在 SQLAlchemy 连接子句中使用函数输出 [英] Using function output in SQLAlchemy join clause

问题描述

我正在尝试将相当短的 SQL 转换为 sqlAlchemy ORM 查询.SQL 使用 Postgres 的 generate_series 制作一组日期，我的目标是制作一组按其中一个列分类的时间序列数组.

I am trying to translate a fairly short bit of SQL into an sqlAlchemy ORM query. The SQL uses Postgres's generate_series to make a set of dates and my goal is to make a set of time series arrays categorized by one of the columns.

表格(简化版)非常简单:

The tables (simplified) are very simple:

counts:
-----------------
count   (Integer)
day     (Date)
placeID (foreign key related to places)

"counts_pkey" PRIMARY KEY (day, placeID)

places:
-----------------
id
name   (varchar)

我所追求的输出是每个地方的计数时间序列，包括一天未报告计数时的空值.例如，这将对应于超过四天的系列:

The output I'm after is a time series of counts for each place including null values when counts are not reported for a day. For example, this would correspond to a series over four days:

    array_agg    |    name
-----------------+-------------------
 {NULL,0,7,NULL} | A Place
 {NULL,1,NULL,2} | Some other place
 {5,NULL,3,NULL} | Yet another

通过对日期范围和地点进行 CROSS JOIN 并将其与计数相结合，我可以很容易地做到这一点:

I can do this fairly easily by taking a CROSS JOIN on a date range and places and joining that with the counts:

SELECT array_agg(counts.count), places.name 
FROM generate_series('2018-11-01', '2018-11-04', interval '1 days') as day 
CROSS JOIN  places 
LEFT OUTER JOIN counts on counts.day = day.day AND counts.PlaceID = places.id 
GROUP BY places.name;

我似乎无法弄清楚如何让 SQLAlchemy 做到这一点.经过大量挖掘，我找到了一个旧的谷歌群组线程 几乎可以导致这个:

What I can't seem to figure out is how to get SQLAlchemy to do this. After a lot of digging, I found an old google groups thread which almost works leading to this:

date_list = select([column('generate_series')])\
.select_from(func.generate_series(backthen, today, '1 day'))\ 
.alias('date_list')

time_series = db.session.query(Place.name, func.array_agg(Count.count))\
.select_from(date_list)\
.outerjoin(Count, (Count.day == date_list.c.generate_series) & (Count.placeID == Place.id ))\
.group_by(Place.name)

这会为时间序列创建一个子选择，但会产生数据库错误:

This creates a sub-select for the time series, but it produces a database error:

表places"有一个条目，但不能从查询的这一部分引用.

There is an entry for table "places", but it cannot be referenced from this part of the query.

所以我的问题是:你会如何在 sqlalchemy 中做到这一点.另外，我认为这很困难，因为我使用 SQL 的方法是愚蠢的.

So my question is: how would you do this in sqlalchemy. Also, I'm open to the idea that this is difficult because my approach with the SQL is bone-headed.

推荐答案

问题是给定的查询构造 SQLAlchemy 生成了一个查询

The problem is that given the query construct SQLAlchemy produces a query along the lines of

SELECT ...
FROM places,
     (...) AS date_list LEFT OUTER JOIN count ON ... AND count."placeID" = places.id
...

有 2 个 FROM-list 项:places 和 join.项目不能相互交叉引用¹，因此由于ON子句中的places.id而导致错误.

There are 2 FROM-list items: places and the join. Items cannot cross-reference each other¹, and hence the error due to places.id in the ON-clause.

SQLAlchemy 不支持显式 CROSS JOIN，但另一方面 CROSS JOIN 等效于 INNER JOIN ON (TRUE).您还可以省略在子查询中包装函数表达式，并按原样使用 给它一个别名:

SQLAlchemy does not support explicit CROSS JOIN, but on the other hand a CROSS JOIN is equivalent to an INNER JOIN ON (TRUE). You could also omit wrapping the function expression in a subquery and use it as is by giving it an alias:

date_list = func.generate_series(backthen, today, '1 day').alias('gen_day')

time_series = session.query(Place.name, func.array_agg(Count.count))\
    .join(date_list, true())\
    .outerjoin(Count, (Count.day == column('gen_day')) &
                      (Count.placeID == Place.id ))\
    .group_by(Place.name)

¹:除了函数调用FROM-items，或者使用LATERAL.

¹: Except function-call FROM-items, or using LATERAL.

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