使用 peewee 创建子查询,在子查询结果上使用 `.select`
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【中文标题】使用 peewee 创建子查询,在子查询结果上使用 `.select`【英文标题】:Create subquery using peewee, using `.select` on the subquery results 【发布时间】:2021-12-25 16:25:04 【问题描述】:我有一个相当复杂的 peewee 查询,看起来像这样:
SolutionAlias = Solution.alias()
fields = [
SolutionAlias.id.alias('solution_id'),
SolutionAlias.solver.id.alias('solver_id'),
SolutionAlias.exercise.id.alias('exercise_id'),
]
query = (
User
.select(*fields)
.join(Exercise, JOIN.CROSS)
.join(Course, JOIN.LEFT_OUTER, on=(Exercise.course == Course.id))
.join(SolutionAlias, JOIN.LEFT_OUTER, on=(
(SolutionAlias.exercise == Exercise.id)
& (SolutionAlias.solver == User.id)
))
.where(
(Exercise.id << self.get_exercise_ids()),
(User.id << self.get_user_ids()),
)
.group_by(Exercise.id, User.id, SolutionAlias.id)
.having(
(SolutionAlias.id == fn.MAX(SolutionAlias.id))
| (SolutionAlias.id.is_null(True))
)
.alias('solutions_subquery')
)
full_query_fields = [
query.c.solver_id,
query.c.exercise_id,
Solution.id.alias('solution_id'),
SolutionAssessment.icon.alias('assessment_icon'),
]
solutions = (
Solution
.select(*full_query_fields)
.join(query, JOIN.RIGHT_OUTER, on=(
Solution.id == query.c.solution_id
))
.join(SolutionAssessment, JOIN.LEFT_OUTER, on=(
(Solution.assessment == SolutionAssessment.id)
))
)
这个确实有效,生成以下 SQL 查询:
SELECT
"solutions_subquery"."solver_id",
"solutions_subquery"."exercise_id",
"t1"."id" AS "solution_id",
"t2"."icon" AS "assessment_icon"
FROM
"solution" AS "t1"
RIGHT OUTER JOIN (
SELECT
"t3"."id" AS "solution_id",
"t4"."id" AS "solver_id",
"t5"."id" AS "exercise_id"
FROM
"user" AS "t4" CROSS
JOIN "exercise" AS "t5"
LEFT OUTER JOIN "course" AS "t6" ON ("t5"."course_id" = "t6"."id")
LEFT OUTER JOIN "solution" AS "t3" ON (
("t3"."exercise_id" = "t5"."id")
AND ("t3"."solver_id" = "t4"."id")
)
WHERE
(
(
"t5"."id" IN (155, 156, 157)
)
AND (
"t4"."id" IN (1, 24, 25, 26, 27, 28)
)
)
GROUP BY
"t5"."id",
"t4"."id",
"t3"."id"
HAVING
(
(
"t3"."id" = MAX("t3"."id")
)
OR ("t3"."id" IS NULL)
)
) AS "solutions_subquery" ON (
"t1"."id" = "solutions_subquery"."solution_id"
)
LEFT OUTER JOIN "solutionassessment" AS "t2" ON ("t1"."assessment_id" = "t2"."id")
但我真的不想使用 RIGHT_JOIN,因为 SQLite 不支持它。
当尝试使用子查询 query
和 JOIN
将 Solution
表查询到子查询的结果中时,我收到来自 peewee 的错误。
新查询:
solutions = (
query
.select(*full_query_fields)
.join(Solution, JOIN.LEFT_OUTER, on=(
Solution.id == query.c.solution_id
))
.join(SolutionAssessment, JOIN.LEFT_OUTER, on=(
(Solution.assessment == SolutionAssessment.id)
))
)
生成的查询:
SELECT
"solutions_subquery"."solver_id",
"solutions_subquery"."exercise_id",
"t1"."id" AS "solution_id",
"t1"."checker_id",
"t1"."state",
"t1"."submission_timestamp",
"t2"."name" AS "assessment",
"t2"."icon" AS "assessment_icon"
FROM
"user" AS "t3" CROSS
JOIN "exercise" AS "t4"
LEFT OUTER JOIN "course" AS "t5" ON ("t4"."course_id" = "t5"."id")
LEFT OUTER JOIN "solution" AS "t6" ON (
("t6"."exercise_id" = "t4"."id")
AND ("t6"."solver_id" = "t3"."id")
)
LEFT OUTER JOIN "solution" AS "t1" ON (
"t1"."id" = "solutions_subquery"."solution_id"
)
LEFT OUTER JOIN "solutionassessment" AS "t2" ON ("t1"."assessment_id" = "t2"."id")
WHERE
(
(
"t4"."id" IN (155, 156, 157)
)
AND (
"t3"."id" IN (1, 24, 25, 26, 27, 28)
)
)
GROUP BY
"t4"."id",
"t3"."id",
"t6"."id"
HAVING
(
(
"t6"."id" = MAX("t6"."id")
)
OR ("t6"."id" IS NULL)
)
结果:
psycopg2.errors.UndefinedTable: missing FROM-clause entry for table "solutions_subquery"
LINE 1: ...EFT OUTER JOIN "solution" AS "t1" ON ("t1"."id" = "solutions...
During handling of the above exception, another exception occurred:
[Truncated for readability...]
loguru.logger.critical(str(list(solutions.dicts().execute())))
[Truncated for readability...]
peewee.ProgrammingError: missing FROM-clause entry for table "solutions_subquery"
LINE 1: ...EFT OUTER JOIN "solution" AS "t1" ON ("t1"."id" = "solutions...
为什么 peewee 会展平查询?还有其他方法可以使用LEFT_JOIN
吗?
【问题讨论】:
【参考方案1】:最终在文档中找到了Select
函数,这让我可以包装上一个查询:
solutions = (
Select(columns=full_query_fields)
.from_(query)
.join(Solution, JOIN.LEFT_OUTER, on=(
Solution.id == query.c.solution_id
))
.join(SolutionAssessment, JOIN.LEFT_OUTER, on=(
(Solution.assessment == SolutionAssessment.id)
))
)
此解决方案有效。
【讨论】:
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