查询以查找 ID 总数和在任何列中具有 null 的 ID 以及两者的百分比（即具有 null 的 ID 计数/ID 总数）

Question

有两张桌子。 Tables A 具有以下结构

ID	Flag	Name
1X	1	Y
2Y	0	Null
3Z	1	Null
4A	1	Y

Table B 有如下结构

B_ID	City	State
1X	Y	Null
2Y	Null	Null
3Z	Null	Y
4A	Y	Y

我想获取所有 ID 的计数以及在任何列（名称、城市、州）中具有 Null 的 ID 的计数，例如，从上表中只有 ID 4A 具有非空值在两个表的所有三列中，所以输出应该像

Total_Count	Ids having null	Percentage missing
4	3	0.75%

Total_count 是 4，因为总共有四个 ID，具有 NULL 的 ID 是 3，因为在三列（即名称、城市、州）中的任何一个中都有 3 个 ID 为空，并且缺失百分比只是具有空值的 ID / Total_Count.

我尝试使用以下几行查询

select (count/total) * 100 pct, count,total 
from (select sum(count) count 
      from(select count(*) count from tableA T1
      where T1.name is null
      union all
      select count(*) count from tableA T1
      join tableB T2 on T1.ID = T2.B_ID 
      where T2.city is null
      union all
      select count(*) count from tableA T1
      join tableB T2 on T1.ID = T2.B_ID 
      where T2.state is null)),
select count(ID) total from tableA);

但是查询没有返回所需的输出，你能给我一个更好的方法吗？谢谢

Answer 1

使用条件聚合：

SELECT COUNT(*) Total_Count,    
       COUNT(CASE WHEN t1.Name IS NULL OR t2.City IS NULL OR t2.State IS NULL THEN 1 END) Ids_having_null,  
       AVG(CASE WHEN COALESCE(t1.Name, t2.City, t2.State) IS NOT NULL THEN 1 ELSE 0 END) Percentage_missing
FROM Table1 t1 INNER JOIN Table2 t2
ON t2.B_ID = t1.ID;

参见demo。

Answer 2

如果您不知道 table 是否拥有所有 ID？
然后我建议 full join 在子查询中使用一些条件聚合。

例如：

select 
  Total as "Total_Count"  
, TotalMissing as "Ids having null" 
, (TotalMissing / Total)*100||'%' as "Percentage missing" 
from
(
    select 
      count(distinct coalesce(a.ID, b.B_ID)) as Total
    , count(distinct case
                     when a.name is null
                       or b.city is null
                       or b.state is null
                      then coalesce(a.ID, b.B_ID) 
                     end) as TotalMissing
    from TableA a
    full outer join TableB b 
      on a.ID = b.B_ID
) q

Total_Count | Ids having null | Percentage missing
----------: | --------------: | :-----------------
          4 |               3 | 75%

db<>fiddle here

Answer 3

试试这个 ->

select total_count, (total_count - cn) as ids_having_null, 
(total_count - cn) *100 / total_count as Percentage_missing 
FROM
(select count(t1.ID) as cn , t1.total_count 
 FROM ( select ID,Name, sum(tmp_col) over ( order by tmp_col) as total_count from  (select ID,Name, 1 as tmp_col  from tableA ) ) t1
 JOIN TableB t2
 ON t1.ID = t2.B_ID
 WHERE t1.Name is not null and t2.City is not null and t2.State is not null );

根据您对百分比或比率的要求，您可以更改 Percentage_missing 列的逻辑

查询以查找 ID 总数和在任何列中具有 null 的 ID 以及两者的百分比（即具有 null 的 ID 计数/ID 总数）

Query to find the count of total IDs, and IDs having null in any column and the percentage of the two (i.e. Count of ID having null /total ID count)

sql

oracle

average

count