Question

如何查询行，其中输出将是列中具有最小null值的行？

我的数据是：

ID         | col1     | col2      | col3      | col4     
-----------+----------+-----------+-----------+-----------
 1         | Null     |Null       | with value| with value
 2         |with value|Null       | with value| with value
 3         |with value|Null       | Null      | Null

结果将是：

 ID         | col1     | col2      | col3      | col4     
 -----------+----------+-----------+-----------+-----------
  2         |with value|Null       | with value| with value

因为id 2是具有最少空值的记录。任何帮助将不胜感激。谢谢

Answer 1

您可以：

按空位数（升序）排列行
将行限制为1（LIMIT 1）

您的代码：

SELECT *
FROM your_table
ORDER BY 
    CASE WHEN col1 IS NULL THEN 1 ELSE 0 END +
    CASE WHEN col2 IS NULL THEN 1 ELSE 0 END +
    CASE WHEN col3 IS NULL THEN 1 ELSE 0 END +
    CASE WHEN col4 IS NULL THEN 1 ELSE 0 END 
LIMIT 1

Answer 2

如果只需要一行，则可以执行以下操作：

select t.*
from t
order by ( (col1 is null)::int + (col2 is null)::int +
           (col3 is null)::int + (col4 is null)::int
         ) asc
fetch first 1 row only;

如果您想要所有此类行，我想我会这样做：

select t.*
from (select t.*,
             dense_rank() over 
                 (order by (col1 is null)::int + (col2 is null)::int +
                           (col3 is null)::int + (col4 is null)::int
                 ) as null_ranking
      from t
     ) t
where null_ranking = 1;

PostgreSQL查询列上具有最小空值的行

2 个答案: