mysql的逆袭:如何做递归层次查询(二)
创建一个function getChildLst, 得到一个由所有子节点号组成的字符串.
mysql> delimiter //
mysql>
mysql> CREATE FUNCTION `getChildLst`(rootId INT)
-> RETURNS varchar(1000)
-> BEGIN
-> DECLARE sTemp VARCHAR(1000);
-> DECLARE sTempChd VARCHAR(1000);
->
-> SET sTemp = '$';
-> SET sTempChd =cast(rootId as CHAR);
->
-> WHILE sTempChd is not null DO
-> SET sTemp = concat(sTemp,',',sTempChd);
-> SELECT group_concat(id) INTO sTempChd FROM treeNodes where FIND_IN_SET(pid,sTempChd)>0;
-> END WHILE;
-> RETURN sTemp;
-> END
-> //
Query OK, 0 rows affected (0.00 sec)
mysql>
mysql> delimiter ;
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使用我们直接利用find_in_set函数配合这个getChildlst来查找
mysql> select getChildLst(1);
+-----------------+
| getChildLst(1) |
+-----------------+
| $,1,2,3,4,5,6,7 |
+-----------------+
1 row in set (0.00 sec)
mysql> select * from treeNodes
-> where FIND_IN_SET(id, getChildLst(1));
+----+----------+------+
| id | nodename | pid |
+----+----------+------+
| 1 | A | 0 |
| 2 | B | 1 |
| 3 | C | 1 |
| 4 | D | 2 |
| 5 | E | 2 |
| 6 | F | 3 |
| 7 | G | 6 |
+----+----------+------+
7 rows in set (0.01 sec)
mysql> select * from treeNodes
-> where FIND_IN_SET(id, getChildLst(3));
+----+----------+------+
| id | nodename | pid |
+----+----------+------+
| 3 | C | 1 |
| 6 | F | 3 |
| 7 | G | 6 |
+----+----------+------+
3 rows in set (0.01 sec)
--------------------------------------------
只要按我的做，百发百中弹无虚发，遇到问题万变不离其宗直接粘贴复制就是。。。
补充：
还可以做嵌套查询：
select id,pid from treeNodes where id in(
select id from treeNodes where FIND_IN_SET(id, getChildLst(3))
);
子查询的结果集是
www.2cto.com
+--------+
id
----
3
6
7
+-------+
然后经过外层查询就是
id pid
3 1
6 3
6 6
---------
好了 Perfect