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django file upload from json
Hi am using the following ajax upload from the template but i do not get a response from django view.What is wrong here..i do not see any alert
function ajax_upload(formid)
{
var form = $(formid);
form.ajaxSubmit({
dataType: 'json',
success: function (data) {
alert("Hereeeeeeee");
if(data.status == '1')
{
alert("Uploaded Successfull");
}
else
{
alert("Uploaded UnSuccessfull :(");
}
}
} ) ;
}
EDIT
Django:
def someview(request):
response_dict={'status':1}
logging.debug("seen") //This is seen in the logs
return HttpResponse(simplejson.dumps(response_dict), mimetype='application/javascript')
EDIT1
Please also for complete source code look at django file upload from json
From your code:
form.ajaxSubmit({
dataType: 'json',
success: function (data) {
alert("Hereeeeeeee");
Your callback executes on 'success', so we can narrow down the failure to the django view side. Assuming everything is syntactically correct, my best guess is that since you're returning json data, your response mimetype should be:
mimetype='application/json'.
If that doesn't work, I would suggest using Firebug on Firefox or Developer Tools on Chrome to look at the server response. You should be able to see a django stacktrace there.
Check that someview added to urls.py.
Check that {% csrf_token %} added to form
It's rather impossible you will see that log, hence this is a python syntax error:
response_dict{'status':1}
and should be:
response_dict = dict(status=1)
or:
response_dict = {'status':1}
or:
response_dict = dict()
response_dict['status'] = 1
or:
response_dict = dict()
response_dict.update({'status':1})
Related
I'm processing a table of banking/statement entries that have been exported from another system via a CSV file. They are imported into a view and checked for duplicates before being presented to the user in a HTML table for final review.
Once checked they are sent via AJAX to the server so they can be added into a Django model. Everything is working OK including CSRF but I cannot access the POSTed variable although I can see it!
Unfortunately making a hidden form isn't viable as there are 80+ rows to process.
My Javascript looks like:
$.ajax({
type: 'POST',
url: '......./ajax/handleImports/',
data: entriesObj,
success: function (data) {
if (data.response && data.response) {
console.log("Update was successful");
console.log(data.entries)
} else { ... }
},
error: function() { ... }
where entriesObj is
var entriesObj = JSON.stringify({ "newentries": newEntries });
console.log(entriesObj)
and when dumped to console.log looks like:
{"newentries":[{"Include":"","Upload ID":"0","Date":"2019-01-09", ... }
Now in view.py when I return the whole request.POST object as data.entries using
context['entries'] = request.POST
return JsonResponse(context)
I get
{"{"newentries":[{"Include":"","Upload ID":"0","Date":"2019-01-09", ... }
but if I try and retrieve newentries with:
entries = request.POST.get('newentries', None)
context['entries'] = entries
return JsonResponse(context)
the console.log(data.entries) will output null?
How am I supposed to access the POSTed entriesObj?
The data is JSON, you need to get the value from request.body and parse it.
data = json.loads(request.body)
entries = data.get('newentries')
Im having a hard time figuring out how to integrate this ajax request into my view. I'm still learning how to integrate django with ajax requests.
My first question would be: Does the ajax request need to have its own dedicated URL?
In my case I am trying to call it on a button to preform a filter(Preforms a query dependent on what is selected in the template). I have implemented this using just django but it needs to make new request everytime the user preforms a filter which I know is not efficient.
I wrote the most basic function using JQuery to make sure the communication is there. Whenever the user changed the option in the select box it would print the value to the console. As you will see below in the view, I would to call the ajax request inside this view function, if this is possible or the correct way of doing it.
JQuery - Updated
$("#temp").change( function(event) {
var filtered = $(this).val();
console.log($(this).val());
$.ajax({
url : "http://127.0.0.1:8000/req/ajax/",
type : "GET",
data : {
'filtered': filtered
},
dataType: 'json',
success: function(data){
console.log(data)
},
error: function(xhr, errmsg, err){
console.log("error")
console.log(error_data)
}
});
Views.py
def pending_action(request):
requisition_status = ['All', 'Created', 'For Assistance', 'Assistance Complete', 'Assistance Rejected']
FA_status = RequisitionStatus.objects.get(status='For Assistance')
current_status = 'All'
status_list = []
all_status = RequisitionStatus.objects.all()
status_list = [status.status for status in all_status]
# This is where I am handling the filtering currently
if request.GET.get('Filter') in status_list:
user_req_lines_incomplete = RequisitionLine.objects.filter(Q(parent_req__username=request.user) & Q(status__status=request.GET.get('Filter')))
current_status = request.GET.get('Filter')
else:
user_req_lines_incomplete = RequisitionLine.objects.filter(parent_req__username=request.user).exclude(status__status='Completed')
user_reqs = Requisition.objects.filter(par_req_line__in=user_req_lines_incomplete).annotate(aggregated_price=Sum('par_req_line__total_price'),
header_status=Max('par_req_line__status__rating'))
return render(request, 'req/pending_action.html', { 'user_reqs':user_reqs,
'user_req_lines_incomplete':user_req_lines_incomplete,
'requisition_status':requisition_status,
'current_status':current_status,
'FA_status':FA_status})
def filter_status(request):
status = request.GET.get('Filter')
data = {
'filtered': RequisitionLine.objects.filter(Q(parent_req__username=request.user) & Q(status__status=status)),
'current_status': status
}
return JsonResponse(data)
Urls.py
path('pending/', views.pending_action, name='pending_action')
First: you have to divide your template to unchangeable part and the part that you want to modify with your filter.
Second: for your goal you can use render_to_string. See the followning link https://docs.djangoproject.com/en/2.1/topics/templates/#usage
code example (views.py):
cont = {
'request': request, #important key-value
'your_models_instances': your_models_instances
}
html = render_to_string('your_filter_template.html', cont)
return_dict = {'html': html}
return JsonResponse(return_dict)
In your js file you need to determine relative url "{% url 'name in yours url file'%}"
And in success you need to add next line:
success: function(data){
$(".filter-block").html(data.html);
}
i hope it will help you! Good luck!
I am getting a missing CSRF_Token error that only occurs in production mode on my server. However everything works great when I am running it from my computer terminal using the runserver command. I've read through many of the other questions pertaining to this with no luck. It seems that my case is slightly different than others, since it works locally but not in production.
I get the error when submitting an Ajax form that submits to the "submit" in views.py. Does anybody know what could be causing this? Also, looking at my cookies in Production mode, the CSRF_Token is not even there to begin with. Locally it is. Thanks for any help.
Here is my views.py
from django.shortcuts import render
from django.http import HttpResponse
def index(request):
return render(request, 'index.html')
def submit(request):
#Receive Request
inputone = request.POST['randominfo']
inputtwo = request.POST['randominfo2']
#Some more code here that setups response.
#Deleted since Im posting to StackOverflow
return response
Code Pertaining to the Ajax Submit
$(function () {
$.ajaxSetup({
headers: { "X-CSRFToken": getCookie("csrftoken") }
});
});
function getCookie(c_name)
{
if (document.cookie.length > 0)
{
c_start = document.cookie.indexOf(c_name + "=");
if (c_start != -1)
{
c_start = c_start + c_name.length + 1;
c_end = document.cookie.indexOf(";", c_start);
if (c_end == -1) c_end = document.cookie.length;
return unescape(document.cookie.substring(c_start,c_end));
}
}
return "";
}
function submitAjax(event){
$.ajax({
type:'POST',
url:'/submit/',
data:{
randominfo:document.getElementById('Random').innerHTML,
randominfo2:document.getElementById('Random2').innerHTML,
},
dateType: 'json',
success:function() {
# Url here
}
})
};
Solution that fixed this problem.
Adding "from django.views.decorators.csrf import ensure_csrf_cookie" in views.py and then "#ensure_csrf_cookie" above the view that returns the html file that contained the ajax form
The error ocurs because you are not setting the csrf token, to prevent this we have to check some details
First of all, you have to set the csrf token to your form, in your html you have to set some as follow:
<form id="id" name="form">
{% csrf_token %}
<!-- Form body here -->
</form>
Second the approach to set the csrf cookie to your request header is ok, i only suggest that instead you set your data field one by one, use method serialize of jquery
data: $("#your-form-id").serialize()
I would like to recommend you to read this post about ajax request with django that is very helpful
There are 2 things you can do:
1.) Submit a CSRF token in your ajax call. You have to use a getCookie() javascript function to get it. Luckily the django documentation has some code you can copy and paste.
javascript
$.ajax({
type:'POST',
url:'/submit/',
data:{
randominfo:document.getElementById('Random').innerHTML,
randominfo2:document.getElementById('Random2').innerHTML,
'csrfmiddlewaretoken': getCookie('csrftoken'), // add this
...
2.) Disable csrf for your /submit view. You can do this with a decorator. Note that this is less secure so make sure there's no confidential data.
views.py:
from django.views.decorators.csrf import csrf_exempt
...
#csrf_exempt
def your_submit_view(request):
#view code
This question already has answers here:
How to get POSTed JSON in Flask?
(13 answers)
Closed 6 years ago.
I've got a google-maps lookup with autocomplete on the client side - I'm trying to transfer the "place" object over to the server once the user selects it - I could parse it down client-side, but I figured it's easier to do server-side. I verified from the browser 'network' inspect that it is sending over the correct json object, but server side I can't get the right object. I've tried every permutation of request.* that I can find and either get None or a <module 'flask.json'> which I know isn't right.
Code:
function sendplace() {
$('placebutton').click(function() {
var add1 = place;
console.log(place);
$.ajax({
url: '/new_place2',
data: JSON.stringify(place),
contentType: 'application/json;charset=UTF-8',
type: 'POST',
success: function(response) {
console.log(response);
},
error: function(error) {
console.log(error);
}
and server-side:
#app.route('/new_place2', methods=['GET', 'POST'])
def new_place2():
place = request.form.get('place')
print "address: ", place
return ("Success, (*&)er!")
You should use request.get_json(). request.form is reserved for the mimetype application/www-form-urlencoded. If get_json() doesn't work look at request.data and ensure that you can run the following on it:
import json
data = json.loads(request.data)
data.get('place')
If that doesn't work you probably have a problem with your JSON
I got a very strange problem, I thought this worked before but it doesn't any more. I dont even remember changing anything. I tried with an older jQuery library.
I got an error that says: http://i.imgur.com/H51wG4G.png on row 68: (anonymous function). which refer to row 68:
var jsondata = $.parseJSON(data);
This is my ajax function
I can't get my alert to work either because of this error. this script by the way is for logging in, so if I refresh my website I will be logged in, so that work. I also return my json object good as you can see in the image. {"success":false,"msg":"Fel anv\u00e4ndarnamn eller l\u00f6senord.","redirect":""}
When I got this, I will check in login.success if I got success == true and get the login panel from logged-in.php.
$('#login_form').submit(function()
{
var login = $.ajax(
{
url: '/dev/ajax/trylogin.php',
data: $(this).serialize(),
type: 'POST',
}, 'json');
login.success(function(data)
{
var jsondata = $.parseJSON(data);
console.log(jsondata);
if(jsondata.success == true)
{
$.get("/dev/class/UI/logged-in.php", function(data) {
$(".login-form").replaceWith(data);
});
}
else
{
alert(jsondata.msg);
$('#pwd').val('');
}
});
return false;
});
Thank you.
If the response you have showed in the attached screenshot is something to go by, you have a problem in your PHP script that's generating the JSON response. Make sure that thePHP script that's generating this response (or any other script included in that file) is not using a constant named SITE_TITLE. If any of those PHP files need to use that constant, make sure that that SITE_TILE is defined somewhere and included in those files.
What might have happened is that one of the PHP files involved in the JSON response generation might have changed somehow and started using the SITE_TITLE costant without defining it first, or without including the file that contains that constant.
Or, maybe none of the files involved in the JSON generation have changed, but rather, your error_reporting settings might have changed and now that PHP interpreter is outputting the notice level texts when it sees some undefined constant.
Solving the problem
If the SITE_TITLE constant is undefined, define it.
If the SITE_TITLE constant is defined in some other file, include that file in the PHP script that's generating the response.
Otherwise, and I am not recommending this, set up your error_reporting settings to ignore the Notice.
Your response is not a valid JSON. You see: "unexpected token <".
It means that your response contains an unexpected "<" and it cannot be converted into JSON format.
Put a console.log(data) before converting it into JSON.
You shoud use login.done() , not login.success() :)
Success is used inside the ajax() funciton only! The success object function is deprecated, you can set success only as Ajax() param!
And there is no need to Parse the data because its in Json format already!
jQuery Ajax
$('#login_form').submit(function()
{
var login = $.ajax(
{
url: '/dev/ajax/trylogin.php',
data: $(this).serialize(),
type: 'POST',
}, 'json');
login.done(function(data)
{
var jsondata = data;
console.log(jsondata);
if(jsondata.success == true)
{
$.get("/dev/class/UI/logged-in.php", function(data) {
$(".login-form").replaceWith(data);
});
}
else
{
alert(jsondata.msg);
$('#pwd').val('');
}
});
return false;
});