subject:"Re\: django filter"

Re: Django-filter customization for rest api

2019-09-05 Thread Nishant Bhanushali

I have an doubt..
There are many field types and many other field option.
How can we make it

On Thu, Sep 5, 2019, 18:56 Abhineet Baranwal 
wrote:

> When we pass fields in Meta class for custom filter then django-filter
> adds 'and' in-between the fields .how to add 'or' operator in between
> the fields.please explain with an example for rest api.
>
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Re: django-filter over models

2018-09-19 Thread brajesh kumar

I think this answers my question , trying to implement it.

On Wednesday, September 19, 2018 at 4:45:02 AM UTC+5:30, Mateusz wrote:
>
> In that answer I assume Jobs are connected anyhow with Persons with a 
> ForeignKey field like so:
>
> *File /appname/models.py:*
>
> from django.db import models
>
>
> class Job(models.Model):
> name = models.CharField(max_length=100)
>
> def __str__(self):
> return self.name
>
>
> class Person(models.Model):
> name = models.CharField(max_length=100)
> surname = models.CharField(max_length=100)
>
> job = models.ForeignKey(Job, on_delete=models.SET_NULL, null=True, 
> default=None)
>
> def __str__(self):
> return self.name + ' ' + self.surname
>
>
> *File /appname/views.py:*
>
> from django.shortcuts import render
> from django.views.generic.list import ListView
>
> from appname.models import Job, Person
>
>
> class JobsWithPersonsView(ListView):
> queryset = Job.objects.all()
> template_name = 'jobs_list.html'
>
>
>
> *File /appname/templates/jobs_list.html:*
>
> 
> 
> 
> 
> Jobs
> 
> 
> {% for job in object_list %}
> {{ job.name }}
> {% for person in job.person_set.all %}
> {{ person.name }} {{ person.surname }}
> {% empty %}
> Nobody works there.
> {% endfor %}
> {% empty %}
> No jobs.
> {% endfor %}
> 
> 
>
>
> And others you probably know how to deal with...
> File /urls.py:
>
> from django.contrib import admin
> from django.urls import path, include
>
> urlpatterns = [
> path('admin/', admin.site.urls),
> path('appname/', include('appname.urls'))
> ]
>
>
> *File /appname/urls.py:*
>
> from django.urls import path
> from appname import views
>
> urlpatterns = [
> path('jobslist/', views.JobsWithPersonsView.as_view(), name='jobslist')
> ]
>
>
> *BTW: *Your model names should be in singular form, if you want to be 
> able to use plural forms in some cases, use verbose_name_plural (docs 
> 
> ).
>
> W dniu czwartek, 13 września 2018 05:26:20 UTC+2 użytkownik René L. 
> Hechavarría napisał:
>>
>> Hello,
>>
>> I have a model, Jobs and Persons, Jobs are listed in ListView, i need to 
>> add a filter (django-filter), if i filter over field in Jobs all work good, 
>> but i need to filter also for Persons. this is and example
>>
>> @property
>> def get_persons(self):
>> return self.persons_all.all()
>>
>>
>> That allow to add persons to Jobs model in de html. How i filter over 
>> persons in Jobs view?
>>
>> Thanks in advance.
>>
>>
>>
>>

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Re: django-filter over models

2018-09-18 Thread Mateusz

Sorry, my previous answer was senseless, I realized that just after adding 
it.

1) You can filter jobs in appname/views.py:

queryset = Job.objects.filter(name__contains="job")


2) and people in template /appname/templates/jobs_list.html:

{% for job in object_list %}
{{ job.name }}
{% for person in job.person_set.all %}
{% if person.name|length <= 4 %}
{{ person.name }} {{ person.surname }}
{% endif %}
{% empty %}
Nobody works there.
{% endfor %}


I still find that 2) inefficient, but could do anything else. Read 
something about registering filters that I do not understand that well, 
sorry.

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Re: django-filter over models

2018-09-18 Thread Mateusz

In that answer I assume Jobs are connected anyhow with Persons with a 
ForeignKey field like so:

*File /appname/models.py:*

from django.db import models


class Job(models.Model):
name = models.CharField(max_length=100)

def __str__(self):
return self.name


class Person(models.Model):
name = models.CharField(max_length=100)
surname = models.CharField(max_length=100)

job = models.ForeignKey(Job, on_delete=models.SET_NULL, null=True, 
default=None)

def __str__(self):
return self.name + ' ' + self.surname


*File /appname/views.py:*

from django.shortcuts import render
from django.views.generic.list import ListView

from appname.models import Job, Person


class JobsWithPersonsView(ListView):
queryset = Job.objects.all()
template_name = 'jobs_list.html'



*File /appname/templates/jobs_list.html:*





Jobs


{% for job in object_list %}
{{ job.name }}
{% for person in job.person_set.all %}
{{ person.name }} {{ person.surname }}
{% empty %}
Nobody works there.
{% endfor %}
{% empty %}
No jobs.
{% endfor %}




And others you probably know how to deal with...
File /urls.py:

from django.contrib import admin
from django.urls import path, include

urlpatterns = [
path('admin/', admin.site.urls),
path('appname/', include('appname.urls'))
]


*File /appname/urls.py:*

from django.urls import path
from appname import views

urlpatterns = [
path('jobslist/', views.JobsWithPersonsView.as_view(), name='jobslist')
]


*BTW: *Your model names should be in singular form, if you want to be able 
to use plural forms in some cases, use verbose_name_plural (docs 

).

W dniu czwartek, 13 września 2018 05:26:20 UTC+2 użytkownik René L. 
Hechavarría napisał:
>
> Hello,
>
> I have a model, Jobs and Persons, Jobs are listed in ListView, i need to 
> add a filter (django-filter), if i filter over field in Jobs all work good, 
> but i need to filter also for Persons. this is and example
>
> @property
> def get_persons(self):
> return self.persons_all.all()
>
>
> That allow to add persons to Jobs model in de html. How i filter over 
> persons in Jobs view?
>
> Thanks in advance.
>
>
>
>

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Re: django filter

2017-08-29 Thread Muhammad M

Please, post your related code (from your models, views and template) here so 
that we can take a look.

Thank you.


> On Aug 29, 2017, at 1:59 AM, sum abiut  > wrote:
> 
> I did that but the approved leave is still showing on the table.
> 
> Cheers
> 
> On Tue, Aug 29, 2017 at 2:44 PM, sum abiut  > wrote:
> Thanks heaps Muhammad.
> Sum
> 
> On Tue, Aug 29, 2017 at 10:13 AM, Muhammad M  > wrote:
> Hi Sum,
> 
> Add an "approved" field of type BooleanField ( ) to your Leave model. 
> 
> So, something like this should work: 
> 
> In app/models.py:
> 
> class Leave (models.Model):
> #other fields go here...
> approved = models.BooleanField ( )
> 
> In app/views.py, handle the query like this:
>  unapproved_leaves = Leave.objects.filter (approved=False)
> 
> Then, do with the returned queryset as you wish. 
> 
> Best of luck.
> 
> Sincerely,
> Muhammad 
> 
> On Aug 28, 2017 7:06 PM, "sum abiut"  > wrote:
> Hi all,
> i am working on an eleave system where staff apply for annual leave and their 
> leave manager approved the leave online.
> 
> currently i have a table that shows all the leave for each department. what i 
> want to accomplished is to only show the leave that are not yet approved on 
> the table. Any leave that have been approved should not be showing on the 
> table.
> 
> Please point me to the right direction.
> 
> 
> Cheers
> 
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>  
> .
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> 
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> 
> 
> 
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Re: django filter

2017-08-29 Thread sum abiut

I did that but the approved leave is still showing on the table.

Cheers

On Tue, Aug 29, 2017 at 2:44 PM, sum abiut  wrote:

> Thanks heaps Muhammad.
> Sum
>
> On Tue, Aug 29, 2017 at 10:13 AM, Muhammad M  wrote:
>
>> Hi Sum,
>>
>> Add an "approved" field of type BooleanField ( ) to your Leave model.
>>
>> So, something like this should work:
>>
>> In app/models.py:
>>
>> class Leave (models.Model):
>> #other fields go here...
>> approved = models.BooleanField ( )
>>
>> In app/views.py, handle the query like this:
>>  unapproved_leaves = Leave.objects.filter (approved=False)
>>
>> Then, do with the returned queryset as you wish.
>>
>> Best of luck.
>>
>> Sincerely,
>> Muhammad
>>
>> On Aug 28, 2017 7:06 PM, "sum abiut"  wrote:
>>
>>> Hi all,
>>> i am working on an eleave system where staff apply for annual leave and
>>> their leave manager approved the leave online.
>>>
>>> currently i have a table that shows all the leave for each department.
>>> what i want to accomplished is to only show the leave that are not yet
>>> approved on the table. Any leave that have been approved should not be
>>> showing on the table.
>>>
>>> Please point me to the right direction.
>>>
>>>
>>> Cheers
>>>
>>> --
>>> You received this message because you are subscribed to the Google
>>> Groups "Django users" group.
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>>> an email to django-users+unsubscr...@googlegroups.com.
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>>> gid/django-users/CAPCf-y6f6%3DWtOW%3DQNW%3DD59gWtAmc3avZ4s3j
>>> HA09Pgj6C4yXEQ%40mail.gmail.com
>>> 
>>> .
>>> For more options, visit https://groups.google.com/d/optout.
>>>
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>> TNHnoF1f7A%40mail.gmail.com
>> 
>> .
>> For more options, visit https://groups.google.com/d/optout.
>>
>
>
>
> --
>



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Re: django filter

2017-08-28 Thread sum abiut

Thanks heaps Muhammad.
Sum

On Tue, Aug 29, 2017 at 10:13 AM, Muhammad M  wrote:

> Hi Sum,
>
> Add an "approved" field of type BooleanField ( ) to your Leave model.
>
> So, something like this should work:
>
> In app/models.py:
>
> class Leave (models.Model):
> #other fields go here...
> approved = models.BooleanField ( )
>
> In app/views.py, handle the query like this:
>  unapproved_leaves = Leave.objects.filter (approved=False)
>
> Then, do with the returned queryset as you wish.
>
> Best of luck.
>
> Sincerely,
> Muhammad
>
> On Aug 28, 2017 7:06 PM, "sum abiut"  wrote:
>
>> Hi all,
>> i am working on an eleave system where staff apply for annual leave and
>> their leave manager approved the leave online.
>>
>> currently i have a table that shows all the leave for each department.
>> what i want to accomplished is to only show the leave that are not yet
>> approved on the table. Any leave that have been approved should not be
>> showing on the table.
>>
>> Please point me to the right direction.
>>
>>
>> Cheers
>>
>> --
>> 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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>> Visit this group at https://groups.google.com/group/django-users.
>> To view this discussion on the web visit https://groups.google.com/d/ms
>> gid/django-users/CAPCf-y6f6%3DWtOW%3DQNW%3DD59gWtAmc3avZ4s3j
>> HA09Pgj6C4yXEQ%40mail.gmail.com
>> 
>> .
>> For more options, visit https://groups.google.com/d/optout.
>>
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> .
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>



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Re: django filter

2017-08-28 Thread Muhammad M

Hi Sum,

Add an "approved" field of type BooleanField ( ) to your Leave model.

So, something like this should work:

In app/models.py:

class Leave (models.Model):
#other fields go here...
approved = models.BooleanField ( )

In app/views.py, handle the query like this:
 unapproved_leaves = Leave.objects.filter (approved=False)

Then, do with the returned queryset as you wish.

Best of luck.

Sincerely,
Muhammad

On Aug 28, 2017 7:06 PM, "sum abiut"  wrote:

> Hi all,
> i am working on an eleave system where staff apply for annual leave and
> their leave manager approved the leave online.
>
> currently i have a table that shows all the leave for each department.
> what i want to accomplished is to only show the leave that are not yet
> approved on the table. Any leave that have been approved should not be
> showing on the table.
>
> Please point me to the right direction.
>
>
> Cheers
>
> --
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> "Django users" group.
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> msgid/django-users/CAPCf-y6f6%3DWtOW%3DQNW%3DD59gWtAmc3avZ4s3jHA09Pgj6C4y
> XEQ%40mail.gmail.com
> 
> .
> For more options, visit https://groups.google.com/d/optout.
>

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Re: Django Filter

2016-06-17 Thread 'David Turner' via Django users

The issue is that one of the fields use choices. If I remove this
everything is fine!

On 17 June 2016 at 16:58, 'David Turner' via Django users <
django-users@googlegroups.com> wrote:

> I am using django-filter
>
> I have create the the filter as follows:
> filters.py
> import django_filters
> from .models import Job
>
>
> class JobFilter(django_filters.FilterSet):
> class Meta:
> model = Job
> fields = ['practice_type']
>
> And I have created the following view:
> from .models import Job
> from .filters import JobFilter
>
>
> def jobs_list(request):
> filter = JobFilter(request.GET, queryset=Job.objects.all())
> return render(request, 'jobs/jobs_list.html', {'filter': filter})
>
> job_list.html
> And then a templater as follows:
> {% block content %}
> 
> {{ filter.form.as_p }}
> 
> 
> {% for job in filter %}
> {{ job.name }} 
> {% endfor %}
> {% endblock %}
>
> Everything is fine and the filter shows on the relevant page
>
> However on pressing submit no results are returned.
>
> I know this is probably something quite simple but I have been staring at
> this for so long now any help would be appreciated.
>
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Re: Django filter hack

2006-03-12 Thread Adrian Holovaty


On 3/8/06, Andu <[EMAIL PROTECTED]> wrote:
> One of the problems I faced when using Django's admin was the filter
> part overlapped the table with data, when I showed lots of fields in
> the admin listing.
>
> I did a quick hack today and made it toggable. If you want to aply a
> filter, show it. If not, leave it there and browse the listings.

Nice improvement! Thanks for sharing.

If we end up rolling this into the framework, we'll opt for the
JavaScript one-liner rather than the overhead of a JavaScript
framework for effects. (What effects does it do, anyway? I'm not in a
position to apply the patch to my local copy.)

Adrian

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Re: Django filter hack

2006-03-08 Thread tonemcd


This is very nice - very helpful!

Cheers,
Tone


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Re: Django filter hack

2006-03-08 Thread oggie rob


> I did a quick hack today and made it toggable. If you want to aply a
filter, show it. If not, leave it there and browse the listings.

This is nice - can you write up a bug?

 -rob


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Re: Django filter hack

2006-03-08 Thread limodou


On 3/8/06, Andu <[EMAIL PROTECTED]> wrote:
>
> Hi all,
>
> One of the problems I faced when using Django's admin was the filter
> part overlapped the table with data, when I showed lots of fields in
> the admin listing.
>
> I did a quick hack today and made it toggable. If you want to aply a
> filter, show it. If not, leave it there and browse the listings.
>
> I documented it here, and you can also see some images on how it looks:
> http://andu.417.ro/blog/django_filter_hack.html
>

I'v seen that already. It's good.

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