Question

我在查询中遇到问题，其中一个CTE没有返回任何行。但这很难注意到，调试还有很长一段时间。

是否可以输出Postgres中的所有CTE而无需注释掉主查询？

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;

在该示例中，CTE centroid逐渐添加到之前运行良好的函数中。如果grid为空，它应该返回行。错误（我已修复）是它没有，因为它从空CTE grid中选择。现在，当我描述问题时，显然它失败了，但是当你编写和调试时，可能会发生各种各样的事情，比如混合几何SRID，错误的SRID等。

Answer 1

EXPLAIN ANALYZE会分别报告CTE。

当我运行它（Postgresql 9.4）时，它会单独显示CTE，并且在结果部分中它确实显示从＆＃34; CTE扫描x＆＃34;上返回的实际行数。是0。

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

返回：

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

我不知道解释会一直显示这样的数据，我怀疑它取决于Postgresql如何决定优化查询，但它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，当其中一个派生表为空时，它不会产生输出：

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

您需要 OUTER CROSS JOIN 。

在SQL Server中有很棒的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;

的 LiveDemo

您可以使用LEFT JOIN LATERAL来模拟此行为，但它看起来有点“丑陋”：

;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;

的 SqlFiddleDemo

输出：

╔═══════════╦═════════╦═══╗ ║ val ║ c ║ d ║ ╠═══════════╬═════════╬═══╣ ║ Base row ║ (null) ║ 2 ║ ╚═══════════╩═════════╩═══╝
在这种情况下，
或简单LEFT JOIN：

;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;

如何调试PostgreSQL中的公用表表达式？

2 个答案: