This topic created in 4689 days ago, the information mentioned may be changed or developed.
例如:
tasks_list=tasks.objects.filter(update_time='today')
假设获得的tasks_list为 [task1,task2,...taskn]
每条记录都包含state这个字段,且state值可以为 'finished', ‘todo’,'delay'
那么如何根据state的值给tasks_list的各条记录排序?
比如优先级为'todo','finished','delay'
6 replies • 1970-01-01 08:00:00 +08:00
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cxe2v Aug 4, 2013
优先级这个没法设置吧,除了你给state赋予数字前缀,前缀就是优先级,或者来个state2字段给定优先级,当然灵活一点的是另建一张表,用state作为外键,另一个字段为优先级,这样就可以实现排序了
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felix021 Aug 4, 2013 1
一句话:
tasks_list.sort(key=lambda x: {'todo': 1, 'finished': 2, 'delay': 3}[x.state])
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z4none Aug 4, 2013
如果是 MySQL : select * from table where update_time='today' order by field (field, 'todo', 'finished', 'delay');
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z4none Aug 4, 2013
好吧 是 order by field (state, 'todo', 'finished', 'delay');
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hahastudio Aug 5, 2013
其实吧,你还是改SQL语句更划算= =我觉得= =
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raquelken Aug 5, 2013 1
你可以这样做 tasks_list=tasks.objects.filter(update_time='today').extra(select={"ordering": "case when state='todo' then 1 when state='finished' then 2 when state='delay' then 3 else 4 end"}).order_by('ordering')
但个人更建议你加字段
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