MySQL从第一列中选​​择在第二列中具有特定值的所有值

Question

我有一个包含2列的mysql表。

+---------+-----------+
| Barcode | StationID |
+---------+-----------+
|   89411 |         1 |
|   89411 |         2 |
|   89411 |         3 |
|   89412 |         1 |
|   89413 |         1 |
+---------+-----------+

我想从“条形码”列中选择所有StationID = 1且不具有不同于1的StationID的值。

如图所示，条形码89411出现了3次，但StationID不同，应该从结果中排除。

您能帮我查询吗？

Answer 1

另一种方法是使用EXISTS查询：

SELECT t1.*
FROM yourTable t1
WHERE
    t1.StationID = 1 AND
    NOT EXISTS (SELECT 1 FROM yourTable t2
                WHERE t1.Barcode = t2.Barcode AND t2.StationID <> 1);

Demo

Answer 2

尝试一下：https://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=641d334c5f9e57bbdde07e4f24365f88

select barcode from tablename
group by barcode
having sum(case when sanctionid=1 then 0 else 1 end)=0

输出：

barcode
89412
89413

Answer 3

使用聚合函数GROUP_CONCAT，并使用HAVING子句过滤掉只有一个StationID即'1'的那些条形码：

SELECT barcode, GROUP_CONCAT(DISTINCT StationID) AS stations 
FROM table_name 
GROUP BY barcode 
HAVING stations = '1';