SQL递归表 [英] SQL Recursive Tables
问题描述
我有以下表，groups表包含按层次排序的组，而group_member表则存储用户所属的组.
groups
---------
id
parent_id
name
group_member
---------
id
group_id
user_id
ID PARENT_ID NAME
---------------------------
1 NULL Cerebra
2 1 CATS
3 2 CATS 2.0
4 1 Cerepedia
5 4 Cerepedia 2.0
6 1 CMS
ID GROUP_ID USER_ID
---------------------------
1 1 3
2 1 4
3 1 5
4 2 7
5 2 6
6 4 6
7 5 12
8 4 9
9 1 10
我想检索给定用户的可见组.就是说用户所属的组以及这些组的子级.例如，使用上面的数据:
USER VISIBLE_GROUPS
9 4, 5
3 1,2,4,5,6
12 5
我正在使用递归和几个数据库查询来获取这些值.但是我想知道是否可以通过单个SQL查询来提高我的应用程序性能.我正在使用MySQL.
我想到两件事:
1-，您可以反复将表外部连接到自身，以递归地沿树走，如:
SELECT *
FROM
MY_GROUPS MG1
,MY_GROUPS MG2
,MY_GROUPS MG3
,MY_GROUPS MG4
,MY_GROUPS MG5
,MY_GROUP_MEMBERS MGM
WHERE MG1.PARENT_ID = MG2.UNIQID (+)
AND MG1.UNIQID = MGM.GROUP_ID (+)
AND MG2.PARENT_ID = MG3.UNIQID (+)
AND MG3.PARENT_ID = MG4.UNIQID (+)
AND MG4.PARENT_ID = MG5.UNIQID (+)
AND MGM.USER_ID = 9
那会给你这样的结果:
UNIQID PARENT_ID NAME UNIQID_1 PARENT_ID_1 NAME_1 UNIQID_2 PARENT_ID_2 NAME_2 UNIQID_3 PARENT_ID_3 NAME_3 UNIQID_4 PARENT_ID_4 NAME_4 UNIQID_5 GROUP_ID USER_ID
4 2 Cerepedia 2 1 CATS 1 null Cerebra null null null null null null 8 4 9
这里的限制是，您必须为要沿树走的每个级别"添加一个新的联接.如果您的树少于20个级别，则可以通过创建一个显示每个用户20个级别的视图来摆脱它.
2-我知道的唯一其他方法是创建一个递归数据库函数，然后从代码中调用它.这样，您仍然会有一些查找开销(即，您的查询数量仍将等于您在树上行走的级别数量)，但是总体而言，它应该更快，因为它们全部在数据库中进行. /p>
我不确定MySql，但是在Oracle中，此功能类似于此功能(您必须更改表名和字段名；我只是复制过去所做的事情):
CREATE OR REPLACE FUNCTION GoUpLevel(WO_ID INTEGER, UPLEVEL INTEGER) RETURN INTEGER
IS
BEGIN
DECLARE
iResult INTEGER;
iParent INTEGER;
BEGIN
IF UPLEVEL <= 0 THEN
iResult := WO_ID;
ELSE
SELECT PARENT_ID
INTO iParent
FROM WOTREE
WHERE ID = WO_ID;
iResult := GoUpLevel(iParent,UPLEVEL-1); --recursive
END;
RETURN iResult;
EXCEPTION WHEN NO_DATA_FOUND THEN
RETURN NULL;
END;
END GoUpLevel;
/
I have the following tables, the groups table which contains hierarchically ordered groups and group_member which stores which groups a user belongs to.
groups
---------
id
parent_id
name
group_member
---------
id
group_id
user_id
ID PARENT_ID NAME
---------------------------
1 NULL Cerebra
2 1 CATS
3 2 CATS 2.0
4 1 Cerepedia
5 4 Cerepedia 2.0
6 1 CMS
ID GROUP_ID USER_ID
---------------------------
1 1 3
2 1 4
3 1 5
4 2 7
5 2 6
6 4 6
7 5 12
8 4 9
9 1 10
I want to retrieve the visible groups for a given user. That it is to say groups a user belongs to and children of these groups. For example, with the above data:
USER VISIBLE_GROUPS
9 4, 5
3 1,2,4,5,6
12 5
I am getting these values using recursion and several database queries. But I would like to know if it is possible to do this with a single SQL query to improve my app performance. I am using MySQL.
Two things come to mind:
1 - You can repeatedly outer-join the table to itself to recursively walk up your tree, as in:
SELECT *
FROM
MY_GROUPS MG1
,MY_GROUPS MG2
,MY_GROUPS MG3
,MY_GROUPS MG4
,MY_GROUPS MG5
,MY_GROUP_MEMBERS MGM
WHERE MG1.PARENT_ID = MG2.UNIQID (+)
AND MG1.UNIQID = MGM.GROUP_ID (+)
AND MG2.PARENT_ID = MG3.UNIQID (+)
AND MG3.PARENT_ID = MG4.UNIQID (+)
AND MG4.PARENT_ID = MG5.UNIQID (+)
AND MGM.USER_ID = 9
That's gonna give you results like this:
UNIQID PARENT_ID NAME UNIQID_1 PARENT_ID_1 NAME_1 UNIQID_2 PARENT_ID_2 NAME_2 UNIQID_3 PARENT_ID_3 NAME_3 UNIQID_4 PARENT_ID_4 NAME_4 UNIQID_5 GROUP_ID USER_ID
4 2 Cerepedia 2 1 CATS 1 null Cerebra null null null null null null 8 4 9
The limit here is that you must add a new join for each "level" you want to walk up the tree. If your tree has less than, say, 20 levels, then you could probably get away with it by creating a view that showed 20 levels from every user.
2 - The only other approach that I know of is to create a recursive database function, and call that from code. You'll still have some lookup overhead that way (i.e., your # of queries will still be equal to the # of levels you are walking on the tree), but overall it should be faster since it's all taking place within the database.
I'm not sure about MySql, but in Oracle, such a function would be similar to this one (you'll have to change the table and field names; I'm just copying something I did in the past):
CREATE OR REPLACE FUNCTION GoUpLevel(WO_ID INTEGER, UPLEVEL INTEGER) RETURN INTEGER
IS
BEGIN
DECLARE
iResult INTEGER;
iParent INTEGER;
BEGIN
IF UPLEVEL <= 0 THEN
iResult := WO_ID;
ELSE
SELECT PARENT_ID
INTO iParent
FROM WOTREE
WHERE ID = WO_ID;
iResult := GoUpLevel(iParent,UPLEVEL-1); --recursive
END;
RETURN iResult;
EXCEPTION WHEN NO_DATA_FOUND THEN
RETURN NULL;
END;
END GoUpLevel;
/
这篇关于SQL递归表的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
