比较select 上同table (mysql)

Question

我真的是 mysql 的新手，我正在尝试做一个简单的 select，但我无法弄清楚。

这就是我所拥有的：

I got this in a Table named Control:
| CODE | OFFICE |
|  1   |   usa  |
|  2   |   usa  |
|  3   |   usa  |
|  4   |   usa  |
|  5   |   usa  |
|  1   | china  |
|  3   | china  |
|  4   | china  |

And I need get this:

| CODE | OFFICE |
|  2   |   usa  |
|  5   |   usa  |

然后，SELECT code,office WHERE codes are still not registered with office = china.

我必须进行自连接或类似的操作或使用 GROUP BY 语句吗？我被困在这里...我真的很感激任何帮助。

Answer 1

我猜这是可行的

 create table Test(id integer, code integer, office varchar(100));

insert into Test(id, code, office) values(1, 1, "usa"),(2, 2, "usa"),(3, 3, "usa"),(4, 4, "usa"),(5, 5, "usa"),(6, 1, "china"),(7, 3, "china"),(8, 4, "china");

select * from Test;

Select * from Test where code NOT IN (select code from Test where office =   "china");

您已经使用了子查询。

Answer 2

您可以在 Code 上执行 "Self-Left-Join"，并仅考虑右侧未找到匹配项的那些行，即，右侧值 IS NULL

SELECT 
  tleft.*
FROM Control AS tleft
LEFT JOIN Control AS tright
  ON tright.Code = tleft.Code AND 
     tright.Office = 'china' 
WHERE tleft.Office = 'usa' AND 
      tright.Code IS NULL -- this filters no matching code found in china

比较select 上同table (mysql)

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