Paginate raw sql queries using LIMIT and OFFSET
It will also supports ORDER BY queries
To install from pypi
pip install django-sqlpaginator To get the latest (and possibly non stable version) from git
pip install git+git://github.com/bulkan/django-sqlpaginator.git You also need to install sqlparser
pip install git+git://github.com/andialbrecht/sqlparse.git In settings.py
INSTALLED_APPS= ( ... 'sqlpaginator', ... )Thats it !!
Pretty much same as django.core.pagination.Paginator
If you have the following models
classAlbum(models.Model): albumid=models.IntegerField(primary_key=True, db_column=u'AlbumId') title=models.TextField(db_column=u'Title') artistid=models.IntegerField(db_column=u'ArtistId') classArtist(models.Model): artistid=models.IntegerField(primary_key=True, db_column=u'ArtistId') name=models.TextField(db_column=u'Name', blank=True) and you want to paginate on Albums, then inside a view;
fromsqlpaginator.paginatorimportSqlPaginatorfrommodelsimportAlbumdefget_albums(request, page=1): sql="select * from %s"%Album._meta.db_tablepaginator=SqlPaginator(sql, Album, page=page, order_by='title') try: albums=paginator.page(page) exceptPageNotAnInteger: # If page is not an integer, deliver first page.albums=paginator.page(1) exceptEmptyPage: # If page is out of range (e.g. 9999), deliver last page of results.albums=paginator.page(paginator.num_pages) returnrender_to_response('albums_list.html',{'albums': albums})In the template albums_list.html
{%foralbuminalbums%}{# Each "album" is a Album model object. #}{{album.title|upper }}<br/>{%endfor%} <divclass="pagination"><spanclass="step-links">{%ifalbums.has_previous%} <ahref="?page={{albums.previous_page_number }}">previous</a>{%endif%} <spanclass="current">Page{{albums.number }} of{{albums.paginator.num_pages }}. </span>{%ifalbums.has_next%} <ahref="?page={{albums.next_page_number }}">next</a>{%endif%} </span></div>- Clone repo
- Change code
- Add tests
- Run the tests
nosetests -s --with-coverage --cover-package=sqlpaginator - Submit pull request