Re: where is the error in pagination?

[leb dev]

>
> 
>>
>> 
>>
>> {% if object_list.has_previous %}
>>
>>  first
>>
>> previous
>>
>> {% endif %}
>>
>>
>>> 
>>
>> Page {{ object_list.number }} of {{ 
>>> object_list.paginator.num_pages }}.
>>
>> 
>>
>>
>>> {% if object_list.has_next %}
>>
>> next
>>
>> last 
>>
>> {% endif %}
>>
>> 
>>
>> 
>>
>>
On Wednesday, September 25, 2019 at 2:39:16 PM UTC+3, laya Mahmoudi wrote:
>
> Would you send your html codes relating to pagination
>
> در تاریخ چهارشنبه ۲۵ سپتامبر ۲۰۱۹،‏ ۱۵:۰۳ leb dev  
> نوشت:
>
>> i tried your answer and display an error :
>>
>> That page number is not an integer
>>
>> -- 
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>> "Django users" group.
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>>  
>> 
>> .
>
>

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Re: where is the error in pagination?

[laya Mahmoudi]

Would you send your html codes relating to pagination

در تاریخ چهارشنبه ۲۵ سپتامبر ۲۰۱۹،‏ ۱۵:۰۳ leb dev 
نوشت:

> i tried your answer and display an error :
>
> That page number is not an integer
>
> --
> 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.
> To view this discussion on the web visit
> https://groups.google.com/d/msgid/django-users/0323b126-726e-4889-beee-b68988b1a3f1%40googlegroups.com
> 
> .
>

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Re: where is the error in pagination?

[leb dev]

i tried your answer and display an error :

That page number is not an integer

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RE: where is the error in pagination?

[laya]

Try this one :
paginator = Paginator(queryset_list,8)  # Show 25 contacts per page
page_request_var = "page"
page = request.GET.get(page_request_var)
try:
queryset = paginator.page(page)
except PageNotAnInteger:
# If page is not an integer, deliver first page.
queryset = paginator.page(1)
except EmptyPage:
# If page is out of range (e.g. ), deliver last page of results.
queryset = paginator.page(paginator.num_pages)


Sent from Mail for Windows 10

From: leb dev
Sent: Wednesday, September 25, 2019 2:32 PM
To: Django users
Subject: where is the error in pagination?

i have a django project  and it includes pagination i do not know where is the 
error in my code but once i tired to go to the next or previous page it crash 
and display the below error :

> can only concatenate str(not ""NoneType) to str

            query=request.GET.get("search")
            paginator = Paginator(queryset_list, 5) 
            page_request_var = "page"
            print("page_request_var===>",page_request_var)
            page = request.GET.get(page_request_var)
            print("page===>",page)
            queryset = paginator.get_page(page)
            print("queryset===>",queryset)

When i print **page** it display None


 In the browser it suppose to display **page=2=sam**
but instead 
it display **/?=2&20search%20=%20sam**
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Re: where is the error in pagination?

[leb dev]

no i am using function based views

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RE: where is the error in pagination?

[laya]

Hi,
Do you use Class Based View ?


Sent from Mail for Windows 10

From: leb dev
Sent: Wednesday, September 25, 2019 2:32 PM
To: Django users
Subject: where is the error in pagination?

i have a django project  and it includes pagination i do not know where is the 
error in my code but once i tired to go to the next or previous page it crash 
and display the below error :

> can only concatenate str(not ""NoneType) to str

            query=request.GET.get("search")
            paginator = Paginator(queryset_list, 5) 
            page_request_var = "page"
            print("page_request_var===>",page_request_var)
            page = request.GET.get(page_request_var)
            print("page===>",page)
            queryset = paginator.get_page(page)
            print("queryset===>",queryset)

When i print **page** it display None


 In the browser it suppose to display **page=2=sam**
but instead 
it display **/?=2&20search%20=%20sam**
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