我使用php制作目录树列表。目录（数据）来自MySQL数据库表，但我不想写sql来获取我想要的顺序的记录：
创建表的sql：
CREATE TABLE section (
section_id tinyint(4) NOT NULL auto_increment,
name varchar(500) default NULL,
parent_id tinyint(4) default NULL,
lineage varchar(45) default NULL,
level tinyint(4) default NULL,
PRIMARY KEY (section_id)
) ENGINE=MyISAM AUTO_INCREMENT=15 DEFAULT CHARSET=utf8;
插入数据的SQL：
INSERT INTO section (
section_id,name,lineage,parent_id,level)
VALUES
('1', 'Dashboard', '1', '0', '1'),
('2', 'Slider', '2', '0', '1'),
('3', 'Column', '3', '0', '1'),
('4', 'Column list', '3-4', '3', '2'),
('5', 'Add column', '3-5', '3', '2'),
('6', 'Permission', '6', '0', '1'),
('7', 'Permission Group', '6-7', '6', '2'),
('8', 'User List', '6-8', '6', '2'),
('9', 'Section permission', '6-9', '6', '2'),
('10', 'Add permission', '6-7-10', '7', '3'),
('11', 'Add user', '6-8-11', '8', '3'),
('12', 'Add section', '6-9-12', '9', '3');
表格的所有记录：
section_id name parent_id level
-----------------------------------------------------
1 Dashboard 0 1
2 Slider 0 1
3 Column 0 1
4 Column list 3 2
5 Add column 3 2
6 Permission 0 1
7 Permission Group 6 2
8 User List 6 2
9 Section permission 6 2
10 Add permission 7 3
11 Add user 8 3
12 Add section 9 3
我想使用/ some SQL按此顺序获取行：
section_id name parent_id level
-----------------------------------------------------
1 Dashboard 0 1
2 Slider 0 1
3 Column 0 1
4 Column list 3 2
5 Add column 3 2
6 Permission 0 1
7 Permission Group 6 2
10 Add permission 7 3
8 User List 6 2
11 Add user 8 3
9 Section permission 6 2
12 Add section 9 3
制作此目录树：
- Dashboard
- Slider
- Column
- Column list
- Add column
- Permission
- Permission Group
- Add permission
- User List
- Add user
- Section permission
- Add Section
我有一个想法，使用许多SQL来获取目录并将数据存储到php对象中。 这是获取子目录的SQL：
SELECT * FROM tbl_section AS a1 WHERE a1.parent_id = 0;
SELECT * FROM tbl_section AS a1 WHERE a1.parent_id = 1;
SELECT * FROM tbl_section AS a1 WHERE a1.parent_id = 2;
SELECT * FROM tbl_section AS a1 WHERE a1.parent_id = 3;
等等。
但是，如果有100个目录，它需要运行select sql 100次， 我认为这不是一个好方法，其他人都不知道吗？
答案 0 :(得分：1)
这个方法（Lineage Column）可能不是列出从MySQL表中获取数据的目录树的最佳和最动态的方法，但这非常容易和快速：
stackoverflow中的相关问题：
What are the options for storing hierarchical data in a relational database?
参考文章：
谱系列（a.k.a.物化路径，路径枚举）
http://www.ferdychristant.com/blog/articles/DOMM-7QJPM7
这是使用＆＃34; Lineage Column＆＃34;选择SQL。方法：
SELECT c.section_id, c.name, c.lineage, c.level, c.parent_id,
(SELECT COUNT(*) FROM section
where section.lineage
LIKE (CONCAT(c.lineage,'%')) AND
section.lineage!=c.lineage) as replies
FROM section as c
order by c.lineage;
它返回包含我想要的记录的记录：
section_id name lineage level parent_id replies
----------------------------------------------------------------------
1 Dashboard 1 1 0 0
2 Slider 2 1 0 0
3 Column 3 1 0 2
4 Column list 3-4 2 3 0
5 Add column 3-5 2 3 0
6 Permission 6 1 0 6
7 Permission Group 6-7 2 6 1
10 Add permission 6-7-10 3 7 0
8 User List 6-8 2 6 1
11 Add user 6-8-11 3 8 0
9 Section permission 6-9 2 6 1
12 Add section 6-9-12 3 9 0
这是回复问题中提到的目录的PHP代码：
<?php
CONST SERVERNAME = ""; //your mysql server name
CONST USERNAME = ""; //your mysql user name
CONST PASSWORD = ""; //your mysql password
CONST DATABASE = ""; //your mysql database name
// Create connection
$conn = new mysqli(SERVERNAME, USERNAME, PASSWORD, DATABASE);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT c.section_id, c.name, c.lineage, c.level, c.parent_id,
(SELECT COUNT(*) FROM section
where section.lineage
LIKE (CONCAT(c.lineage,'%')) AND
section.lineage!=c.lineage) as replies
FROM section as c
order by c.lineage";
$html = "";
foreach ($conn->query($sql) as $row) {
switch ($row['level']){
case "1": $html.="-".$row['name'] . "<br />"; break;
case "2": $html.=" -".$row['name'] . "<br />"; break;
case "3": $html.=" -".$row['name'] . "<br />"; break;
}
}
print $html;
?>
php代码的结果：
-Dashboard
-Slider
-Column
-Column list
-Add column
-Permission
-Permission Group
-Add permission
-User List
-Add user
-Section permission
-Add section