姜戈prefetch_related

Question

我有一个关系模型集，它有5个表，每个表都有一个到多个关系。在第二张表之后，我很难访问模型信息。例如。

class TableA(models.Model):
  pass

class TableB(models.Model):
  tablea = models.ForeignKey(TableA, oncascade...)

class TableC(models.Model):
  tableb = models.ForeignKey(TableB, oncas...)

class TableD(models.Model):
  tablec = models.ForeignKey(TableC, ...)

class TableE(models.Model):
  tabled = models.ForeignKey(TableE, ...)

在我的views.py中，我在TableA上使用了预取相关的方法：

from . models import *
data = TableA.objects.all().prefetch_related('tableb_set__tablec_set__tabled_set')

我似乎无法从TableC、TableD或TableE获取信息。

for a in data:
  print(a.id)
  for b in a.tableb_set.values():
    print(b['id])
    for c in b.tablec_set.values(): #This is where I get an error message stating the model doesn't have a relatedmanager

如何访问深入到2+关系中的模型信息prefetch_related表？

提前感谢

Answer 1

这对我有用。检查您的代码示例是否与您的源代码对齐，并且您在源代码中没有任何输入。

for a in TableA.objects.prefetch_related(
    "tableb_set__tablec_set__tabled_set__tablee_set"
):
    print(f"+ {a.id}")
    for b in a.tableb_set.all():
        print(f"+- {b.id}")
        for c in b.tablec_set.all():
            print(f"+-- {c.id}")
            for d in c.tabled_set.all():
                print(f"+--- {d.id}")
                for e in d.tablee_set.all():
                    print(f"+--- {e.id}")

for query in connection.queries:
    print(query["sql"])

# prints:

+ 1                                                                                                                                                                        
+ 2                                                                                                                                                                        
+- 2                                                                                                                                                                       
+-- 2                                                                                                                                                                      
+--- 2                                                                                                                                                                     
+--- 11                                                                                                                                                                    
+---- 3
+-- 4
+- 5
+-- 1
+--- 4
+--- 6

[...]

SELECT "main_tablea"."id" FROM "main_tablea"
SELECT "main_tableb"."id", "main_tableb"."a_id" FROM "main_tableb" WHERE "main_tableb"."a_id" IN (1, 2, 3, 4, 5)
SELECT "main_tablec"."id", "main_tablec"."tableb_id" FROM "main_tablec" WHERE "main_tablec"."tableb_id" IN (2, 5, 6, 4, 1, 3)
SELECT "main_tabled"."id", "main_tabled"."tablec_id" FROM "main_tabled" WHERE "main_tabled"."tablec_id" IN (5, 9, 2, 4, 6, 7, 1, 3, 8)
SELECT "main_tablee"."id", "main_tablee"."tabled_id" FROM "main_tablee" WHERE "main_tablee"."tabled_id" IN (4, 6, 2, 11, 3, 7, 8, 1, 10, 5, 9)

Answer 2

使用.all()方法在_set上访问信息！