Re: Django phone number validation

[Andréas Kühne]

Hi,

https://github.com/stefanfoulis/django-phonenumber-field

You could probably use that library - it works against googles phonenumber
library, which is the library that is used on andoid phones as well.

Regards,

Andréas

2017-06-21 8:18 GMT+02:00 Santosh Yelamarthi :

> Hi All,
>
> Can anyone please share how to validate phone number in django.
>
> Thanks in advance.
>
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User Based role in Django

[Mannu Gupta]

I want to make a role based user in Django like the following :- 

User with Subscription level 0 will have some feature and some basic fields 
like first name , last name etc. 
User with Subscription level 1 will have some more feature and some extra 
fields  like tax-id , national id , nationality, is_disable etc . 
User with Subscription level 2 will have more features with all the fields 
with the fields with same as fields of Subscription level 1 since he has 
already subscribed only he is upgrading.
User with Subscription level 3 will have more feature with all the fields 
with the fields same above .

How do i make the user model and related this relate to its role , and how 
do i be able to set permission for different user ?

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Re: User Based role in Django

[Jani Tiainen]

Hi,

We have been using user groups for that. And then we check and filter out
based on given group(s).

21.6.2017 10.45 "Mannu Gupta"  kirjoitti:

> I want to make a role based user in Django like the following :-
>
> User with Subscription level 0 will have some feature and some basic
> fields like first name , last name etc.
> User with Subscription level 1 will have some more feature and some extra
> fields  like tax-id , national id , nationality, is_disable etc .
> User with Subscription level 2 will have more features with all the fields
> with the fields with same as fields of Subscription level 1 since he has
> already subscribed only he is upgrading.
> User with Subscription level 3 will have more feature with all the fields
> with the fields same above .
>
> How do i make the user model and related this relate to its role , and how
> do i be able to set permission for different user ?
>
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> .
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Re: User Based role in Django

[Mannu Gupta]

Hi ,
Will I be able to trace if user upgrde the subscription level ?

Manni

On Wed, Jun 21, 2017 at 1:41 PM, Jani Tiainen  wrote:

> Hi,
>
> We have been using user groups for that. And then we check and filter out
> based on given group(s).
>
> 21.6.2017 10.45 "Mannu Gupta"  kirjoitti:
>
>> I want to make a role based user in Django like the following :-
>>
>> User with Subscription level 0 will have some feature and some basic
>> fields like first name , last name etc.
>> User with Subscription level 1 will have some more feature and some extra
>> fields  like tax-id , national id , nationality, is_disable etc .
>> User with Subscription level 2 will have more features with all the
>> fields with the fields with same as fields of Subscription level 1 since he
>> has already subscribed only he is upgrading.
>> User with Subscription level 3 will have more feature with all the fields
>> with the fields same above .
>>
>> How do i make the user model and related this relate to its role , and
>> how do i be able to set permission for different user ?
>>
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>> 
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Re: User Based role in Django

[Jani Tiainen]

Well its your code, so yes if you write code that does it.

It doesn't happen magically for you.

21.6.2017 11.15 "Mannu Gupta"  kirjoitti:

> Hi ,
> Will I be able to trace if user upgrde the subscription level ?
>
> Manni
>
> On Wed, Jun 21, 2017 at 1:41 PM, Jani Tiainen  wrote:
>
>> Hi,
>>
>> We have been using user groups for that. And then we check and filter out
>> based on given group(s).
>>
>> 21.6.2017 10.45 "Mannu Gupta"  kirjoitti:
>>
>>> I want to make a role based user in Django like the following :-
>>>
>>> User with Subscription level 0 will have some feature and some basic
>>> fields like first name , last name etc.
>>> User with Subscription level 1 will have some more feature and some
>>> extra fields  like tax-id , national id , nationality, is_disable etc .
>>> User with Subscription level 2 will have more features with all the
>>> fields with the fields with same as fields of Subscription level 1 since he
>>> has already subscribed only he is upgrading.
>>> User with Subscription level 3 will have more feature with all the
>>> fields with the fields same above .
>>>
>>> How do i make the user model and related this relate to its role , and
>>> how do i be able to set permission for different user ?
>>>
>>> --
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>>> Groups "Django users" group.
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>>> gid/django-users/4bdc5c54-5c90-4515-b977-30fcb612b3ff%40googlegroups.com
>>> 
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>
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Re: User Based role in Django

[Mannu Gupta]

I meant that using your method i.e by making different groups .
When he upgrade the group .
How about making another table having user, his subscription level and
changed_at field(which traces when he /upgrade the group )

Just want to know your suggestion .

Mannu

On Jun 21, 2017 1:48 PM, "Jani Tiainen"  wrote:

> Well its your code, so yes if you write code that does it.
>
> It doesn't happen magically for you.
>
> 21.6.2017 11.15 "Mannu Gupta"  kirjoitti:
>
>> Hi ,
>> Will I be able to trace if user upgrde the subscription level ?
>>
>> Manni
>>
>> On Wed, Jun 21, 2017 at 1:41 PM, Jani Tiainen  wrote:
>>
>>> Hi,
>>>
>>> We have been using user groups for that. And then we check and filter
>>> out based on given group(s).
>>>
>>> 21.6.2017 10.45 "Mannu Gupta"  kirjoitti:
>>>
 I want to make a role based user in Django like the following :-

 User with Subscription level 0 will have some feature and some basic
 fields like first name , last name etc.
 User with Subscription level 1 will have some more feature and some
 extra fields  like tax-id , national id , nationality, is_disable etc .
 User with Subscription level 2 will have more features with all the
 fields with the fields with same as fields of Subscription level 1 since he
 has already subscribed only he is upgrading.
 User with Subscription level 3 will have more feature with all the
 fields with the fields same above .

 How do i make the user model and related this relate to its role , and
 how do i be able to set permission for different user ?

 --
 You received this message because you are subscribed to the Google
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 gid/django-users/4bdc5c54-5c90-4515-b977-30fcb612b3ff%40goog
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 .
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Re: url patterns question

[Melvyn Sopacua]

On Tuesday 20 June 2017 18:08:02 James Schneider wrote:

> > And his problem is that it does *not* match. Not that it does.
> 
> And for that I think my statement still stands that Django is
> stripping the last portion of the URL as a GET argument. I'm betting
> that
> requests.GET.get('abc') will return '12/' per the last example from
> the OP.

Yes, but for the first problem, r'^blank/.*$' should match blank/abc and 
generates a 
404. I think the matching isn't the problem and the 404 is caused either by 
ordering 
issue in urlpatterns (first match wins, no others are tried) or with the view 
itself 
(get_object_or_404).

You are correct that urlpatterns are only matched against request.path. This 
excludes the query string.

-- 
Melvyn Sopacua

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Django Import/Export ManyTo Many

['dtdave' via Django users]

I have a model with a manytomany field as follows:
  
  contact = ChainedManyToManyField(
'contacts.Contact',
chained_field="practice",
chained_model_field="practice",
)

Within my admin I have the following for import/export:

contact = fields.Field(column_name='contact name', attribute='contact',
   widget=ManyToManyWidget(Contact, 'contact'))

However, this only returns the contact id.
I am stuck on how I could return the name of the contact instead. Within 
the contact model contact equates to the name of the contact.

Any assistance would be appreciated and apologies in advance if I have 
overlooked the obvious!

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Re: User Based role in Django

[Jani Tiainen]

Hi,

User can't upgrade his group. It is your code that does it. And it is your
code responsibility to track changes etc.

User just may have a button on ui that invokes your code (view) but rest is
really up to you.

21.6.2017 11.24 "Mannu Gupta"  kirjoitti:

> I meant that using your method i.e by making different groups .
> When he upgrade the group .
> How about making another table having user, his subscription level and
> changed_at field(which traces when he /upgrade the group )
>
> Just want to know your suggestion .
>
> Mannu
>
> On Jun 21, 2017 1:48 PM, "Jani Tiainen"  wrote:
>
>> Well its your code, so yes if you write code that does it.
>>
>> It doesn't happen magically for you.
>>
>> 21.6.2017 11.15 "Mannu Gupta"  kirjoitti:
>>
>>> Hi ,
>>> Will I be able to trace if user upgrde the subscription level ?
>>>
>>> Manni
>>>
>>> On Wed, Jun 21, 2017 at 1:41 PM, Jani Tiainen  wrote:
>>>
 Hi,

 We have been using user groups for that. And then we check and filter
 out based on given group(s).

 21.6.2017 10.45 "Mannu Gupta"  kirjoitti:

> I want to make a role based user in Django like the following :-
>
> User with Subscription level 0 will have some feature and some basic
> fields like first name , last name etc.
> User with Subscription level 1 will have some more feature and some
> extra fields  like tax-id , national id , nationality, is_disable etc .
> User with Subscription level 2 will have more features with all the
> fields with the fields with same as fields of Subscription level 1 since 
> he
> has already subscribed only he is upgrading.
> User with Subscription level 3 will have more feature with all the
> fields with the fields same above .
>
> How do i make the user model and related this relate to its role , and
> how do i be able to set permission for different user ?
>
> --
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> 
> .
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 .

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Re: Non-primary auto-incrementing field with Postgres

[Melvyn Sopacua]

On Monday 19 June 2017 16:29:48 Thomas Hauk wrote:
> I am working on a project that uses Django 1.10.5 with Postgres 9.6
> (and Python 3.6.1).
> 
> I am currently migrating historical data from an old system into the
> new system. This historical data has a table with a (non-primary key)
> "ID" column. I would like to migrate these rows into the new
> database, and have that column be auto-incrementing (starting from,
> say, 10) for new rows inserted.
> 
> I think the way to do this with raw Postgres SQL would be to create a
> serial field, and then run a SQL command like "SELECT
> SETVAL('some_table_field_id_seq', 10)". I think.  :)
> 
> Does Django expose functionality that lets me accomplish this?

First and foremost, verify if ID meets your requirements:
- All values are unique
- They are integers
- Their values are smaller then your new cut-off (10)

Then you can simply transform the field explicitly to primary key and then to 
BigAutoField. 
Setting the next value can be done using RunSQL[1].

-- 
Melvyn Sopacua


[1] https://docs.djangoproject.com/en/1.11/ref/migration-operations/#runsql

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Re: import different entries using filter

[Derek]

Your clue is in the word "variable" that you used.  The "variables" for a 
Django model are its fields; so what you are looking for is a value stored 
in your model's field.

e.g. 

service = 
food.objects.get(restaurant-name__icontains='burger-place').values_list('type_of_service',
 
flat=True)

This assumes that your Django "Food" model has a  field called 
'type_of_service'.

Another way to do this would be to first get the restaurant object:

place = food.objects.get(restaurant-name__icontains='burger-place')

Then you can more get to the fields that you need quite easily:

print(place.type_of_service)
print(place.product)

But you may need to think about your database design to ensure the right 
data is being stored in the right models/fields e.g.

https://stackoverflow.com/questions/32366294/database-design-for-food-ordering-system-mysql
http://www.wellho.net/resources/ex.php4?item=s154/sql


On Tuesday, 20 June 2017 06:11:23 UTC+2, jon stan wrote:
>
> hey im trying to import one set of info from a database based on the name 
> of something else if that makes sense.
>
> basically in the database its setup like this:
>
>food-
>   restaurant-name   products type of service
>   
>  (burger-place)(burgers)   (fast food)
>---
>
> im trying to display the products and service type based on the name but i 
> cant figure out how todo that. here's what i have:
>
>  views.py
>   name = 
> food.objects.get(restaurant-name__icontains='burger-place')
>   prod = 
> food.objects.filter(restaurant-name__icontains='burger-place').filter(products__icontains='burgers')
>   service = 
> food.objects.filter(restaurant-name__icontains='burger-place').filter(service__icontains='fast
>  
> food')
>
> but i cant get it to work correctly. i get 'burger-place' for the name, 
> prod, and service variables for some reason. would i have to create 
> completely separate models or is there a way i can get these variables 
> based on the restaurant name?
>

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Re: Non-primary auto-incrementing field with Postgres

[thauk]

On Tuesday, June 20, 2017 at 10:40:05 PM UTC-7, Scot Hacker wrote:
>
> One strategy might be to:
>
> 1) Bring in the data under a different column name ('old_id' ?)
> 2) In a single migration, drop the default ID column,  rename old_id to 
> id, and give it primary_key=True
>

I can't change the primary key for this table, unfortunately.

T
 

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django_rq or django settings issue ?

[bonet]

Hi all,

I'm facing a strange behavior when trying to create Django_rq queues 
configuration dynamically from an environment variable.

my env var is :

CUSTOM_QUEUES=default:q1:q2

into my settings.py I have:

from getenv import env



# Cache
CACHES_DEFAULT = "redis://127.0.0.1:6379/1"
CACHES = {
"default": {
"BACKEND": "django_redis.cache.RedisCache",
"LOCATION": env("CACHE_URL", CACHES_DEFAULT),
"OPTIONS": {
"CLIENT_CLASS": "django_redis.client.DefaultClient",
},
'TIMEOUT': 3600
},
}


RQ_QUEUES_REDIS = {'USE_REDIS_CACHE': 'default'}
RQ_QUEUES = {k:RQ_QUEUES_REDIS for k in env('CUSTOM_QUEUES', "default").
split(':')}



I'm expecting the denerated RQ_QUEUES dict to be like this one which is 
working :

RQ_QUEUES = {
'default': {
'USE_REDIS_CACHE': 'default',
},
'q1': {
'USE_REDIS_CACHE': 'default',
},
'q2': {
'USE_REDIS_CACHE': 'default',
},
}


even though the configuration seems to run well and I can see the queues 
from django_rq web page, my workers cannot connect and actually throw this 
error as like there's no queue key in redis


Traceback (most recent call last):
  File "manage.py", line 10, in 
execute_from_command_line(sys.argv)
  File 
"/home/work/virtualenv/runaway_dev/lib/python3.6/site-packages/django/core/management/__init__.py"
, line 350, in execute_from_command_line
utility.execute()
  File 
"/home/work/virtualenv/runaway_dev/lib/python3.6/site-packages/django/core/management/__init__.py"
, line 342, in execute
self.fetch_command(subcommand).run_from_argv(self.argv)
  File 
"/home/work/virtualenv/runaway_dev/lib/python3.6/site-packages/django/core/management/base.py"
, line 348, in run_from_argv
self.execute(*args, **cmd_options)
  File 
"/home/work/virtualenv/runaway_dev/lib/python3.6/site-packages/django/core/management/base.py"
, line 399, in execute
output = self.handle(*args, **options)
  File 
"/home/work/virtualenv/runaway_dev/lib/python3.6/site-packages/django_rq/management/commands/rqworker.py"
, line 79, in handle
queues = get_queues(*args)
  File 
"/home/work/virtualenv/runaway_dev/lib/python3.6/site-packages/django_rq/queues.py"
, line 166, in get_queues
queue_params = QUEUES[queue_names[0]]
KeyError: 'default'

By The Way: if I use the Static configuration above it just works well, so 
my suspect is more focused on Django_rq not creating the queues but maybe 
I'm also doing something unproper on the settings of Django. 

Any help would be really appreciated. thanks.

F.

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Re: Raw SQL - How to approach paging and lazy fetching

[Kevin Yu]

Thanks Jani. I was able to solve my original issue using your suggestion.

On Friday, June 16, 2017 at 12:54:26 PM UTC-7, Jani Tiainen wrote:
>
> Yes, even there isn't real foreign keys in  the database you can still 
> join models.
>
> Have been doing that few times with legacy database. When defining fkey 
> you need to point it to corresponding field in fkey attributes in your 
> model if it isnt your target model pk.
>
> And you can join nonmanged models with managed just fine.
>
> 16.6.2017 19.26 "Kevin Yu" > kirjoitti:
>
> Hi Jani,
>
> Thanks for the link. I was aware of this link and actually followed the 
> steps when I started the project. I tried using model but then found out 
> when i need to query a join, it doesn't seem to work unless there's a 
> foreign key between the two models, however, given it's a legacy database, 
> i can't touch the schema at all. Is it possible to query join results with 
> managed=False?
>
> Thanks
>
> On Thursday, June 15, 2017 at 11:31:42 PM UTC-7, Jani Tiainen wrote:
>
>> Hi,
>>
>> Even you do have legacy database, you can use unmanaged models and then 
>> leverage full power of ORM [1].
>>
>> Just create models and in their Meta set managed = False and there you 
>> go. 
>>
>> And what comes to paging - you need to somehow pass offset and limit to 
>> your query.
>>
>> [1] https://docs.djangoproject.com/en/1.11/howto/legacy-databases/
>>
>> On 16.06.2017 02:05, Kevin Yu wrote:
>>
>> Hi All, 
>>
>> I am using raw sql to connect to database. The reason we used raw sql 
>> instead of the Django model is because the database is legacy and is being 
>> shared by multiple applications...
>>
>> I have one use case that I'm struggling right now. Basically I have a 
>> page that fetch more than 1000 results. My query is like this:
>>
>> cursor = connection.cursor()
>> cursor.execute('''
>> SELECT br.id, br.name, br.created_at, br.updated_at,
>> br.branchpoint_str, br.source
>> FROM branches as br
>> LEFT JOIN branches_projects as bp
>> ON br.id = bp.branch_id 
>> WHERE bp.project_id = "%s" AND source != "other"
>> ORDER BY updated_at DESC
>>''', [int(project_id)]
>>)
>>
>> Then in my template, I have this:
>> {% for br in special_branches %}
>>
>>   > link="/files/{{build.image_path}}/build{{build.build_number}}/">
>> {{br.name}}
>> {{br.branchpoint}}
>> {{ br.source|upper }}
>> {{br.updated_at}}
>> {{br.built_at}}
>>   
>> {% endfor %}
>>
>> The current problem is this would create a very long page... I am 
>> wondering how to approach this problem so that I can have different page on 
>> the template, and say 100 result per page, when i click the second page, 
>> then  django will fetch result 100-200.
>>
>> Thanks!
>> -- 
>> 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...@googlegroups.com.
>> To post to this group, send email to django...@googlegroups.com.
>>
>> Visit this group at https://groups.google.com/group/django-users.
>> To view this discussion on the web visit 
>> https://groups.google.com/d/msgid/django-users/f6a3df97-8218-4fb8-b200-f4535797e135%40googlegroups.com
>>  
>> 
>> .
>> For more options, visit https://groups.google.com/d/optout.
>>
>>
>> -- 
>> Jani Tiainen
>>
>> -- 
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>  
> 
> .
>
> For more options, visit https://groups.google.com/d/optout.
>
>
>

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Need help for defining foreign key based on value from another table

[Kevin Yu]

I'm working with a legacy database so I have to set managed=False in the 
model. Here's the 3 related tables:


class Branches(models.Model):
name = models.CharField(max_length=128)
branchpoint_str = models.CharField(max_length=255)
dev_lead_id = models.IntegerField(blank=True, null=True)
source = models.CharField(max_length=255)
state = models.CharField(max_length=255)
kind = models.CharField(max_length=255)
desc = models.TextField(blank=True, null=True)
approved = models.IntegerField()
for_customer = models.IntegerField()
deactivated_at = models.DateTimeField(blank=True, null=True)
created_at = models.DateTimeField(blank=True, null=True)
updated_at = models.DateTimeField(blank=True, null=True)
codb_id = models.IntegerField(blank=True, null=True)
pm_lead_id = models.IntegerField(blank=True, null=True)
version = models.CharField(max_length=20, blank=True, null=True)
path_id = models.IntegerField(blank=True, null=True)
branchpoint_type = models.CharField(max_length=255, blank=True, 
null=True)
branchpoint_id = models.IntegerField(blank=True, null=True)

class Meta:
managed = False
db_table = 'branches'
verbose_name_plural = 'Branches'

class Projects(models.Model):
id = models.AutoField(primary_key=True)
name = models.CharField(max_length=40, primary_key=True)
status = models.CharField(max_length=255)
platform = models.CharField(max_length=255)
enabled = models.IntegerField()
path = models.CharField(max_length=128, blank=True, null=True)
tag_prefix = models.CharField(max_length=64, blank=True, null=True)
created_at = models.DateTimeField(blank=True, null=True)
updated_at = models.DateTimeField(blank=True, null=True)
codb_id = models.IntegerField(blank=True, null=True)
template = models.CharField(max_length=64, blank=True, null=True)
image_path = models.CharField(max_length=128, blank=True, null=True)
repository_id = models.IntegerField(blank=True, null=True)
number_scheme = models.CharField(max_length=32)
special_dir = models.CharField(max_length=32, blank=True, null=True)
project_family_id = models.IntegerField()
class Meta:
managed = False
db_table = 'projects'
verbose_name_plural = 'projects'

class BranchesProjects(models.Model):
# project_id = models.IntegerField()
# branch_id = models.IntegerField()
project = models.ForeignKey(Projects, on_delete=models.CASCADE)
branch = models.ForeignKey(Branches, on_delete=models.CASCADE)

class Meta:
managed = False
db_table = 'branches_projects'

I have been able to do the join using raw(). However, the return object is 
rawqueryset. What I want is queryset so that I can use django-filter to 
process it. My current raw sql is like this:
Branches.objects.raw(
'''SELECT br.id, br.name, br.created_at, br.updated_at,
br.branchpoint_str, br.source
FROM branches as br
LEFT JOIN branches_projects as bp
ON br.id = bp.branch_id 
WHERE bp.project_id = "%s" AND source != "other"
ORDER BY updated_at DES'''

 
My question is, is there a way to achieve the same result using Django's 
queryset? I've also explored  the idea of using the extra() in django extra 

 but 
it doesn't really work for me.

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Re: Need help for defining foreign key based on value from another table

[Kevin Yu]

I just found out on Django document, there's an example that's for 
manytomany, using 'through' when defining the foreign key. I think this is 
what i need.

Here's the example:

from django.db import models

class Person(models.Model):
name = models.CharField(max_length=128)

def __str__(self):  # __unicode__ on Python 2
return self.name

class Group(models.Model):
name = models.CharField(max_length=128)
members = models.ManyToManyField(Person, through='Membership')

def __str__(self):  # __unicode__ on Python 2
return self.name

class Membership(models.Model):
person = models.ForeignKey(Person, on_delete=models.CASCADE)
group = models.ForeignKey(Group, on_delete=models.CASCADE)
date_joined = models.DateField()
invite_reason = models.CharField(max_length=64)




On Wednesday, June 21, 2017 at 2:49:40 PM UTC-7, Kevin Yu wrote:
>
> I'm working with a legacy database so I have to set managed=False in the 
> model. Here's the 3 related tables:
>
>
> class Branches(models.Model):
> name = models.CharField(max_length=128)
> branchpoint_str = models.CharField(max_length=255)
> dev_lead_id = models.IntegerField(blank=True, null=True)
> source = models.CharField(max_length=255)
> state = models.CharField(max_length=255)
> kind = models.CharField(max_length=255)
> desc = models.TextField(blank=True, null=True)
> approved = models.IntegerField()
> for_customer = models.IntegerField()
> deactivated_at = models.DateTimeField(blank=True, null=True)
> created_at = models.DateTimeField(blank=True, null=True)
> updated_at = models.DateTimeField(blank=True, null=True)
> codb_id = models.IntegerField(blank=True, null=True)
> pm_lead_id = models.IntegerField(blank=True, null=True)
> version = models.CharField(max_length=20, blank=True, null=True)
> path_id = models.IntegerField(blank=True, null=True)
> branchpoint_type = models.CharField(max_length=255, blank=True, 
> null=True)
> branchpoint_id = models.IntegerField(blank=True, null=True)
>
> class Meta:
> managed = False
> db_table = 'branches'
> verbose_name_plural = 'Branches'
>
> class Projects(models.Model):
> id = models.AutoField(primary_key=True)
> name = models.CharField(max_length=40, primary_key=True)
> status = models.CharField(max_length=255)
> platform = models.CharField(max_length=255)
> enabled = models.IntegerField()
> path = models.CharField(max_length=128, blank=True, null=True)
> tag_prefix = models.CharField(max_length=64, blank=True, null=True)
> created_at = models.DateTimeField(blank=True, null=True)
> updated_at = models.DateTimeField(blank=True, null=True)
> codb_id = models.IntegerField(blank=True, null=True)
> template = models.CharField(max_length=64, blank=True, null=True)
> image_path = models.CharField(max_length=128, blank=True, null=True)
> repository_id = models.IntegerField(blank=True, null=True)
> number_scheme = models.CharField(max_length=32)
> special_dir = models.CharField(max_length=32, blank=True, null=True)
> project_family_id = models.IntegerField()
> class Meta:
> managed = False
> db_table = 'projects'
> verbose_name_plural = 'projects'
>
> class BranchesProjects(models.Model):
> # project_id = models.IntegerField()
> # branch_id = models.IntegerField()
> project = models.ForeignKey(Projects, on_delete=models.CASCADE)
> branch = models.ForeignKey(Branches, on_delete=models.CASCADE)
>
> class Meta:
> managed = False
> db_table = 'branches_projects'
>
> I have been able to do the join using raw(). However, the return object is 
> rawqueryset. What I want is queryset so that I can use django-filter to 
> process it. My current raw sql is like this:
> Branches.objects.raw(
> '''SELECT br.id, br.name, br.created_at, br.updated_at,
> br.branchpoint_str, br.source
> FROM branches as br
> LEFT JOIN branches_projects as bp
> ON br.id = bp.branch_id 
> WHERE bp.project_id = "%s" AND source != "other"
> ORDER BY updated_at DES'''
>
>  
> My question is, is there a way to achieve the same result using Django's 
> queryset? I've also explored  the idea of using the extra() in django 
> extra 
> 
>  but 
> it doesn't really work for me.
>
>

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Re: Abridged summary of django-users@googlegroups.com - 23 updates in 11 topics

[Ranjith Kumar]

Hello All,
On Django admin when I tried to update the existing data I’m getting an error,

AttributeError: 'str' object has no attribute '__iter__'


models.py

class SSPDrive(models.Model):

drive_year  = 
models.DateField(verbose_name='Drive year', null=True, blank=True, 
help_text="Enter drive year.")

class Meta:
verbose_name= "SSP Drive"
verbose_name_plural = "SSP Drives"

def __str__(self):
return 
str('SSP'+str(self.drive_year.year)[-2:]+'T'+str(self.drive_year.year+1)[-2:])

class SSPDriveAdmin(admin.ModelAdmin):

save_as = True

class Meta:
model = SSPDrive

admin.site.register(SSPDrive, SSPDriveAdmin)

Can someone help me out, why am I getting this error…

> On 21-Jun-2017, at 6:37 PM, django-users@googlegroups.com wrote:
> 
> django-users@googlegroups.com 
> 
> Google Groups 
> 
>
> 
>
> Today's topic summary  <>
> View all topics 
> 
> import different entries using filter  - 2 Updates
> Non-primary auto-incrementing field with Postgres 
>  - 2 Updates
> User Based role in Django  - 6 Updates
> Django Import/Export ManyTo Many  - 1 Update
> url patterns question  - 3 Updates
> Django phone number validation  - 2 Updates
> If you have multiple custom users - what should you set AUTH_USER_MODEL to? 
>  - 1 Update
> [django] Excluir un ítem de un dropdown en el admin de django 
>  - 1 Update
> Django app to manage PHP-composer repositories  - 
> 2 Updates
> Django website  - 2 Updates
> [django-channels] Issues with model post_save and Group.Send() 
>  - 1 Update
>  <>import different entries using filter  
> 
>   
> Melvyn Sopacua : Jun 20 10:37PM +0200 
> 
> On Monday 19 June 2017 21:11:23 jon stan wrote:
>  
> > service__icontains='fast food')
>  
> > but i cant get it to work correctly. i get 'burger-place' for the
> > name, prod, and service variables for ...more 
> > 
> >  
> Derek : Jun 21 03:12AM -0700 
> 
> Your clue is in the word "variable" that you used. The "variables" for a 
> Django model are its fields; so what you are looking for is a value stored 
> in your model's field.
>  
> e.g. 
>  
> service = ...more 
> 
>  
> Back to top  <>Non-primary auto-incrementing field 
> with Postgres  
> 
>   
> Scot Hacker : Jun 20 10:40PM -0700 
> 
> One strategy might be to:
>  
> 1) Bring in the data under a different column name ('old_id' ?)
> 2) In a single migration, drop the default ID column, rename old_id to id, 
> and give it primary_key=True ...more 
> 
>  
> Melvyn Sopacua : Jun 21 11:43AM +0200 
> 
> On Monday 19 June 2017 16:29:48 Thomas Hauk wrote:
> > serial field, and then run a SQL command like "SELECT
> > SETVAL('some_table_field_id_seq', 10)". I think. :)
>  
> > Does Django expose ...more 
> > 
> >  
> Back to top  <>User Based role in Django  
> 
>   
> Mannu Gupta : Jun 21 12:45AM -0700 
> 
> I want to make a role based user in Django like the following :- 
>  
> User with Subscription level 0 will have some feature and some basic fields 
> like first name , last name etc. 
> ...more 
> 
>  
> Jani Tiainen : Jun 21 11:11AM +0300 
> 
> Hi,
>  
> We have been using user groups for that. And then we check and filter out
> based on given group(s).
>  
> ...more 
> 
>  
> Mannu Gupta : Jun 21 01:44PM +0530 
> 
> Hi ,
> Will I be able to trace if user upgrde the subscription level ?
>  
> Manni
>  
> ...more 
> 
>  
> Jani Tiainen : Jun 21 11:18AM +0300 
> 
> Well its your code, so yes if you write code that does it.
>  
> It doesn't happen magically for you.
>  
> ...more 
> 

Error running ML Code on Django

[A.S. Khangura]

I am trying to run a function from views.py which carries some
Machine-Learning code. While I am loading ML model using Joblib from
sklearn It throws error related to tfidf.

Here is traceback: http://dpaste.com/1FPN16G

-- 
Thanks
Arshpreet Singh
Python Developer
Web Development/Data Science/Systems Integration
https://www.linkedin.com/in/arsh840

Life is a journey, not a destination.

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