我有两个型号：
class Author(models.Model);
name = models.CharField(max_length=255)
class Book(models.Model):
title = models.CharField(max_length=255)
authors = models.ManyToManyField(Author, null=True, blank=True)
现在，我想要所有书籍的信息。所以我做了：
book_info = Book.objects.all().values('title', 'authors__name')
并且，它给出了一个输出（对于1本书有2位作者）：
[{'title': u'book1', 'authors__name': u'author1'},{'title': u'book1', 'authors__name': u'author2'}]
我想要的是：
[{'title': u'book1', 'authors': [{'name':u'author1'},{'name':u'author2'}]}]
我可能在作者模型中有更多字段，因此也希望获得这些字段。
我可以在一个查询中执行此操作吗？
我能做些什么来获得所需的结果？
Django 1.4
好问题，使用 prefetch_related：
In [3]: [{'name': b.name, 'authors': [a.name for a in b.authors.all()]} for b in Book.objects.prefetch_related('authors')]
(0.000) SELECT "test_app_book"."id", "test_app_book"."name" FROM "test_app_book"; args=()
(0.000) SELECT ("test_app_book_authors"."book_id") AS "_prefetch_related_val", "test_app_author"."id", "test_app_author"."name" FROM "test_app_author" INNER JOIN "test_app_book_authors" ON ("test_app_author"."id" = "test_app_book_authors"."author_id") WHERE "test_app_book_authors"."book_id" IN (1, 2); args=(1, 2)
Out[3]:
[{'authors': [u'a', u'b'], 'name': u'book'},
{'authors': [u'b'], 'name': u'test'}]
Django 1.3
prefetch_related是在Django 1.4中引入的。对于Django 1.3，您需要 Django的selectreverse：
In [19]: [{'name': b.name, 'authors': [a.name for a in b.authors_prefetch]} for b in Book.objects.select_reverse({'authors_prefetch': 'authors'})]
(0.000) SELECT "test_app_book"."id", "test_app_book"."name" FROM "test_app_book"; args=()
(0.001) SELECT (test_app_book_authors.book_id) AS "main_id", "test_app_author"."id", "test_app_author"."name" FROM "test_app_author" INNER JOIN "test_app_book_authors" ON ("test_app_author"."id" = "test_app_book_authors"."author_id") WHERE "test_app_book_authors"."book_id" IN (1, 2); args=(1, 2)
Out[19]:
[{'authors': [u'a', u'b'], 'name': u'book'},
{'authors': [u'b'], 'name': u'test'}]
运用 Django的selectreverse：
class Book(models.Model):
name = models.CharField(max_length=100)
authors = models.ManyToManyField(Author, null=True, blank=True)
objects = ReverseManager()
为了避免像@jpic中的描述性答案那样进行两次查询，您可以在之后手动合并结果。对我来说感觉有点笨拙，但它确实有效。
def merge_values(values):
grouped_results = itertools.groupby(values, key=lambda value: value['id'])
merged_values = []
for k, g in grouped_results:
groups = list(g)
merged_value = {}
for group in groups:
for key, val in group.iteritems():
if not merged_value.get(key):
merged_value[key] = val
elif val != merged_value[key]:
if isinstance(merged_value[key], list):
if val not in merged_value[key]:
merged_value[key].append(val)
else:
old_val = merged_value[key]
merged_value[key] = [old_val, val]
merged_values.append(merged_value)
return merged_values
book_info = marge_values(Book.objects.all().values('title', 'authors__name'))
merge_values函数取自 这个要点