一句话解决查找MySQL中已知记录的上一条和下一条记录
糊涂粥 于 2007-11-08 00:47:42
设MySQL中有一个形如下表的数据表,表名为test:
+------+------+---------------+
| ID | No | Other |
+------+------+---------------+
| 1 | 1 | Some others 1 |
| 2 | 2 | Some others 2 |
| 3 | 3 | Some others 3 |
| 10 | 4 | Some others 4 |
| 15 | 5 | Some others 5 |
| 16 | 6 | Some others 6 |
| 19 | 7 | Some others 7 |
+------+------+---------------+
其中,"ID"为每条记录的唯一ID(一般为自增字段),"No"为每条记录在表中对应的位置(为直观起见),"Other"为表中的其他信息(可为N个字段).
现在我们要查找"ID=3"的记录的"上一条"和"下一条"记录.
1.仅列出字段"ID"的值:
SELECT CASE WHEN SIGN(ID - 3) > 0 THEN 'Next' ELSE 'Prev' END AS DIR, CASE WHEN SIGN(ID - 3) > 0 THEN MIN(ID) WHEN SIGN(ID - 3) < 0 THEN MAX(ID) END AS ID FROM test WHERE ID <> 3 GROUP BY SIGN(ID - 3) ORDER BY SIGN(ID - 3);
结果:
+------+------+
| DIR | ID |
+------+------+
| PREV | 2 |
| NEXT | 10 |
+------+------+
2.列出所有字段值:
SELECT * FROM Test WHERE ID IN (SELECT CASE WHEN SIGN(ID - 3) > 0 THEN MIN(ID) WHEN SIGN(ID - 3) < 0 THEN MAX(ID) END AS ID FROM test WHERE ID <> 3 GROUP BY SIGN(ID - 3) ORDER BY SIGN(ID - 3)) ORDER BY ID ASC;
结果:
+------+------+---------------+
| ID | No | Other |
+------+------+---------------+
| 2 | 2 | Some others 2 |
| 10 | 4 | Some others 4 |
+------+------+---------------+
+------+------+---------------+
| ID | No | Other |
+------+------+---------------+
| 1 | 1 | Some others 1 |
| 2 | 2 | Some others 2 |
| 3 | 3 | Some others 3 |
| 10 | 4 | Some others 4 |
| 15 | 5 | Some others 5 |
| 16 | 6 | Some others 6 |
| 19 | 7 | Some others 7 |
+------+------+---------------+
其中,"ID"为每条记录的唯一ID(一般为自增字段),"No"为每条记录在表中对应的位置(为直观起见),"Other"为表中的其他信息(可为N个字段).
现在我们要查找"ID=3"的记录的"上一条"和"下一条"记录.
1.仅列出字段"ID"的值:
SELECT CASE WHEN SIGN(ID - 3) > 0 THEN 'Next' ELSE 'Prev' END AS DIR, CASE WHEN SIGN(ID - 3) > 0 THEN MIN(ID) WHEN SIGN(ID - 3) < 0 THEN MAX(ID) END AS ID FROM test WHERE ID <> 3 GROUP BY SIGN(ID - 3) ORDER BY SIGN(ID - 3);
结果:
+------+------+
| DIR | ID |
+------+------+
| PREV | 2 |
| NEXT | 10 |
+------+------+
2.列出所有字段值:
SELECT * FROM Test WHERE ID IN (SELECT CASE WHEN SIGN(ID - 3) > 0 THEN MIN(ID) WHEN SIGN(ID - 3) < 0 THEN MAX(ID) END AS ID FROM test WHERE ID <> 3 GROUP BY SIGN(ID - 3) ORDER BY SIGN(ID - 3)) ORDER BY ID ASC;
结果:
+------+------+---------------+
| ID | No | Other |
+------+------+---------------+
| 2 | 2 | Some others 2 |
| 10 | 4 | Some others 4 |
+------+------+---------------+
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