Question

Feb 2, 1917

以上代码生成以下查询：

class Sentence(Model):
    name = CharField()

class Tokens(Model):
   token = CharField()
   sentence = ForeignKey(Sentence, related_name='tokens')

Sentence.objects.annotate(n=Count('tokens', distinct=True)).filter(n=5).filter(tokens__name__in=['se']).annotate(n0=F('tokens')).filter(tokens__name__in=['faire']).annotate(n1=F('tokens')).filter(tokens__name__in=['faire']).annotate(n2=F('tokens')).filter(tokens__name__in=['un']).annotate(n3=F('tokens')).filter(tokens__name__in=['avoir']).annotate(n4=F('tokens'))

为什么编号如此奇怪（以SELECT "sentence"."id", "sentence"."name", COUNT(DISTINCT "token"."id") AS "n", T3."id" AS "n0", T4."id" AS "n1", T4."id" AS "n2", T6."id" AS "n3", T6."id" AS "n4" FROM "sentence" LEFT OUTER JOIN "token" ON ("sentence"."id" = "token"."sentence_id") INNER JOIN "token" T3 ON ("sentence"."id" = T3."sentence_id") INNER JOIN "token" T4 ON ("sentence"."id" = T4."sentence_id") INNER JOIN "token" T5 ON ("sentence"."id" = T5."sentence_id") INNER JOIN "token" T6 ON ("sentence"."id" = T6."sentence_id") INNER JOIN "token" T7 ON ("sentence"."id" = T7."sentence_id") WHERE (T3."name" IN (se) AND T4."name" IN (faire) AND T5."name" IN (un) AND T6."name" IN (avoir) AND T7."name" IN (faire)) GROUP BY "word_frword"."id", T3."id", T4."id", T6."id" HAVING COUNT(DISTINCT "token"."id") = 5开头）？但是，为什么T3分配给n2，而不是T4？ T5和n4也是如此。看起来数字是2。

我想要完成的是在内连接的每一步上捕获令牌ID。它在有一个连接时有效，但随后会中断。

有什么建议吗？

Django：外键查找和注释F表达式的奇怪行为

0 个答案: