PostgreSQL 查询列上空值最少的行

Question

如何查询输出为列中空值最少的行的行？

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 是空值最少的记录。任何帮助将不胜感激。谢谢

Answer 1

您可以：

按空值数（升序）对行进行排序
行数限制为 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

如果你只想要一行，那么你可以这样做：

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;

如果你想要所有这样的行，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 查询列上空值最少的行

PostgreSQL query rows with least null value on columns

sql

postgresql

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