相关子查询上的过滤器返回 null oracle

Question

我有以下查询，它将子表记录组合为逗号分隔的字符串。但是当我尝试将过滤器应用于相关子查询时，值将替换为 null 而不是删除记录。

父表：员工

身份证	姓名
I1BF91860BCB711EDBF4EF0FBCBCA9E17	蒂姆
I1BF91860BCB711EDBF4EF0FBCBCA9E18	山姆
I1BF91860BCB711EDBF4EF0FBCBCA9E19	罗汉

子表：培训

身份证	姓名	员工编号
I1BF91860BCB711EDBF4EF0FBCBCA9P10	Java 基础知识	I1BF91860BCB711EDBF4EF0FBCBCA9E17
I1BF91860BCB711EDBF4EF0FBCBCA9P11	角度	I1BF91860BCB711EDBF4EF0FBCBCA9E17
I1BF91860BCB711EDBF4EF0FBCBCA9P12	点网	I1BF91860BCB711EDBF4EF0FBCBCA9E18

下面是查询，其输出如下所示

select e."Id" as Id
, e."Name" as Name
, (SELECT listagg(tr."Name", ',') within group(order by tr."Id") 
  FROM trainings tr WHERE e."Id" = tr."Employee Id")
  trainingss
from employee e
 

输出：

如图所示，子值按预期分组为逗号分隔的字符串，但我在过滤子值时遇到问题。

如何使用子查询对子列值应用过滤器。假设我想用 JavaBasics 和 Angular 显示行，预期应该是第一行。但在下面的输出图像中，我可以看到所有行都显示为空值。

我尝试了什么？

select e."Id" as Id
, e."Name" as Name
, (SELECT listagg(tr."Name", ',') within group(order by tr."Id") 
  FROM trainings tr WHERE e."Id" = tr."Employee Id" and tr."Id" = 'I1BF91860BCB711EDBF4EF0FBCBCA9P10')
  trainingss
from employee e

输出：

请注意，上面的示例只是我们需求的一个示例，在我们的实际案例场景中，我们有超过 150 列，而且都是动态的。因此我们不想使用 GroupBy

Answer 1

您可以对每个条件使用存在，或者计算接受过培训次数的员工（如果他们与您搜索的培训次数完全相同）

select e."Id" as Id
, e."Name" as Name
, (SELECT listagg(tr."Name", ',') within group(order by tr."Id") 
  FROM trainings tr WHERE e."Id" = tr."Employee Id"
  )
  trainings
from employee e 
WHERE EXISTS (SELECT 1 FROM trainings tr1 WHERE e."Id" = tr1."Employee Id"
  AND tr1."Id" IN ('I1BF91860BCB711EDBF4EF0FBCBCA9P10'))

身份证	姓名	培训
I1BF91860BCB711EDBF4EF0FBCBCA9E17	蒂姆	Java 基础知识，Angular

select e."Id" as Id
, e."Name" as Name
, (SELECT listagg(tr."Name", ',') within group(order by tr."Id") 
  FROM trainings tr WHERE e."Id" = tr."Employee Id"
  )
  trainings
from employee e 
WHERE EXISTS (SELECT 1 FROM trainings tr1 WHERE e."Id" = tr1."Employee Id"
  AND tr1."Id" IN ('I1BF91860BCB711EDBF4EF0FBCBCA9P11'))
AND EXISTS (SELECT 1 FROM trainings tr1 WHERE e."Id" = tr1."Employee Id"
  AND tr1."Id" IN ('I1BF91860BCB711EDBF4EF0FBCBCA9P10'))

身份证	姓名	培训
I1BF91860BCB711EDBF4EF0FBCBCA9E17	蒂姆	Java 基础知识，Angular

select e."Id" as Id
, e."Name" as Name
, (SELECT listagg(tr."Name", ',') within group(order by tr."Id") 
  FROM trainings tr WHERE e."Id" = tr."Employee Id"
  )
  trainings
from employee e 
WHERE e."Id" IN
  (SELECT "Employee Id" FROM trainings 
  WHERE "Id" IN ('I1BF91860BCB711EDBF4EF0FBCBCA9P10','I1BF91860BCB711EDBF4EF0FBCBCA9P11')
  GROUP BY "Employee Id"
HAVING COUNT(*) = 2)

身份证	姓名	培训
I1BF91860BCB711EDBF4EF0FBCBCA9E17	蒂姆	Java 基础知识，Angular

小提琴