sql - 合并两个 SQL 查询

Question

我需要一个模仿以下形式的 if-then-else 语句的 SQL 查询:

if (query1 != null)
  return query1
else
  return query2

由于 COALESCE 不适用于结果集，因此我创建了一个联合查询来完成这项工作:

SELECT * FROM obs WHERE cond1  --query1
UNION
SELECT * FROM obs WHERE (NOT EXISTS(query1)) AND cond2

在 SQL 中:

  ( SELECT * FROM obs WHERE src = @id AND tstart <= @instant AND tend >= @instant )
  UNION
  ( SELECT * FROM obs WHERE (NOT EXISTS (SELECT 1 FROM obs WHERE src = @id AND tstart <= @instant AND tend >= @instant )) AND src = @id AND tstart <= @instant ORDER BY tend DESC LIMIT 1);

表 obs 具有字段 ( src | tstart | tent | ... )。我想选择与@instant 重叠的那些行。如果没有找到重叠行，则应返回 @instant 之前最接近的行。

SQL UNION 语句可以工作，但它非常笨拙，我正在寻找更短、更清晰的语句。具有 COALESCE ( query1, query2 ) 精神的东西会很好。我的数据库是Postgresql。

Answer 1

首先，在这种情况下，union all 可能比 union 更合适。

其次，您可以使用 with 来表达这一点以简化查询:

WITH t1 as (
      SELECT *
      FROM obs
      WHERE src = @id AND tstart <= @instant AND tend >= @instant
     )
SELECT t1.*
FROM t1
UNION ALL
(SELECT *
 FROM obs
 WHERE NOT EXISTS (SELECT 1 FROM t1) AND
       src = @id AND tstart <= @instant
 ORDER BY tend DESC
 LIMIT 1
);

但是，如果您正在查找单行，这会更简单:

 SELECT *
 FROM obs
 WHERE src = @id
 ORDER BY (CASE WHEN  tstart <= @instant AND tend >= @instant THEN 1
                ELSE 2
           END),
          tend DESC
 LIMIT 1;

而且，如果不是单行，也可以使用窗口函数:

SELECT o.*
FROM (SELECT o.*,
             DENSE_RANK() OVER (PARTITION BY src
                                ORDER BY (CASE WHEN tstart <= @instant AND tend >= @instant THEN 1
                                               ELSE 2
                                          END),
                                         (CASE WHEN tstart <= @instant AND tend >= @instant THEN NULL
                                               ELSE tend
                                          END) DESC
                               ) as seqnum
      FROM obs o
      WHERE src = @id
     ) o
WHERE seqnum = 1;