如何调试 PostgreSQL 中常见的 table 表达式？

Question

我在查询中遇到问题，其中一个 CTE return 没有行。但这很难被注意到，并且调试了很长时间。

是否可以在不注释掉主查询的情况下在 Postgres 中输出所有 CTE？

create or replace function generate_grid(
    poly geometry, step double precision)
    returns setof geometry as
$$ */

with 
    initial as (select st_xmin(poly) x0, st_ymin(poly) y0),
    ...(another 3 CTE skipped here)...
    grid as (select point from points where st_intersects(point, poly)),
    centroid as (select st_centroid(poly) point from grid where (select count(*)=0 from grid))
select * from grid
union all
select * from centroid;
$$ language sql;

在示例中，CTE centroid 被增量添加到之前运行良好的函数中。它应该 return 行以防 grid 为空。错误（我已修复）是它没有，因为它是从空 CTE grid 中选择的。现在，当我描述问题时，它失败的原因很明显，但是当您编写并调试时，可能会发生各种各样的事情，例如混合几何 SRID、错误 SRID 等。

Answer 1

EXPLAIN ANALYZE 希望单独报告 CTE。

当我运行那个 (Postgresql 9.4) 单独显示 CTE 时，在结果部分它确实显示从 "CTE Scan on x" 返回的实际行数是 0。

explain analyze
with x as (select 1 where false),
     y as (select 2 where true)
select * from x, y;

Returns:

Nested Loop  (cost=0.02..0.07 rows=1 width=8) (actual time=0.002..0.002 rows=0 loops=1)
  Output: x."?column?", y."?column?"
  CTE x
    ->  Result  (cost=0.00..0.01 rows=1 width=0) (actual time=0.000..0.000 rows=0 loops=1)
          Output: 1
          One-Time Filter: false
  CTE y
    ->  Result  (cost=0.00..0.01 rows=1 width=0) (never executed)
          Output: 2
  ->  CTE Scan on x  (cost=0.00..0.02 rows=1 width=4) (actual time=0.002..0.002 rows=0 loops=1)
        Output: x."?column?"
  ->  CTE Scan on y  (cost=0.00..0.02 rows=1 width=4) (never executed)
        Output: y."?column?"
Planning time: 0.034 ms
Execution time: 0.018 ms

我不知道 explain 会一直显示这样的数据，我怀疑这取决于 Postgresql 如何决定优化查询，但这应该是一个很好的起点。

在 http://www.postgresql.org/docs/current/static/sql-explain.html

解释文档

Answer 2

CROSS JOIN 的问题是当派生表之一为空时它不会产生任何输出：

with x as (select 1 where false),
     y as (select 2 where true)
select * from x, y;

你需要 OUTER CROSS JOIN.

在SQL Server有很棒的OUTER APPLY:

with x(c) as (select 1 where 1=0),
     y(d) as (select 2 where 1=1)
select * 
FROM (values ('Base')) AS s(val)   -- always one row
OUTER APPLY x
OUTER APPLY y;

LiveDemo

您可以使用 LEFT JOIN LATERAL 模拟此行为，但它看起来有点 "ugly":

;WITH x(c) AS (SELECT 1 WHERE false),
      y(d) AS (SELECT 2 WHERE true)
SELECT * 
FROM (VALUES ('Base row')) AS s(val)
LEFT JOIN LATERAL (SELECT * FROM x) AS x(c) ON true
LEFT JOIN LATERAL (SELECT * FROM y) AS y(d) ON true;

SqlFiddleDemo

输出：

╔═══════════╦═════════╦═══╗
║   val     ║   c     ║ d ║
╠═══════════╬═════════╬═══╣
║ Base row  ║ (null)  ║ 2 ║
╚═══════════╩═════════╩═══╝

或简单的 LEFT JOIN 在这种情况下：

;WITH x(c) AS (SELECT 1 WHERE false),
     y(d) AS (SELECT 2 WHERE true)
SELECT * 
FROM (VALUES ('Base row')) AS s(val)
LEFT JOIN x ON true
LEFT JOIN y ON true;

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