How to build django template GET request arguments?
Django version >= 2.0.0
for example.
in the urls.py, I have a register a view function:
path('detail/<str:db_name>/<str:collection_name>', views.detail, name='detail')
and in the django template, I want to use the url function to build this url: {% url 'detail' db_name collection_name %}, this will product url: /detail/db_name/collection, but I want the url is /detail/db_name/collection?per_page_count=15.
so, How can I build the GET request arguments by url.
thanks.
python django
asked Nov 16 '18 at 1:51
ltoddyltoddy
64
add a comment |
Django version >= 2.0.0
for example.
in the urls.py, I have a register a view function:
path('detail/<str:db_name>/<str:collection_name>', views.detail, name='detail')
and in the django template, I want to use the url function to build this url: {% url 'detail' db_name collection_name %}, this will product url: /detail/db_name/collection, but I want the url is /detail/db_name/collection?per_page_count=15.
so, How can I build the GET request arguments by url.
thanks.
python django
asked Nov 16 '18 at 1:51
ltoddyltoddy
64
A solution is{% url 'detail' db_name collection_name %}?per_page_count=15, but I dislike this ugly methods.
– ltoddy
Nov 16 '18 at 1:52
Please check this answer: stackoverflow.com/a/51377425/2696165
– ruddra
Nov 16 '18 at 3:17
add a comment |
Django version >= 2.0.0
for example.
in the urls.py, I have a register a view function:
path('detail/<str:db_name>/<str:collection_name>', views.detail, name='detail')
and in the django template, I want to use the url function to build this url: {% url 'detail' db_name collection_name %}, this will product url: /detail/db_name/collection, but I want the url is /detail/db_name/collection?per_page_count=15.
so, How can I build the GET request arguments by url.
thanks.
python django
asked Nov 16 '18 at 1:51
ltoddyltoddy
64
Django version >= 2.0.0
for example.
in the urls.py, I have a register a view function:
path('detail/<str:db_name>/<str:collection_name>', views.detail, name='detail')
and in the django template, I want to use the url function to build this url: {% url 'detail' db_name collection_name %}, this will product url: /detail/db_name/collection, but I want the url is /detail/db_name/collection?per_page_count=15.
so, How can I build the GET request arguments by url.
thanks.
python django
python django
asked Nov 16 '18 at 1:51
ltoddyltoddy
64
asked Nov 16 '18 at 1:51
ltoddyltoddy
64
asked Nov 16 '18 at 1:51
ltoddyltoddy
64
asked Nov 16 '18 at 1:51
ltoddyltoddy
64
asked Nov 16 '18 at 1:51
ltoddyltoddy
64
64
A solution is{% url 'detail' db_name collection_name %}?per_page_count=15, but I dislike this ugly methods.
– ltoddy
Nov 16 '18 at 1:52
Please check this answer: stackoverflow.com/a/51377425/2696165
– ruddra
Nov 16 '18 at 3:17
add a comment |
A solution is{% url 'detail' db_name collection_name %}?per_page_count=15, but I dislike this ugly methods.
– ltoddy
Nov 16 '18 at 1:52
Please check this answer: stackoverflow.com/a/51377425/2696165
– ruddra
Nov 16 '18 at 3:17
A solution is
{% url 'detail' db_name collection_name %}?per_page_count=15, but I dislike this ugly methods.– ltoddy
Nov 16 '18 at 1:52
A solution is
{% url 'detail' db_name collection_name %}?per_page_count=15, but I dislike this ugly methods.– ltoddy
Nov 16 '18 at 1:52
Please check this answer: stackoverflow.com/a/51377425/2696165
– ruddra
Nov 16 '18 at 3:17
Please check this answer: stackoverflow.com/a/51377425/2696165
– ruddra
Nov 16 '18 at 3:17
add a comment |
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A solution is
{% url 'detail' db_name collection_name %}?per_page_count=15, but I dislike this ugly methods.– ltoddy
Nov 16 '18 at 1:52
Please check this answer: stackoverflow.com/a/51377425/2696165
– ruddra
Nov 16 '18 at 3:17