Question

鉴于此表：

+----+-----------+--------+
| id | condition | values |
+----+-----------+--------+
| 1  | a         | 1      |
+----+-----------+--------+
| 2  | a         | 2      |
+----+-----------+--------+
| 3  | a         | 3      |
+----+-----------+--------+
| 4  | a         | 4      |
+----+-----------+--------+
| 5  | b         | 5      |
+----+-----------+--------+
| 6  | b         | 6      |
+----+-----------+--------+

如何获得一个以id=3开头的新表（包括），直到condition = b（不包括）：

+----+-----------+--------+
| id | condition | values |
+----+-----------+--------+
| 3  | a         | 3      |
+----+-----------+--------+
| 4  | a         | 4      |
+----+-----------+--------+

添加了小提琴： http://sqlfiddle.com/#!2/9882f7

基本上我想要一个匹配的第一个条件（通过特定的列 - id）和第二个（通过不同的列 - 条件）之间的表

Answer 1

您需要停止将SQL数据视为有任何订单。考虑集合中的SQL数据;您必须按值搜索，而不是按位置搜索。

SELECT t1.*
FROM t AS t1
JOIN (
    SELECT MIN(id) AS id FROM t
    WHERE id >= 3 AND `condition` = 'b'
) AS t2
WHERE t1.id >= 3 AND t1.id < t2.id
ORDER BY t1.id

Answer 2

这样的事情：

select t.*
from table t
where id >= 3 and id < (select min(t2.id) from table t2 where t2.condition = 'b');

编辑：

此查询在SQL Fiddle上运行良好：

select t.*
from  t
where id >= 3 and id < (select min(t2.id) from t t2 where t2.condition = 'b');

Answer 3

如果我理解你的要求，我相信这对你有用：

SELECT id, condition, values
FROM tableName
WHERE id > 2
AND condition != b
ORDER BY id

我希望这适合你。

MySQL从特定ID中选择直到匹配条件

3 个答案: