postgresql - 在 SQLAlchemy 中加入集合返回函数 (SRF) 和访问列

Question

假设我有一个activity 表和一个subscription 表。每个事件都有一组对其他对象的通用引用，每个订阅都有一个对同一集合中其他对象的单个通用引用。

CREATE TABLE activity (
    id serial primary key,
    ob_refs UUID[] not null
);

CREATE TABLE subscription (
    id UUID primary key,
    ob_ref UUID,
    subscribed boolean not null
);

我想加入设置返回函数unnest，这样我就可以找到“最深”的匹配订阅，如下所示:

SELECT id
FROM (
    SELECT DISTINCT ON (activity.id)
        activity.id,
        x.ob_ref, x.ob_depth,
        subscription.subscribed IS NULL OR subscription.subscribed = TRUE
            AS subscribed,
    FROM activity

    LEFT JOIN subscription
        ON activity.ob_refs @> array[subscription.ob_ref]

    LEFT JOIN unnest(activity.ob_refs)
        WITH ORDINALITY AS x(ob_ref, ob_depth)
        ON subscription.ob_ref = x.ob_ref

    ORDER BY x.ob_depth DESC
) sub
WHERE subscribed = TRUE;

但我不知道如何进行第二次连接并访问列。我试过创建一个 FromClause like this :

act_ref_t = (sa.select(
        [sa.column('unnest', UUID).label('ob_ref'),
         sa.column('ordinality', sa.Integer).label('ob_depth')],
        from_obj=sa.func.unnest(Activity.ob_refs))
    .suffix_with('WITH ORDINALITY')
    .alias('act_ref_t'))

...

query = (query
    .outerjoin(
        act_ref_t,
        Subscription.ob_ref == act_ref_t.c.ob_ref))
    .order_by(activity.id, act_ref_t.ob_depth)

但这会导致此 SQL 带有另一个子查询:

LEFT OUTER JOIN (
        SELECT unnest AS ob_ref, ordinality AS ref_i
        FROM unnest(activity.ob_refs) WITH ORDINALITY
    ) AS act_ref_t
    ON subscription.ob_refs @> ARRAY[act_ref_t.ob_ref]

... 由于缺少和 unsupported 而失败LATERAL 关键字:

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

那么，如何在不使用子查询的情况下为此 SRF 创建 JOIN 子句？还是我还缺少其他东西？

编辑 1 使用 sa.text与 TextClause.columns而不是 sa.select 让我更接近:

act_ref_t = (sa.sql.text(
        "unnest(activity.ob_refs) WITH ORDINALITY")
    .columns(sa.column('unnest', UUID),
             sa.column('ordinality', sa.Integer))
    .alias('act_ref'))

但是生成的 SQL 失败了，因为它将子句括在括号中:

LEFT OUTER JOIN (unnest(activity.ob_refs) WITH ORDINALITY)
    AS act_ref ON subscription.ob_ref = act_ref.unnest

错误是“)”处或附近的语法错误。我可以让 TextAsFrom 不包含在括号中吗？

Answer 1

事实证明这不是 SA 直接支持的，但是可以通过 ColumnClause 实现正确的行为。和一个 FunctionElement .第一次进口this recipe如 zzzeek 所述在 this SA issue .然后创建一个包含 WITH ORDINALITY 修饰符的特殊 unnest 函数:

class unnest_func(ColumnFunction):
    name = 'unnest'
    column_names = ['unnest', 'ordinality']

@compiles(unnest_func)
def _compile_unnest_func(element, compiler, **kw):
    return compiler.visit_function(element, **kw) + " WITH ORDINALITY"

然后您可以像这样在连接、排序等中使用它:

act_ref = unnest_func(Activity.ob_refs)
query = (query
    .add_columns(act_ref.c.unnest, act_ref.c.ordinality)
    .outerjoin(act_ref, sa.true())
    .outerjoin(Subscription, Subscription.ob_ref == act_ref.c.unnest)
    .order_by(act_ref.c.ordinality.desc()))