对另一个问题也是这样吗

Mysql选择递归获取所有子级别

查询将是:

SELECT GROUP_CONCAT(lv SEPARATOR ',') FROM (
  SELECT @pv:=(
    SELECT GROUP_CONCAT(id SEPARATOR ',')
    FROM table WHERE parent_id IN (@pv)
  ) AS lv FROM table 
  JOIN
  (SELECT @pv:=1)tmp
  WHERE parent_id IN (@pv)
) a;

这是一个有点棘手的问题，检查一下它是否适合你

select a.id,if(a.parent = 0,@varw:=concat(a.id,','),@varw:=concat(a.id,',',@varw)) as list from (select * from recursivejoin order by if(parent=0,id,parent) asc) a left join recursivejoin b on (a.id = b.parent),(select @varw:='') as c  having list like '%19,%';

SQL小提琴链接http://www.sqlfiddle.com/#!2 / e3cdf / 2

用字段名和表名替换。

我发现更容易做到:

1)创建一个函数，检查一个项目是否在另一个项目的父层次结构中的任何地方。就像这样(我不会写函数，用WHILE DO):

is_related(id, parent_id);

在你的例子中

is_related(21, 19) == 1;
is_related(20, 19) == 1;
is_related(21, 18) == 0;

2)使用子选择，就像这样:

select ...
from table t
join table pt on pt.id in (select i.id from table i where is_related(t.id,i.id));

我向你提出了一个问题。这将给你递归类别与一个单一的查询:

SELECT id,NAME,'' AS subName,'' AS subsubName,'' AS subsubsubName FROM Table1 WHERE prent is NULL
UNION 
SELECT b.id,a.name,b.name AS subName,'' AS subsubName,'' AS subsubsubName FROM Table1 AS a LEFT JOIN Table1 AS b ON b.prent=a.id WHERE a.prent is NULL AND b.name IS NOT NULL 
UNION 
SELECT c.id,a.name,b.name AS subName,c.name AS subsubName,'' AS subsubsubName FROM Table1 AS a LEFT JOIN Table1 AS b ON b.prent=a.id LEFT JOIN Table1 AS c ON c.prent=b.id WHERE a.prent is NULL AND c.name IS NOT NULL 
UNION 
SELECT d.id,a.name,b.name AS subName,c.name AS subsubName,d.name AS subsubsubName FROM Table1 AS a LEFT JOIN Table1 AS b ON b.prent=a.id LEFT JOIN Table1 AS c ON c.prent=b.id LEFT JOIN Table1 AS d ON d.prent=c.id WHERE a.prent is NULL AND d.name IS NOT NULL 
ORDER BY NAME,subName,subsubName,subsubsubName

这是一把小提琴。

这是一个分类表。

SELECT  id,
        NAME,
        parent_category 
FROM    (SELECT * FROM category
         ORDER BY parent_category, id) products_sorted,
        (SELECT @pv := '2') initialisation
WHERE   FIND_IN_SET(parent_category, @pv) > 0
AND     @pv := CONCAT(@pv, ',', id)

输出:

使用BlueM/tree php类在mysql中创建一个自关系表的树。

Tree和Tree\Node是PHP类，用于处理使用父ID引用分层结构的数据。一个典型的例子是关系数据库中的一个表，其中每个记录的“父”字段引用另一个记录的主键。当然，Tree不能只使用来自数据库的数据，而是使用任何数据:您提供数据，Tree使用它，而不管数据来自何处以及如何处理。阅读更多

下面是一个使用BlueM/tree的例子:

<?php 
require '/path/to/vendor/autoload.php'; $db = new PDO(...); // Set up your database connection 
$stm = $db->query('SELECT id, parent, title FROM tablename ORDER BY title'); 
$records = $stm->fetchAll(PDO::FETCH_ASSOC); 
$tree = new BlueM\Tree($records); 
...