Hi!

You can only use a list of int in this case, because a *__in* option in
exclude expect a list of field type:

Try replace with *pk__in*, like this:

user_list = user_list.exclude(*pk__in*=user_exclude_list)

Cheers.


On Tue, Apr 15, 2014 at 9:04 AM, nobody <jupiter....@gmail.com> wrote:

> Thanks Lucas, but it is the same error:
>
> int() argument must be a string or a number, not 'list'
>
> The id__in is an int argument, but the user_list is a list, I don't see it 
> matchs. I tried to use *user_list, did not work either.
>
>
>
> On Friday, April 11, 2014 8:50:55 PM UTC+10, Lucas Klassmann wrote:
>
>> Hi,
>>
>> Sorry, you must specify the field *ID*, try this:
>>
>> for user in user_object_list:
>>     if user.is_locked():
>>         user_exclude_list.append(user.id)
>>
>> user_list = user_list.exclude(*id__in*=user_exclude_list)  # [1, 3, 4,
>> 5,]  IDs, is more fast
>>
>>
>> Cheers.
>>
>>
>>  On Fri, Apr 11, 2014 at 7:47 AM, nobody <jupit...@gmail.com> wrote:
>>
>>>  Hi Lucas,
>>>
>>>
>>> On Friday, April 11, 2014 12:07:02 PM UTC+10, Lucas Klassmann wrote:
>>>
>>>> Hi,
>>>>
>>>> You can use *ID* of object for compare with a list and *IN* :
>>>>
>>>> for user in user_object_list:
>>>>     if user.is_locked():
>>>>         user_exclude_list.append(user.*id*)
>>>>
>>>> user_list = user_list.exclude(*user__in*=*user_exclude_list*)  # [1,
>>>> 3, 4, 5,]  IDs, is more fast
>>>>
>>>
>>>  Hmm, I actually got an exception:
>>>
>>> Exception Type: TypeError  Exception Value:
>>>
>>> int() argument must be a string or a number, not 'list'
>>>
>>>
>>> Thanks.
>>>
>>>
>>>
>>>
>>>>
>>>> Cheers.
>>>>
>>>>
>>>> On Thu, Apr 10, 2014 at 9:23 PM, nobody <jupit...@gmail.com> wrote:
>>>>
>>>>>
>>>>> Thanks Lucas, sorry for not being clear, I like not to include the
>>>>> object when the name is matching, I figured out that can be done to use
>>>>> user_list.exclude, not user_list.filter. It is working currently in an
>>>>> inefficient way:
>>>>>
>>>>> for user in user_object_list:
>>>>>     if user.is_locked():
>>>>>         user_list = user_list.exclude(user__name=user.name)
>>>>>
>>>>> Is there anyway to filter out all users to match the user exclude list
>>>>> in one statement such as following example (wrong syntax I guess):
>>>>>
>>>>>
>>>>> for user in user_object_list:
>>>>>     if user.is_locked():
>>>>>         user_exclude_list.append(user.name)
>>>>>
>>>>> user_list = user_list.exclude(user__name.icontain=user_exclude_list)
>>>>>
>>>>>
>>>>> Thank you.
>>>>>
>>>>>
>>>>>
>>>>>
>>>>> On Friday, April 11, 2014 4:26:23 AM UTC+10, Lucas Klassmann wrote:
>>>>>
>>>>>> Hi,
>>>>>>
>>>>>> If you want exact match:
>>>>>>
>>>>>> user_list.filter(user__name__iexact=user.name)
>>>>>>
>>>>>>
>>>>>> Or containment test:
>>>>>> user_list.filter(user__name__icontains=user.name)
>>>>>>
>>>>>>
>>>>>> Note that "i" on start is for case-insensitive filtering.
>>>>>>
>>>>>> Look here for more information:
>>>>>> https://docs.djangoproject.com/en/dev/topics/db/queries/
>>>>>>
>>>>>>
>>>>>> Cheers
>>>>>>
>>>>>>
>>>>>> On Thu, Apr 10, 2014 at 2:38 PM, carlos <croch...@gmail.com> wrote:
>>>>>>
>>>>>>> hi, maybe you filter you queryset
>>>>>>> user_list = Model.objects.filter(user__name=user.name
>>>>>>> ).filter(other__field=user.email)
>>>>>>>
>>>>>>> Cheers
>>>>>>>
>>>>>>>
>>>>>>> On Thu, Apr 10, 2014 at 9:44 AM, nobody <jupit...@gmail.com> wrote:
>>>>>>>
>>>>>>>> Hi,
>>>>>>>>
>>>>>>>> How can I filter an item from the queryset object list if the name
>>>>>>>> is matching? The following won't work.
>>>>>>>>
>>>>>>>> user_list.filter(user__name=user.name)
>>>>>>>>
>>>>>>>> Thank you.
>>>>>>>>
>>>>>>>>
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>>>>>>
>>>>>>
>>>>>> --
>>>>>> Lucas Klassmann
>>>>>> Desenvolvedor de Software
>>>>>>
>>>>>> Email: lucaskl...@gmail.com
>>>>>> Web site: http://www.lucasklassmann.com
>>>>>>
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>>>>>
>>>>
>>>>
>>>>
>>>> --
>>>> Lucas Klassmann
>>>> Desenvolvedor de Software
>>>>
>>>> Email: lucaskl...@gmail.com
>>>> Web site: http://www.lucasklassmann.com
>>>>
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>>
>>
>>
>> --
>> Lucas Klassmann
>> Desenvolvedor de Software
>>
>> Email: lucaskl...@gmail.com
>> Web site: http://www.lucasklassmann.com
>>
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-- 
Lucas Klassmann
Desenvolvedor de Software

Email: lucasklassm...@gmail.com
Web site: http://www.lucasklassmann.com

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