Django：构造dyanamic查询 [英] Django: constructing dyanamic query

问题描述

我有一个表类型在名称列下具有值Drana，Thriller，Sci-fi等等。另一个表电影。
电影表格类型是ManyTomanyField类型表

i have a table Genre having values Drana, Thriller, Sci-fi, etc under "name" column. and another table Movie. In Movie table genre is ManyTOManyField for Genre table

情景

movie_1 = {D,S, T}
movie_2 = {S,T}
movie_3 = {D}
movie_4 = {D, T}

D：戏剧，S：科幻，T：惊悚

D: Drama, S: Sci-Fi, T: Thriller

案例1：
说说用户选择D和T（这是复选框选项）

case 1: Let say user choses D and T (which are checkbox options)

all D movies : movie_1, movie_3, movie_4  ---(a)
all T movies : movie_1, movie_2, movie_4  ---(b)

当用户选择这些选项时：
i想要显示（a）和（b）$ b $的交集的movie_1，movie_2，movie_3，movie_4 b对于上述情况，可以像下列情况一样执行以下查询：

when user selects those options : i want to show movie_1, movie_2, movie_3, movie_4 which is a intersection of both (a) and (b) for above case following query can be done like:

        qs_d = Genre.objects.get(name="Drama")
        qs_t = Genre.objects.get(name="Thriller")
        m_d = Movies.objects.filter( Q(gen__id=qs_d.id) | Q(gen__id=qs_t.id)).distinct()
        for movie_name in m_d:
            print movie_name

在上述查询中，我已经硬编码了戏剧和惊悚片（for解释），但是

根据用户选择的选项数量，我将在{戏剧，喜剧，惊悚片等列表中获得电影类型，在这种情况下构造查询？

in above queries i have hard-coded the genre "Drama" and "Thriller" (for explanation) , but
I'll get the movie genre in a list like {"Drama", "Comedy" , "Thriller"} depending upon the number of options user choses , in that case how can i construct the query ?

推荐答案

| 是 __或__ 函数，或者您可以使用 operator.or _ 函数。

所以你可以这样做：

from operator import or_
#Uncomment for Python3
#from functools import reduce
...
qs = [Q(gen__id=Genre.objects.get(name=name).id) for name in names]
m_d = Movies.objects.filter(reduce(or_, qs)).distinct()

应该做你想要的。我不太熟悉Django查询，所以可能会有更好的方法。当然，你可以将它与以下的东西组合起来：

That should do what you're looking for. I'm not terribly familiar with Django queries so there may be a better way. Of course you could combine it with something like:

m_d = Movies.objects.filter(reduce(or_, (Q(gen__id=Genre.objects.get(name=name).id) for name in names))).distinct()

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