Question

我有一个这样的表A：

col1 col2 {{1 }}
col3 1 0
null null null
null 3 null
null null 5

我想要在 Oracle 10G 中输出如下内容：

1 column_name
null_count col1
2 col2
2 col3

我已经使用 UNION ALL 做到了这一点：

工作正常，但要花很多时间，因为有近100个 UNION ALL 。我想在不使用 UNION ALL 的情况下实现相同的输出。

是否可以通过不使用 UNION ALL 来实现这一目标？

Answer 1

您可以为此使用UNPIVOT（我不确定古老的Oracle 10是否已支持该功能-十多年来我一直没有使用它）

select colname, count(*) - count(val) as num_nulls
from t1
  UNPIVOT include nulls 
      (val for colname in (col1 as 'C1', 
                           col2 as 'C2', 
                           col3 as 'C3'))
group by colname
order by colname;

不确定那是否更快。

在线示例：https://dbfiddle.uk/?rdbms=oracle_11.2&fiddle=4e807b8b2d8080abac36574f776dbf04

Answer 2

我会推荐此查询：

SELECT 'col1' AS col_name, COUNT(*) - COUNT(col1) AS col_null_count FROM t UNION ALL
SELECT 'col2' AS col_name, COUNT(*) - COUNT(col2) AS col_null_count FROM t UNION ALL
SELECT 'col3' AS col_name, COUNT(*) - COUNT(col3) AS col_null_count FROM t

但是您可以尝试这样写：

WITH cte AS (
    -- 1 rows by n columns
    SELECT COUNT(*) - COUNT(col1) AS col1_null_count
         , COUNT(*) - COUNT(col2) AS col2_null_count
         , COUNT(*) - COUNT(col3) AS col3_null_count
    FROM t
)
-- n rows by 2 columns
SELECT 'col1' AS col_name, col1_null_count AS col_null_count FROM cte UNION ALL
SELECT 'col2' AS col_name, col2_null_count AS col_null_count FROM cte UNION ALL
SELECT 'col3' AS col_name, col3_null_count AS col_null_count FROM cte

Answer 3

Oracle 10g不支持UNPIVOT和PIVOT运算符，因此要执行10g中的操作，您需要使用伪表（包含与未透视表相同的行数），即3），就像这样：

WITH your_table AS (SELECT 1 col1, 0 col2, NULL col3 FROM dual UNION ALL
                    SELECT NULL col1, NULL col2, NULL col3 FROM dual UNION ALL
                    SELECT 3 col1, NULL col2, NULL col3 FROM dual UNION ALL
                    SELECT NULL col1, 5 col2, 1 col3 FROM dual)
SELECT CASE WHEN dummy.id = 1 THEN 'col1'
            WHEN dummy.id = 2 THEN 'col2'
            WHEN dummy.id = 3 THEN 'col3'
       END column_name,
       COUNT(CASE WHEN dummy.id = 1 THEN CASE WHEN col1 IS NULL THEN 1 END
                  WHEN dummy.id = 2 THEN CASE WHEN col2 IS NULL THEN 1 END
                  WHEN dummy.id = 3 THEN CASE WHEN col3 IS NULL THEN 1 END
             END) null_count
FROM   your_table
       CROSS JOIN (SELECT LEVEL ID
                   FROM   dual
                   CONNECT BY LEVEL <= 3) dummy
GROUP BY dummy.id;

COLUMN_NAME NULL_COUNT
----------- ----------
col1                 2
col2                 2
col3                 3

如果您认为写大量列会花费很多时间，则始终可以编写一个查询，该查询将自己生成大量case语句，例如：

SELECT 'when dummy.id = '||row_number() OVER (PARTITION BY owner, table_name ORDER BY column_id)||' then '''||LOWER(column_name)||'''' first_part,
       'when dummy.id = '||row_number() OVER (PARTITION BY owner, table_name ORDER BY column_id)||' then case when '||column_name||' is null then 1 end' second_part
FROM   all_tab_columns a
WHERE  owner = ...
AND    table_name = ...
-- and column_name in (...)
ORDER BY column_id;

（我包括了row_number()分析函数，而不是使用column_id，因为如果要排除某些列，column_id列将不再是从1开始的连续数字。）

如何计算每列中存在多少个空值？

3 个答案: