Suppose I have such a code: def products(): if len(request.args): page=int(request.args[0]) else: page=0 items_per_page=20 limitby=(page*items_per_page,(page+1)*(items_per_page+1)) qset = db(db['products']['name']!='None') grid = qset.select(limitby=limitby) return dict(grid=grid,page=page,items_per_page=items_per_page)
My view: {{extend 'layout.html'}}<h2>Browse available products:</h2> <table class="products">{{for i, row in enumerate(grid):}}{{if i==items_per_page:break}}<tr><td>{{=row.name}}</td>{{pass}}</tr></table> {{if page:}}<a href="{{=URL(args=[page-1])}}">previous</a>{{pass}} {{if len(grid)>items_per_page:}}<a href="{{=URL(args=[page+1])}}">next</a>{{pass}} I have 'next' button and 'previous' button. How do I get the 'last' button(page with the last items on the list? Regards -- Resources: - http://web2py.com - http://web2py.com/book (Documentation) - http://github.com/web2py/web2py (Source code) - https://code.google.com/p/web2py/issues/list (Report Issues) --- You received this message because you are subscribed to the Google Groups "web2py-users" group. To unsubscribe from this group and stop receiving emails from it, send an email to web2py+unsubscr...@googlegroups.com. For more options, visit https://groups.google.com/d/optout.