如何为具有多对多字段的Django模型创建对象？

你不能从未保存的对象创建m2m关系。如果有pk，请尝试以下操作：

sample_object = Sample()sample_object.save()sample_object.users.add(1,2)

更新：阅读了saverio的答案后，我决定对这个问题进行更深入的研究。这是我的发现。

这是我最初的建议。它可以工作，但不是最佳选择。（注意：我使用

Bar

s

a Foo

User

a Sample

bar1 = Bar.objects.get(pk=1)bar2 = Bar.objects.get(pk=2)foo = Foo()foo.save()foo.bars.add(bar1)foo.bars.add(bar2)

它总共产生7个查询：

SELECt "app_bar"."id", "app_bar"."name" FROM "app_bar" WHERe "app_bar"."id" = 1SELECt "app_bar"."id", "app_bar"."name" FROM "app_bar" WHERe "app_bar"."id" = 2INSERT INTO "app_foo" ("name") VALUES ()SELECt "app_foo_bars"."bar_id" FROM "app_foo_bars" WHERe ("app_foo_bars"."foo_id" = 1  AND "app_foo_bars"."bar_id" IN (1))INSERT INTO "app_foo_bars" ("foo_id", "bar_id") VALUES (1, 1)SELECt "app_foo_bars"."bar_id" FROM "app_foo_bars" WHERe ("app_foo_bars"."foo_id" = 1  AND "app_foo_bars"."bar_id" IN (2))INSERT INTO "app_foo_bars" ("foo_id", "bar_id") VALUES (1, 2)

我相信我们可以做得更好。你可以将多个对象传递给该

add()

bar1 = Bar.objects.get(pk=1)bar2 = Bar.objects.get(pk=2)foo = Foo()foo.save()foo.bars.add(bar1, bar2)

如我们所见，传递多个对象可以节省一个SELECt：

SELECT "app_bar"."id", "app_bar"."name" FROM "app_bar" WHERe "app_bar"."id" = 1SELECt "app_bar"."id", "app_bar"."name" FROM "app_bar" WHERe "app_bar"."id" = 2INSERT INTO "app_foo" ("name") VALUES ()SELECt "app_foo_bars"."bar_id" FROM "app_foo_bars" WHERe ("app_foo_bars"."foo_id" = 1  AND "app_foo_bars"."bar_id" IN (1, 2))INSERT INTO "app_foo_bars" ("foo_id", "bar_id") VALUES (1, 1)INSERT INTO "app_foo_bars" ("foo_id", "bar_id") VALUES (1, 2)

我不知道你还可以分配对象列表：

bar1 = Bar.objects.get(pk=1)bar2 = Bar.objects.get(pk=2)foo = Foo()foo.save()foo.bars = [bar1, bar2]

不幸的是，这又增加了一个SELECt：

SELECT "app_bar"."id", "app_bar"."name" FROM "app_bar" WHERe "app_bar"."id" = 1SELECt "app_bar"."id", "app_bar"."name" FROM "app_bar" WHERe "app_bar"."id" = 2INSERT INTO "app_foo" ("name") VALUES ()SELECt "app_foo_bars"."id", "app_foo_bars"."foo_id", "app_foo_bars"."bar_id" FROM "app_foo_bars" WHERe "app_foo_bars"."foo_id" = 1SELECt "app_foo_bars"."bar_id" FROM "app_foo_bars" WHERe ("app_foo_bars"."foo_id" = 1  AND "app_foo_bars"."bar_id" IN (1, 2))INSERT INTO "app_foo_bars" ("foo_id", "bar_id") VALUES (1, 1)INSERT INTO "app_foo_bars" ("foo_id", "bar_id") VALUES (1, 2)

让我们尝试分配一个pks 列表，如saverio建议的那样：

foo = Foo()foo.save()foo.bars = [1,2]

因为我们不获取两个

Bars

INSERT INTO "app_foo" ("name") VALUES ()SELECT "app_foo_bars"."id", "app_foo_bars"."foo_id", "app_foo_bars"."bar_id" FROM "app_foo_bars" WHERe "app_foo_bars"."foo_id" = 1SELECt "app_foo_bars"."bar_id" FROM "app_foo_bars" WHERe ("app_foo_bars"."foo_id" = 1  AND "app_foo_bars"."bar_id" IN (1, 2))INSERT INTO "app_foo_bars" ("foo_id", "bar_id") VALUES (1, 1)INSERT INTO "app_foo_bars" ("foo_id", "bar_id") VALUES (1, 2)

最终获胜者是：

foo = Foo()foo.save()foo.bars.add(1,2)

路过

pks

add()

INSERT INTO "app_foo" ("name") VALUES ()SELECt "app_foo_bars"."bar_id" FROM "app_foo_bars" WHERe ("app_foo_bars"."foo_id" = 1  AND "app_foo_bars"."bar_id" IN (1, 2))INSERT INTO "app_foo_bars" ("foo_id", "bar_id") VALUES (1, 1)INSERT INTO "app_foo_bars" ("foo_id", "bar_id") VALUES (1, 2)