Re: modelform DateField rendering as text

[Ruchit Bhatt]

Is there any bootstrap hack to fix this ??

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Re: modelform DateField rendering as text

[Ruchit Bhatt]

yes, it should be


ok i will check that link

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Re: How do I set the value of a model object's property after using `get_field()`?

[Dylan Reinhold]

Andrew,
   You do not to get the field with _meta.get_field just use setattr which
takes a string as the field name.
Now if you are creating a new instance of your model and are just passing a
single field all other fields would need to have defaults or be null=True.

In your do stuff area you can just run this to create a new object with the
string 'Test Text':

model_instance = model()
setattr(model_instance, column_name, 'Test Text')
model_instance.save()


Dylan


On Mon, Jan 22, 2018 at 5:32 PM, Tom Tanner 
wrote:

> Darn, is this possible with a new object of the model? My idea is to in
> the end let the user input information that will be made into a new record
> to add to the model. But I need the user to type in the model they want to
> use...
>
>
> On Monday, January 22, 2018 at 1:09:53 PM UTC-5, Andrew Standley wrote:
>
>> Hey Tom,
>> First you'll need to create or get a particular instance of your
>> model using one of it's managers  `model.objects` and a query. Ex for a
>> model with a unique 'name' charfield: `model_obj =
>> model.objects.get(name='MyObject')`
>> You can then use the column_name to set that attribute on the instance
>> `setattr(model_obj, column_name) = 100` and finally save those changes
>> `model_obj.save()`
>> See https://docs.djangoproject.com/en/1.11/topics/db/queries/#
>> -Andrew
>> On 1/21/2018 9:44 PM, Tom Tanner wrote:
>>
>> I'm making a terminal command for my Django app:
>>
>> from django.core.management.base import BaseCommand, CommandError
>> from django.core.exceptions import FieldDoesNotExist
>> from django.apps import apps
>>
>>
>> class Command(BaseCommand):
>> def add_arguments(self, parser):
>> parser.add_argument(
>> "--app",
>> dest="app",
>> required=True,
>> )
>>
>> parser.add_argument(
>> "--model",
>> dest="model",
>> required=True,
>> )
>>
>> parser.add_argument(
>> "--col",
>> dest="col",
>> required=True,
>> )
>>
>> def handle(self, *args, **options):
>> app_label = options.get('app')
>> model_name = options.get('model')
>> column_name = options.get('col')
>>
>> try:
>> model = apps.get_model(app_label=app_label, model_name=
>> model_name)
>> except LookupError as e:
>> msg = 'The model "%s" under the app "%s" does not exist!'
>> \
>>   % (model_name, app_label)
>> raise CommandError(msg)
>> try:
>> column = model._meta.get_field(column_name)
>> except FieldDoesNotExist as e:
>> msg = 'The column "%s" does not match!' % column_name
>> raise CommandError(msg)
>> else:
>> print(column, type(column))
>> # Do stuff here with the column, model.
>>
>>
>> Right now, `column` is `> column_name>`. I want this instance of `model` to have `column_name` set to
>> `100`. How can I set and save this instance in this manner?
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Re: modelform DateField rendering as text

[Costja Covtushenko]

Hi,

Do you mean that it should be rendered like:
?

If so check that documentation page 
.
It is said that DateInput renders as type=’text’.

If you are curios why it is done in that way, please ask at: 
django-develop...@googlegroups.com 

Regards,
Constantine C.

> On Jan 22, 2018, at 10:33 AM, Ruchit Bhatt  
> wrote:
> 
> forms.ModelForm

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Re: How do I set the value of a model object's property after using `get_field()`?

[Tom Tanner]

Darn, is this possible with a new object of the model? My idea is to in the 
end let the user input information that will be made into a new record to 
add to the model. But I need the user to type in the model they want to 
use...

On Monday, January 22, 2018 at 1:09:53 PM UTC-5, Andrew Standley wrote:
>
> Hey Tom,
> First you'll need to create or get a particular instance of your model 
> using one of it's managers  `model.objects` and a query. Ex for a model 
> with a unique 'name' charfield: `model_obj = 
> model.objects.get(name='MyObject')`
> You can then use the column_name to set that attribute on the instance 
> `setattr(model_obj, column_name) = 100` and finally save those changes 
> `model_obj.save()`
> See https://docs.djangoproject.com/en/1.11/topics/db/queries/#
> -Andrew
> On 1/21/2018 9:44 PM, Tom Tanner wrote:
>
> I'm making a terminal command for my Django app:
>
> from django.core.management.base import BaseCommand, CommandError
> from django.core.exceptions import FieldDoesNotExist
> from django.apps import apps
> 
> 
> class Command(BaseCommand):
> def add_arguments(self, parser):
> parser.add_argument(
> "--app",
> dest="app",
> required=True,
> )
> 
> parser.add_argument(
> "--model",
> dest="model",
> required=True,
> )
> 
> parser.add_argument(
> "--col",
> dest="col",
> required=True,
> )
> 
> def handle(self, *args, **options):
> app_label = options.get('app')
> model_name = options.get('model')
> column_name = options.get('col')
> 
> try:
> model = apps.get_model(app_label=app_label, model_name=
> model_name)
> except LookupError as e:
> msg = 'The model "%s" under the app "%s" does not exist!' 
> \
>   % (model_name, app_label)
> raise CommandError(msg)
> try:
> column = model._meta.get_field(column_name)
> except FieldDoesNotExist as e:
> msg = 'The column "%s" does not match!' % column_name
> raise CommandError(msg)
> else:
> print(column, type(column))
> # Do stuff here with the column, model.
>
>
> Right now, `column` is ` column_name>`. I want this instance of `model` to have `column_name` set to 
> `100`. How can I set and save this instance in this manner?
> -- 
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Re: Django 1.5 tutorial error: 'module' has no attribute 'StackedInLine'

[Richard Roberts]

Awesome Spot Karen. 

On Wednesday, 27 March 2013 15:48:18 UTC, Karen Tracey wrote:
>
> On Wed, Mar 27, 2013 at 11:32 AM, +Emmanuel  > wrote:
>
>> Hello there,
>>
>> I am working through the django tutorial at djangoproject.com. So far 
>> everything works out fine until Section 2.4.6 (customizing the admin form). 
>> Django throws an error:  *'module' has no attribute 'StackedInLine'. *
>>
>
> Right, the L in StackedInLine should NOT be capitalized, it should be just 
> an initial capital letter on Inline: StackedInline 
>

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Re: django-payslip

[sum abiut]

Thanks for response. Manage to load the page this time.

Cheers

On Mon, Jan 22, 2018 at 3:59 PM, Ramiro Morales  wrote:

> The detailed error page is giving you all the information you need to
> diagnose the issue.
>
> It tells you you have two paths defined in your URL map: /admin/ and
> /payslip/ as per their respective documentation. So far so good.
>
> But you are accessing / with your browser. Hence the 404 error. Try
> accessing /payslip/
>
> Best regards,
>
> On Jan 22, 2018 12:29 AM, "sum abiut"  wrote:
>
> Hi,
> Has anyone from list have any experience with django-payslip before. I
> follow direction from https://pypi.python.org/pypi/django-payslip/0.2.2
>
> But i got the error below trying to load the page. Please advise
>
> [image: Inline image 1]
>
> cheer,
>
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Re: How do I set the value of a model object's property after using `get_field()`?

[Andrew Standley]

Hey Tom,
    First you'll need to create or get a particular instance of your 
model using one of it's managers  `model.objects` and a query. Ex for a 
model with a unique 'name' charfield: `model_obj = 
model.objects.get(name='MyObject')`
You can then use the column_name to set that attribute on the instance 
`setattr(model_obj, column_name) = 100` and finally save those changes 
`model_obj.save()`

See https://docs.djangoproject.com/en/1.11/topics/db/queries/#

-Andrew
On 1/21/2018 9:44 PM, Tom Tanner wrote:

I'm making a terminal command for my Django app:

|
fromdjango.core.management.baseimportBaseCommand,CommandError
fromdjango.core.exceptions importFieldDoesNotExist
fromdjango.apps importapps


classCommand(BaseCommand):
defadd_arguments(self,parser):
            parser.add_argument(
"--app",
                dest="app",
                required=True,
)

            parser.add_argument(
"--model",
                dest="model",
                required=True,
)

            parser.add_argument(
"--col",
                dest="col",
                required=True,
)

defhandle(self,*args,**options):
            app_label =options.get('app')
            model_name =options.get('model')
            column_name =options.get('col')

try:
                model 
=apps.get_model(app_label=app_label,model_name=model_name)

exceptLookupErrorase:
                msg ='The model "%s" under the app "%s" does not exist!'\
%(model_name,app_label)
raiseCommandError(msg)
try:
                column =model._meta.get_field(column_name)
exceptFieldDoesNotExistase:
                msg ='The column "%s" does not match!'%column_name
raiseCommandError(msg)
else:
print(column,type(column))
# Do stuff here with the column, model.
|


Right now, `column` is `column_name>`. I want this instance of `model` to have `column_name` 
set to `100`. How can I set and save this instance in this manner?

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Re: Need to create an edit box connected to one table field

[eileen]

Yes, I use Form. Here's it's contents:
from django import forms
from .widgets import ChainedSelectWidget
from .models import Child


class SponsorForm(forms.Form):
child = forms.IntegerField()


class FilterForm(forms.Form):
gender = forms.ChoiceField(choices=[(x, x) for x in ('-', 'MALE', 
'FEMALE')], required=False)
age = forms.ChoiceField(choices=[(x, x) for x in range(1, 18)], 
required=False)
#orphaned = forms.BooleanField(initial=False,required=False)
#extreme_need = forms.BooleanField(initial=False,required=False)
handicapped = forms.ChoiceField(choices=[(x, x) for x in ('---', 
'Mental', 'Physcal')], required=False)

def __init__(self, *args, **kwargs):
super(FilterForm, self).__init__(*args, **kwargs)

if 0 == len(self.data):
self.fields['age'].queryset = Child.objects.none()

# assign a widget to second select field
self.fields['age'].widget = ChainedSelectWidget(
parent_name='gender', # the name of parent field
app_name='sponsorship',# the name of model's 
application
model_name='child',  # the name of a model with the 
method
method_name='get_children',  # the name of queryset method
)

It is for finding a child who may or may not have one of two types of 
handicaps.


-Eileen

On Friday, January 19, 2018 at 9:54:15 PM UTC-5, Costja Covtushenko wrote:
>
> Hi Eileen,
>
> Can you please elaborate a little bit?
> Do you use Form? Can you provide its code?
>
> Also sorry but it is not clear what are you trying to achieve with those 
> value?
> Is it for searching data in DB?
>
> Regards,
> Constantine C.
>
> On Jan 19, 2018, at 5:08 PM, eil...@themaii.org  wrote:
>
> handicapped
>
>
>

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Re: Django escapes a string partially

[Julio Biason]

Hi,

You should wrap your  `data-content={{ v.error_msg }}` in quotes, like
this: `data-content="{{ v.error_msg }}"`.

Otherwise you'll generate the template as `data-content='NoneType' object
has not attribute "rfind"`, which is a valid HTML (data-content will have
the string 'NoneType' and the node will have a bunch of tags that it
doesn't know what to do with it, so they just sit there: "object", "has",
"not", "attribute" and ""rfind"".)

On Mon, Jan 22, 2018 at 1:47 PM, Ron Moran  wrote:

> Hi there!
>
> I have a problem with escaping an error.
> I have this unicode in python(2.7) which isn't escaped well in a span
> element:
>
> u"'NoneType' object has no attribute 'rfind'"
> The template is defined as follows:
>
>data-content={{ v.error_msg }} data-trigger="hover" rel="popover"
>   data-original-title="Error Title" {% endif %}
>   class="label extra-label label-pill label-{{ v.show_class | safe }}">
> {{ v.get_human_status }}
>
>
> where v.error_msg is the string above. It should be noted that all other 
> attributes works just fine
>
>
> I wrapped the template in
> {% autoescape on %}
> ...template content...
>
> {% endautoescape %}
>
>
> Then I tried using the escape/force_escape tag on the problematic string:
>
> {{ v.error_msg | force_escape }}
>
> Nothing works, the output is only escaped once:
>
> data-content="'NoneType'" object has not attribute 'rfind'
> Which causes the resulting popover to show only with the message 'NoneType'.
>
> What am I doing wrong? Why isn't the string properly escaped?
>
>
> I'm using Djagno 1.5.11, but this syntax was defined well before version 
> 1.5.11, so I don't think it's a version issue.
>
> It's a minor issue but it's driving me mad.
>
>
> Thanks,
>
> Ron
>
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Re: Django escapes a string partially

[Daniel Roseman]

On Monday, 22 January 2018 16:02:30 UTC, Ron Moran wrote:
>
> Hi there!
>
> I have a problem with escaping an error.
> I have this unicode in python(2.7) which isn't escaped well in a span 
> element:
>
> u"'NoneType' object has no attribute 'rfind'"
> The template is defined as follows:
>
>data-content={{ v.error_msg }} data-trigger="hover" rel="popover"
>   data-original-title="Error Title" {% endif %}
>   class="label extra-label label-pill label-{{ v.show_class | safe }}">
> {{ v.get_human_status }}
>
>
> where v.error_msg is the string above. It should be noted that all other 
> attributes works just fine
>
>
> I wrapped the template in 
> {% autoescape on %}
> ...template content...
>
> {% endautoescape %}
>
>
> Then I tried using the escape/force_escape tag on the problematic string:
>
> {{ v.error_msg | force_escape }}
>
> Nothing works, the output is only escaped once:
>
> data-content="'NoneType'" object has not attribute 'rfind'
> Which causes the resulting popover to show only with the message 'NoneType'.
>
> What am I doing wrong? Why isn't the string properly escaped?
>
>
> I'm using Djagno 1.5.11, but this syntax was defined well before version 
> 1.5.11, so I don't think it's a version issue.
>
> It's a minor issue but it's driving me mad.
>
>
> Thanks,
>
> Ron
>
>
Firstly, you **must** upgrade; Django 1.5 is seven major versions old and 
totally insecure. 

For your actual problem, you need to wrap your template variable in quotes:
data-content="{{ v.error_msg }}"
--
DR.

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Django escapes a string partially

[Ron Moran]

Hi there!

I have a problem with escaping an error.
I have this unicode in python(2.7) which isn't escaped well in a span 
element:

u"'NoneType' object has no attribute 'rfind'"
The template is defined as follows:


{{ v.get_human_status }}


where v.error_msg is the string above. It should be noted that all other 
attributes works just fine


I wrapped the template in 
{% autoescape on %}
...template content...

{% endautoescape %}


Then I tried using the escape/force_escape tag on the problematic string:

{{ v.error_msg | force_escape }}

Nothing works, the output is only escaped once:

data-content="'NoneType'" object has not attribute 'rfind'
Which causes the resulting popover to show only with the message 'NoneType'.

What am I doing wrong? Why isn't the string properly escaped?


I'm using Djagno 1.5.11, but this syntax was defined well before version 
1.5.11, so I don't think it's a version issue.

It's a minor issue but it's driving me mad.


Thanks,

Ron

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modelform DateField rendering as text

[Ruchit Bhatt]

*models.py*
class HumanUser(AbstractUser):
birth_date = models.DateField(null=True, blank=True, verbose_name=u
"DOB")

*forms.py*
class profile_form(forms.ModelForm):

def __init__(self, *args, **kwargs):
kwargs.setdefault('label_suffix', '')
super(profile_form, self).__init__(*args, **kwargs)
for field in self.fields:
self.fields[field].widget.attrs.update({'class': 'form-control 
form-control-line', })

class Meta:
model = User
fields = [
'full_name',
'birth_date',
'mobile',
'bio',
'gender',
]


This is my code where birth_date type is showing as text instead of date in 
HTML..any idea why ?





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Re: Calling api from django view error. forbidden (csrf token is missing or incorrect)

[Yungjae Kim]

You will have to either get CSRF token and send it or ignore it completely. 
csrf_exempt from django.views.decorators.csrf will be helpful. For 
HttpResponse to returning a json, pass in a stringfied dict.

On Monday, January 22, 2018 at 5:06:35 AM UTC-5, chern...@gmail.com wrote:
>
> The reason why im using this is because i am using integrating 2 different 
> project. The api call is suppose to call the api from another app. Further 
> more i also need to save some of the details the was get from both post and 
> get method
>
> On Monday, January 22, 2018 at 5:03:18 PM UTC+9, chern...@gmail.com wrote:
>>
>> I seen alot of other solution, tried it but problem still persist.
>>
>> When i do a requests.get, it works fine but when i'm doing requests.post. 
>> I got this forbidden (csrf token is missing or incorrect) error.
>>
>>
>> Here is my code
>>
>> *models.py*
>>
>> class TestPost(models.Model):
>> # reminderId = models.AutoField()
>> book = models.CharField(max_length=10, blank=True, null=True)
>> author = models.CharField(max_length=10, blank=True, null=True)
>> date = models.DateTimeField(blank=True, null=True)
>>
>> *serializer.py*
>>
>> class TestPostSerializer(serializers.ModelSerializer):
>> # valid_time_formats = ['%H:%M', '%I:%M%p', '%I:%M %p']
>> # time = serializers.TimeField(format='%I:%M %p', 
>> input_formats=valid_time_formats, allow_null=True)
>> date = serializers.DateTimeField(format="%Y-%m-%d %I:%M %p")
>>
>> class Meta:
>> model = TestPost
>> fields = ('id', 'book', 'author', 'date')
>>
>> *views.py*
>>
>> from django.http import HttpResponseimport requests
>> def my_django_view(request):
>> if request.method == 'POST':
>> r = requests.post('http://127.0.0.1:8000/api/test/', 
>> params=request.POST)
>> else:
>> r = requests.get('http://127.0.0.1:8000/api/test/', 
>> params=request.GET)
>> if r.status_code == 200:
>> return HttpResponse('Yay, it worked')
>> return HttpResponse('Could not save data')
>> class TestPostViewSet(viewsets.ModelViewSet):
>> permission_classes = [AllowAny]
>> queryset = TestPost.objects.all()
>> serializer_class = TestPostSerializer
>>
>>
>> I did a POST method on the url of the function but error
>>
>>
>> Forbidden (CSRF token missing or incorrect.): /test/ [22/Jan/2018 
>> 16:59:09] "POST /test/ HTTP/1.1" 403 2502
>>
>>
>> Also, how do i make the HttpResponse to display the json data from my get 
>> and post method ?
>>
>

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Re: Calling api from django view error. forbidden (csrf token is missing or incorrect)

[cherngyorng]

The reason why im using this is because i am using integrating 2 different 
project. The api call is suppose to call the api from another app. Further 
more i also need to save some of the details the was get from both post and 
get method

On Monday, January 22, 2018 at 5:03:18 PM UTC+9, chern...@gmail.com wrote:
>
> I seen alot of other solution, tried it but problem still persist.
>
> When i do a requests.get, it works fine but when i'm doing requests.post. 
> I got this forbidden (csrf token is missing or incorrect) error.
>
>
> Here is my code
>
> *models.py*
>
> class TestPost(models.Model):
> # reminderId = models.AutoField()
> book = models.CharField(max_length=10, blank=True, null=True)
> author = models.CharField(max_length=10, blank=True, null=True)
> date = models.DateTimeField(blank=True, null=True)
>
> *serializer.py*
>
> class TestPostSerializer(serializers.ModelSerializer):
> # valid_time_formats = ['%H:%M', '%I:%M%p', '%I:%M %p']
> # time = serializers.TimeField(format='%I:%M %p', 
> input_formats=valid_time_formats, allow_null=True)
> date = serializers.DateTimeField(format="%Y-%m-%d %I:%M %p")
>
> class Meta:
> model = TestPost
> fields = ('id', 'book', 'author', 'date')
>
> *views.py*
>
> from django.http import HttpResponseimport requests
> def my_django_view(request):
> if request.method == 'POST':
> r = requests.post('http://127.0.0.1:8000/api/test/', 
> params=request.POST)
> else:
> r = requests.get('http://127.0.0.1:8000/api/test/', 
> params=request.GET)
> if r.status_code == 200:
> return HttpResponse('Yay, it worked')
> return HttpResponse('Could not save data')
> class TestPostViewSet(viewsets.ModelViewSet):
> permission_classes = [AllowAny]
> queryset = TestPost.objects.all()
> serializer_class = TestPostSerializer
>
>
> I did a POST method on the url of the function but error
>
>
> Forbidden (CSRF token missing or incorrect.): /test/ [22/Jan/2018 
> 16:59:09] "POST /test/ HTTP/1.1" 403 2502
>
>
> Also, how do i make the HttpResponse to display the json data from my get 
> and post method ?
>

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Django view to call api

[cherngyorng]

So i'm using django view to call api post method but its giving me 400 error

As i encounter csrf forbidden error b4, im using the @csrf_exempt for now. 
I tried doing post method on both the API itself and also using the view to 
call the api.

However when using the view to call, i got 400 error when both are using 
the same post value to post.

Posting using api: QueryDict: {'book': ['Second '], 'author': ['Ban'], 
'date': ['2018-01-10 08:00AM']} [22/Jan/2018 18:56:09] "POST /api/test/ 
HTTP/1.1" 200 61

Posting using view: QueryDict: {} [22/Jan/2018 18:56:12] "POST 
/api/test/?book=Second+&author=Ban&date=2018-01-10+08%3A00AM HTTP/1.1" 400 
36 [22/Jan/2018 18:56:12] "POST /test/ HTTP/1.1" 200 19

Here is my code

*models.py*

class TestPost(models.Model):
book = models.CharField(max_length=10, blank=True, null=True)
author = models.CharField(max_length=10, blank=True, null=True)
date = models.DateTimeField(blank=True, null=True)

*serializer.py*

class TestPostSerializer(serializers.ModelSerializer):
date = serializers.DateTimeField(format="%Y-%m-%d %I:%M %p")

class Meta:
model = TestPost
fields = ('id', 'book', 'author', 'date')

*views.py*

from django.http import HttpResponseimport requests
def my_django_view(request):
if request.method == 'POST':
r = requests.post('http://127.0.0.1:8000/api/test/', 
params=request.POST)
else:
r = requests.get('http://127.0.0.1:8000/api/test/', params=request.GET)
if r.status_code == 200:
return HttpResponse('Yay, it worked')
return HttpResponse('Could not save data')
class TestPostViewSet(viewsets.ModelViewSet):
permission_classes = [AllowAny]
queryset = TestPost.objects.all()
serializer_class = TestPostSerializer

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Re: Calling api from django view error. forbidden (csrf token is missing or incorrect)

[Andréas Kühne]

Hi,

You seem to be doing a very complicated setup. You are creating both the
api viewset and another view. First of all - why?

Secondly, I am not sure that the csrf token will work when chaining your
posts like that.

So my main issue would be, can't you just post directly to the /api/test/
viewset?

To answer you second question, if you just use the viewset directly it will
automatically display the JSON data. That is a core function in the django
restframework.

Regards,

Andréas

2018-01-22 9:03 GMT+01:00 :

> I seen alot of other solution, tried it but problem still persist.
>
> When i do a requests.get, it works fine but when i'm doing requests.post.
> I got this forbidden (csrf token is missing or incorrect) error.
>
>
> Here is my code
>
> *models.py*
>
> class TestPost(models.Model):
> # reminderId = models.AutoField()
> book = models.CharField(max_length=10, blank=True, null=True)
> author = models.CharField(max_length=10, blank=True, null=True)
> date = models.DateTimeField(blank=True, null=True)
>
> *serializer.py*
>
> class TestPostSerializer(serializers.ModelSerializer):
> # valid_time_formats = ['%H:%M', '%I:%M%p', '%I:%M %p']
> # time = serializers.TimeField(format='%I:%M %p', 
> input_formats=valid_time_formats, allow_null=True)
> date = serializers.DateTimeField(format="%Y-%m-%d %I:%M %p")
>
> class Meta:
> model = TestPost
> fields = ('id', 'book', 'author', 'date')
>
> *views.py*
>
> from django.http import HttpResponseimport requests
> def my_django_view(request):
> if request.method == 'POST':
> r = requests.post('http://127.0.0.1:8000/api/test/', 
> params=request.POST)
> else:
> r = requests.get('http://127.0.0.1:8000/api/test/', 
> params=request.GET)
> if r.status_code == 200:
> return HttpResponse('Yay, it worked')
> return HttpResponse('Could not save data')
> class TestPostViewSet(viewsets.ModelViewSet):
> permission_classes = [AllowAny]
> queryset = TestPost.objects.all()
> serializer_class = TestPostSerializer
>
>
> I did a POST method on the url of the function but error
>
>
> Forbidden (CSRF token missing or incorrect.): /test/ [22/Jan/2018
> 16:59:09] "POST /test/ HTTP/1.1" 403 2502
>
>
> Also, how do i make the HttpResponse to display the json data from my get
> and post method ?
>
> --
> You received this message because you are subscribed to the Google Groups
> "Django users" group.
> To unsubscribe from this group and stop receiving emails from it, send an
> email to django-users+unsubscr...@googlegroups.com.
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> msgid/django-users/2b343676-12d3-47e7-9c2d-592580256e1a%40googlegroups.com
> 
> .
> For more options, visit https://groups.google.com/d/optout.
>

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Calling api from django view error. forbidden (csrf token is missing or incorrect)

[cherngyorng]

I seen alot of other solution, tried it but problem still persist.

When i do a requests.get, it works fine but when i'm doing requests.post. I 
got this forbidden (csrf token is missing or incorrect) error.


Here is my code

*models.py*

class TestPost(models.Model):
# reminderId = models.AutoField()
book = models.CharField(max_length=10, blank=True, null=True)
author = models.CharField(max_length=10, blank=True, null=True)
date = models.DateTimeField(blank=True, null=True)

*serializer.py*

class TestPostSerializer(serializers.ModelSerializer):
# valid_time_formats = ['%H:%M', '%I:%M%p', '%I:%M %p']
# time = serializers.TimeField(format='%I:%M %p', 
input_formats=valid_time_formats, allow_null=True)
date = serializers.DateTimeField(format="%Y-%m-%d %I:%M %p")

class Meta:
model = TestPost
fields = ('id', 'book', 'author', 'date')

*views.py*

from django.http import HttpResponseimport requests
def my_django_view(request):
if request.method == 'POST':
r = requests.post('http://127.0.0.1:8000/api/test/', 
params=request.POST)
else:
r = requests.get('http://127.0.0.1:8000/api/test/', params=request.GET)
if r.status_code == 200:
return HttpResponse('Yay, it worked')
return HttpResponse('Could not save data')
class TestPostViewSet(viewsets.ModelViewSet):
permission_classes = [AllowAny]
queryset = TestPost.objects.all()
serializer_class = TestPostSerializer


I did a POST method on the url of the function but error


Forbidden (CSRF token missing or incorrect.): /test/ [22/Jan/2018 16:59:09] 
"POST /test/ HTTP/1.1" 403 2502


Also, how do i make the HttpResponse to display the json data from my get 
and post method ?

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