回答編集履歴
2
修正
answer
CHANGED
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@@ -10,7 +10,7 @@
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10
10
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(3,'タイトルC',null),
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11
11
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(4,'タイトルA-1',1),
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12
12
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(5,'タイトルA-1-1',4),
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13
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-
(6,'タイトルA-1-2',
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13
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+
(6,'タイトルA-1-2',4),
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14
14
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(7,'タイトルA-2',1);
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15
15
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```
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16
16
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# procedure
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1
参考
answer
CHANGED
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@@ -1,2 +1,60 @@
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1
1
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いまのデータ管理方法では何をやっても無駄でしょう
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2
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-
ちゃんとやるなら入れ子集合モデルという方式をつかいます。
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2
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+
ちゃんとやるなら入れ子集合モデルという方式をつかいます。
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3
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+
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4
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+
# sample
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5
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+
```SQL
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6
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+
create table tbl (id int primary key,name varchar(30),parent_id int,level int,l int,r int);
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7
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+
insert into tbl(id,name,parent_id) values
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8
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+
(1,'タイトルA',null),
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9
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+
(2,'タイトルB',null),
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10
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+
(3,'タイトルC',null),
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11
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+
(4,'タイトルA-1',1),
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12
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+
(5,'タイトルA-1-1',4),
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13
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+
(6,'タイトルA-1-2',5),
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14
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+
(7,'タイトルA-2',1);
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15
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+
```
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16
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+
# procedure
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17
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+
```SQL
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18
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DROP PROCEDURE IF EXISTS SET_LR;
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19
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DELIMITER //
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20
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CREATE PROCEDURE SET_LR()
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21
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BEGIN
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22
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DECLARE a INT DEFAULT 0;
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23
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DECLARE done INT DEFAULT 0;
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24
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DECLARE CUR CURSOR FOR
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25
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SELECT id FROM tbl WHERE level=0 ORDER BY parent_id ASC,id DESC;
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26
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+
DECLARE CONTINUE HANDLER FOR SQLSTATE '02000' SET done = 1;
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27
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+
UPDATE tbl SET level=0,l=0,r=0;
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28
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+
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29
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+
UPDATE tbl SET level=1,l=(SELECT @a:=@a+1 FROM (SELECT @a:=0) AS sub),r=@a:=@a+1
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30
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+
WHERE parent_id IS NULL
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31
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+
ORDER BY id;
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32
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+
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33
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+
OPEN CUR;
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34
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+
REPEAT
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35
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FETCH CUR INTO a;
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36
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+
IF NOT done THEN
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37
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SET @id=a;
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38
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+
SET @sql='UPDATE tbl as a1,tbl as a2,tbl as a3 SET a1.l=a2.l+1,a1.r=a2.l+2,a1.level=a2.level+1,a2.r=a2.r+2,a3.r=a3.r+2 WHERE a1.parent_id=a2.id AND a2.l<a3.r AND a1.id=?';
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39
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+
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40
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+
PREPARE stmt from @sql;
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41
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+
EXECUTE stmt USING @id;
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42
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+
SET @sql='UPDATE tbl as a1,tbl as a2,tbl as a3 SET a3.l=a3.l+2 WHERE a1.parent_id=a2.id AND a2.l<a3.l and a3.id!=a1.id AND a1.id=?';
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43
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+
|
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44
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+
PREPARE stmt from @sql;
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45
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+
EXECUTE stmt USING @id;
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46
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+
END IF;
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47
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+
UNTIL done END REPEAT;
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48
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+
CLOSE CUR;
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49
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+
END
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50
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+
//
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51
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+
DELIMITER ;
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52
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+
```
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53
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+
# 実行
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54
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+
```SQL
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55
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+
call set_lr;
|
|
56
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+
```
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57
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+
# 表示
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|
58
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+
```SQL
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|
59
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+
select * from tbl order by l;
|
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60
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+
```
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