},
success:function(data){
console.log(data);
}
});
});
In your controller's getData() function just use Laravel Request to fetch the data.
public function getData(Request $request)
{
// You can return the ID to see if the ajax is working
return $request->test_id;
}
Make it post from Get for easier
At Web.php
Route::post('/list/plots', 'PlotController#getData')->name('getData');
At Blade file Ajax Request :
$(document).ready(function() {
$('#sel_test').change(function() {
var testId = $(this).val();
var url = '{{ route("getData")}}';
var token = "{{ csrf_token()}}";
$.ajax({
method:"post",
url: url,
data:{testId:testId,_token:token}
dataType: 'json',
success: function(response) {
console.log("success",response);
}
});
});
});
At Controller :
public function getData(Request $request){
$testId = $request->testId;
// Write your logic here
}
Try this. Hopefully work for you
The question is pretty straightforward: I use #Html.EditorForModel() to generate fields for my model. Then user fills all these fields and I want to send this field via AJAX, becuase I should do several server's services without page reload.
I googled several approaches, but it seems that there is no standard way to do such things. I mean I do not have an object on client-side that represent model. I found one single library calls JSModel (link) but it seems to be not working. My code for now is:
#model Student
<script src="#Url.Content("~/scripts/jquery-1.12.2.min.js")" type="text/javascript" async="async"></script>
<script src="#Url.Content("~/scripts/Requester.js")" type="text/javascript" async="async"></script>
<script src="#Url.Content("~/scripts/jsmodel.js")" type="text/javascript"></script>
<script type="text/javascript">
var requester = new Requester(#Html.Raw(Json.Encode(new Student())));
function SendSignupRequest() {
requester.SendSignupRequest();
}
</script>
<h2>Student</h2>
<div>
#Html.EditorForModel()
</div>
<input type="button" value="Send" onclick="SendSignupRequest()"/>
Requester.js:
function Requester(rawModel) {
this.modelObj = new JSModel(rawModel);
this.SendSignupRequest = function() {
var model = modelObj.refresh();
var val = model.prop("Name");
alert(val);
}
}
Is there any easy way to serialize a model object in JSON and send it to server, without manually constructing an object with millions of document.getElementById?
View
#using (Html.BeginForm("action", "controller", FormMethod.Post, new { #class = "form-horizontal form-compact ", role = "form", id = "form1" }))
{
}
Java Script
var formdata = $("#form1").serializeArray();
$.ajax({
url: url,
type: 'POST',
data: formdata,
success: function (data) {
}
Controller
public ActionResult action(Model model)
{
//access data here
}
You can serialize your form to a JSON object with jQuery:
var data = $('form').serialize();
(This would, of course, mean wrapping your form elements in a form, which really should be happening anyway.)
Then just pass that data object to the server in the POST request. Something as simple as:
$.post('some/url', data, function(response) {
// success callback
});
without manually constructing an object with millions of document.getElementById
Note that if your object has millions of fields then you may very well encounter other problems here.
Yes you can use form serialize using Jquery
var formData = $('#form').serializeObject();
$.extend(formData, { Contacts : myContacts});
$.extend(formData, { Address : myAddress});
var result = JSON.stringify(formData);
$('#formHiddenField').val(result);
then submit form using:
$.ajax(
url: #Url.Action("post url")
data: myForm.serialize(),
dataType: 'json',
type: 'POST',
success: function(){
})
Why not Ajax.BeginForm() for your purposes. I believe model binding works automatically.
How can i submit a hidden form to php using ajax when the page loads?
I have a form with one hidden value which i want to submit without refreshing the page or any response message from the server. How can implement this in ajax? This is my form. I also have another form in the same page.
<form id = "ID_form" action = "validate.php" method = "post">
<input type = "hidden" name = "task_id" id = "task_id" value = <?php echo $_GET['task_id'];?>>
</form>
similar to Zafar's answer using jQuery
actually one of the examples on the jquery site https://api.jquery.com/jquery.post/
$(document).ready(function() {
$.post("validate.php", $("#ID_form").serialize());
});
you can .done(), .fail(), and .always() if you want to do anything with the response which you said you did not want.
in pure javascript
body.onload = function() {
var xmlhttp = new XMLHttpRequest();
xmlhttp.open("POST","validate.php",true);
xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded");
xmlhttp.send("task_id=" + document.getElementById("task_id").value);
};
I think you have doubts invoking ajax submit at page load. Try doing this -
<script type="text/javascript">
$(document).ready(function(){
$.ajax({
"url": "validate.php",
"type": "post"
"data": {"task_id": $("#task_id").val();},
"success": function(){
// do some action here
}
})
})
</script>
If you're using jQuery you should be able to get the form and then call submit() on it.
E.g.:
var $idForm = $('#ID_form');
$idForm.submit();
Simple solution - jQuery AJAX post the value as others have suggested, but embed the PHP value directly. If you have multiple forms, you can add more key:value pairs as needed. Add a success/error handler if needed.
<script type="text/javascript">
$(document).ready(function(){
$.post( "validate.php", { task_id: "<?=$_GET['task_id']?>" } );
})
</script>
As others have said, no need for a form if you want to send the data in the background.
validate.php
<?php
$task_id = $_POST['task_id'];
//perform tasks//
$send = ['received:' => $task_id]; //json format//
echo json_encode($send);
JQuery/AJAX:
$(function() { //execute code when DOM is ready (page load)//
var $task = $("#task_id").val(); //store hidden value//
$.ajax({
url: "validate.php", //location to send data//
type: "post",
data: {task_id: $task},
dataType: "json", //specify json format//
success: function(data){
console.log(data.received); //use data received from PHP//
}
});
});
HTML:
<input type="hidden" name="task_id" id="task_id" value=<?= $_GET['task_id'] ?>>
I am working with Concrete-5 CMS, I have an issue in passing value form view to controller.In my application I am using following code for displaying employee role.
foreach($rd as $data){
echo "<tr><td>".$data[role_name]."</td><td>".$data[role_description]."</td><td>Edit</td><td>".$ih->button_js(t('Delete'), "deleteRole('".$data['role_id']."')", 'left', 'error')."</td></tr>";
}
<input type="hidden" name="rno" id="rno" />
script:
$delConfirmJS = t('Are you sure you want to remove this Role?'); ?>
<script type="text/javascript">
function deleteRole(myvar) {
var role = document.getElementById('rno');
role.value = myvar;
if (confirm('<?php echo $delConfirmJS ?>')) {
$('#rolelist').submit();
//location.href = "<?php echo $this->url('/role/add_role/', 'delete', 'myvar')?>";
}
}
</script>
html code
I did edit operation by passing role_id through edit action. But, In case of delete i should ask for a conformation, so I use java script to conform it and call the href location and all.
But i don't know how to pass the role_id to script and pass to my controller. how to achieve this task?
thanks
Kumar
You can pass value to server using ajax calls.
See the following code. Here We use a confirm box to get user confirmation.
function deleteEmployee(empId){
var confirm=confirm("Do you want to delete?");
if (confirm)
{
var url = "path/to/delete.php";
var data = "emp_id="+empId;
$.ajax({
type: "POST",
url: "otherfile.php",
data: data ,
success: function(){
alert("Employee deleted successfully.");
}
});
}
}
In delete.php you can take the employee id by using $_POST['emp_id']
You can do it easily by using jquery
var dataString = 'any_variable='+ <?=$phpvariable?>;
$.ajax({
type: "POST",
url: "otherfile.php",
data: dataString,
success: function(msg){
// msg is return value of your otherfile.php
}
}); //END $.ajax
I would add an extra variable in to the delete link address. Preferrably the ID of the row that you need to be deleted.
I don't know Concrete-5 CMS. But, i am giving you the general idea
I think, you are using some button on which users can click if they want to delete role.
<td>".$ih->button_js(t('Delete'), "deleteRole('".$data['role_id']."')", 'left', 'error')."</td>
My suggestion,
add onClick to button
onClick="deleteEmployee(roleId);" // roleId - dynamic id of the role by looping over
Frankly speaking dude, i dont know how you will add this to your button that i guess there would surely be some way to simply add this to existing html.
And now, simply use Sajith's function
// Sajith's function here
function deleteEmployee(empId){
var confirm=confirm("Do you want to delete?");
if (confirm){
var url = "path/to/delete.php";
var data = "emp_id="+empId;
$.ajax({
type: "POST",
url: "otherfile.php",
data: data ,
success: function(){
alert("Employee deleted successfully.");
}
});
}
}