MySQL不在哪里

Question

我有一个名为“ Project”的表，其中包含以下项目数据。

+--------+------------+------------+----------+--------+---------------+--------+
| In Num | pro_no_old | pro_no_new | pro_name | amount | budget_status | status |
+--------+------------+------------+----------+--------+---------------+--------+
|      1 | pro1       |            | project1 | 200000 | original      |      1 |
|      2 | pro2       |            | project2 | 150000 | original      |      1 |
|      3 | pro3       |            | project3 | 300000 | original      |      1 |
|      4 | pro2       | pro4       | project4 | 100000 | revised       |      0 |
+--------+------------+------------+----------+--------+---------------+--------+

02）pro_name“ Project4”已通过修改从Project2变为。然后，我需要获得没有修订项目的项目，例如Project2。换句话说，无需获得先前项目的修订版。期望的投放可能如下。

+--------+------------+------------+----------+--------+---------------+--------+
| In Num | pro_no_old | pro_no_new | pro_name | amount | budget_status | status |
+--------+------------+------------+----------+--------+---------------+--------+
|      1 | pro1       |            | project1 | 200000 | original      |      1 |
|      3 | pro3       |            | project3 | 300000 | original      |      1 |
|      4 | pro2       | pro4       | project4 | 100000 | revised       |      0 |
+--------+------------+------------+----------+--------+---------------+--------+

03）我使用以下查询来获得所需的输出。

SELECT * FROM project WHERE('pro_no_old in (SELECT  pro_no_old FROM project where budget_status = "original"))

04）但是没有得到预期的结果。谁能帮我 ？

编辑后所需的输出

+--------+------------+------------+----------+--------+---------------+--------+
| In Num | pro_no_old | pro_no_new | pro_name | amount | budget_status | status |
+--------+------------+------------+----------+--------+---------------+--------+
|      1 | pro1       |            | project1 | 200000 | original      |      1 |
|      3 | pro3       |            | project3 | 300000 | original      |      1 |
|      5 | pro4       | pro5       | project5 |  50000 | revised       |      0 |
+--------+------------+------------+----------+--------+---------------+--------+

Answer 1

我不确定您的数据模型，但是如果您只对pro_no_old的最后一个条目感兴趣

select * 
from t
where innum = (select max(innum) from t t1 where t1.pro_no_old = t.pro_no_old); 

+-------+------------+------------+----------+--------+---------------+--------+
| InNum | pro_no_old | pro_no_new | pro_name | amount | budget_status | status |
+-------+------------+------------+----------+--------+---------------+--------+
|     1 | pro1       | NULL       | project1 | 200000 | original      |      1 |
|     3 | pro3       | NULL       | project3 | 300000 | original      |      1 |
|     4 | pro2       | pro4       | project4 | 100000 | revised       |      0 |
+-------+------------+------------+----------+--------+---------------+--------+
3 rows in set (0.00 sec)