员工培训分析查询条件查询答案

【问题标题】：Conditional Query for Employee Training Analysis Query员工培训分析查询条件查询
【发布时间】：2011-07-07 03:18:43
【问题描述】：

我需要运行培训查询以返回以下问题的结果：“谁完成了这次培训，但没有完成那次培训？”

在下面的简化表中，我想知道哪个员工已完成 training_id 1（如 completed_date 字段中的日期所示），但尚未完成 training_id 7。

+-------------+-------------+----------------+
| emp_id      | training_id | completed_date |
+-------------+-------------+----------------+
| 1           | 1           | 2010-04-02     |
+-------------+-------------+----------------+
| 1           | 7           | Null           |
+-------------+-------------+----------------+
| 2           | 1           | Null           |
+-------------+-------------+----------------+
| 2           | 7           | Null           |
+-------------+-------------+----------------+

期望的结果是 emp_id 1，我们希望根据查询参数返回他/她完成的训练和未完成的训练：

+-------------+-------------+----------------+
| emp_id      | training_id | completed_date |
+-------------+-------------+----------------+
| 1           | 1           | 2010-04-02     |
+-------------+-------------+----------------+
| 1           | 7           | Null           |
+-------------+-------------+----------------+

我不知道如何使用常规查询来执行此操作，因为它似乎需要 IF 逻辑。例如：返回第一次训练完成的行并返回第二次训练未完成的行，但仅在第一次训练完成时返回。

我如何在 SQL 中表达类似的东西？

【问题讨论】：

标签： mysql sql conditional

【解决方案1】：

您可以使用 EXISTS 子句

SELECT t.*
FROM training t

# which employee has completed training_id 1
WHERE t.training_id = 1 and t.completed_date is not null

#but has not finished training_id 7.
AND NOT EXISTS (
     SELECT * FROM training t2
     where t2.emp_id=t.emp_id
       and t2.training_id = 7
       and t2.completed_date is not null)

如果你想测试更复杂的东西，比如completed (4,5,6) but not (1,9)，那么你可以使用计数：

SELECT t.emp_id
FROM training t
WHERE t.training_id in (4,5,6) and t.completed_date is not null
group by t.emp_id
having count(distinct emp_id) = 3

AND NOT EXISTS (
     SELECT * FROM training t2
     where t2.emp_id=t.emp_id
       and t2.training_id in (1,9)
       and t2.completed_date is not null)

最后，如果您需要完整的员工培训记录

SELECT e.*
FROM
(
    SELECT t.emp_id
    FROM training t
    WHERE t.training_id in (4,5,6) and t.completed_date is not null
    group by t.emp_id
    having count(distinct emp_id) = 3

    AND NOT EXISTS (
         SELECT * FROM training t2
         where t2.emp_id=t.emp_id
           and t2.training_id in (1,9)
           and t2.completed_date is not null)
) search
inner join training e on e.emp_id = search.emp_id
order by e.emp_id

【讨论】：

【解决方案2】：

您可以在此表上进行自联接（“假设它是两个相同的表并加入它们”）：

SELECT t1.emp_id, t1.training_id, t1.completed_date,
       t2.training_id, t2.completed_date
    FROM training AS t1 /* the aliases are needed for the self-join */
    JOIN training AS t2
        ON t1.emp_id = t2.emp_id
        AND t2.training_id = 7 /* second training  ID */
    WHERE t1.training_id = 1 /* first training ID */

这应该会给你这样的结果：

t1.emp_id | t1.training_id | t1.completed_date | t2.training_id | t2.completed_date
 1            1               2010-04-02          7                NULL
 2            1               NULL                7                NULL

然后您可以进一步限制 WHERE 查询，例如与

         AND t1.completed_date IS NOT NULL
         AND t2.completed_date IS NULL

这将为您提供所需的集合 - 训练 1 完成，训练 7 未完成。

【讨论】：

【解决方案3】：

select * from training where emp_id in
(
  select distinct emp_id from training where completed_Date is not null
)

【讨论】：