MySQL选择所有行，但排除相同的值

Question

我在MySQL中的所有表格如下。

+---------+-------------+---------------+
| item_no | item_name   | item_location |
+---------+-------------+---------------+
| ITM145  |  laptop     | USA           |
| ITM146  |  camera     | USA           |
| ITM147  |  cd         | USA           |
| ITM148  |  phone      | USA           |
| ITM149  |  cd         | France        |
| ITM150  |  phone      | France        |
+---------+-------------+---------------+

最后一个拖曳项目的位置是法国，而我在美国拥有的相同项目，我希望查询返回给我，因为波纹管排除了位置法国的相同值。

+---------+-------------+---------------+
| item_no | item_name   | item_location |
+---------+-------------+---------------+
| ITM145  |  laptop     | USA           |
| ITM146  |  camera     | USA           |
+---------+-------------+---------------+

第一次尝试

SELECT * FROM table
WHERE item_location !='France'
GROUP BY item_name

失败第二次尝试

SELECT * FROM table
WHERE item_location NOT IN('France')
GROUP BY item_name

Answer 1

尝试使用<>运算符代替它（这是为了省略“法国”）

SELECT * FROM table
WHERE item_location <> 'France'
GROUP BY item_name

但是如果你想得到结果

SELECT * FROM table
WHERE item_location <> 'France' AND item_name IN ('laptop','camera')
GROUP BY item_name

要获得在法国排除item_name的结果，您可以尝试以下操作

SELECT * FROM table
WHERE item_location <> 'France' AND item_name NOT IN (Select item_name FROM 
table where item_location = 'France')
GROUP BY item_name

Answer 2

尝试这个：

SELECT * FROM table
GROUP BY item_name
Having count(*) = 1

Answer 3

将LEFT JOIN与self表一起使用，如下所示：

SELECT t.* FROM Table1 t
LEFT JOIN
(
  SELECT item_name FROM Table1 WHERE item_location = 'France'
) t2
ON t.item_name = t2.item_name
WHERE t2.item_name IS NULL

输出：

| item_no | item_name | item_location |
|---------|-----------|---------------|
|  ITM145 |    laptop |           USA |
|  ITM146 |    camera |           USA |

请参阅此SQLFiddle 。

要了解联接，请参阅《 SQL联接的直观说明》 。

Answer 4

您只想在其中使用“笔记本电脑”和照相机，请使用SELECT * FROM country WHERE item_location ="usa" AND (item_name = "laptop" OR item_name = "camera") GROUP BY item_no如果您想选择