带有完整子对象的 SQLAlchemy 分组依据 [英] SQLAlchemy Group By With Full Child Objects

问题描述

想象下面的媒体表:

| site       | show_id | time |
| ---------------------|-------|
| CNN        | 1       | 'a'   |
| ABC        | 2       | 'b'   |
| ABC        | 5       | 'c'   |
| CNN        | 3       | 'd'   |
| NBC        | 4       | 'e'   |
| NBC        | 5       | 'f'   |
--------------------------------

我想遍历按 show_id 分组的查询结果并尝试过此查询:

I would like to iterate over query results grouped by show_id and have tried this query:

listings = session.query(Media).filter(Media.site == "CNN").group_by(Media.show_id).all()

以下是我想对结果进行迭代的方式:

Here's how I would like to iterate over the results:

for showtimes in listings:
    for show in showtimes:
        print(show.time)

但是那个查询并没有给我所有分组的子对象.我错过了什么?

But that query doesn't give me all of the grouped child objects. What am I missing?

推荐答案

在 SQL 中，GROUP BY 子句根据分组表达式将分组的行压缩为一行.我的猜测是您使用的是 SQLite 或较旧版本的 MySQL，因为您可以选择非聚合而不是 功能上依赖 分组表达式.在这种情况下，结果包含来自每组未指定行的值，这很少(如果有的话)有用.

In SQL the GROUP BY clause condenses the grouped rows into a single row, based on the grouping expressions. My guess is that you are using SQLite, or an older version of MySQL, since you are allowed to select non-aggregates without them being functionally dependent on the grouping expressions. The results contain values from an unspecified row per group in that case, which is seldom — if ever — useful.

解决方案是在 SQL 中使用 ORDER BY 而不是 GROUP BY，然后 Python中的分组基于订单表达式:

A solution is to ORDER BY instead of GROUP BY in SQL and then group in Python based on the order expressions:

from itertools import groupby
from operator import attrgetter

listings = session.query(Media).\
    filter(Media.site == "CNN").\
    order_by(Media.show_id).\
    all()

# Materialize the subiterators to lists
listings = [list(g) for k, g in groupby(listings, attrgetter('show_id'))]

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