问题
I would like to get all IDs from children in a tree with MySQL only.
I have a table like this:
ID parent_id name
1 0 cat1
2 1 subcat1
3 2 sub-subcat1
4 2 sub-subcat2
5 0 cat2
Now I'm trying to get all child IDs for cat1 (2,3,4) recursively. Is there any way how to achieve that?
回答1:
There are two basic methods for doing this: adjacency lists and nested lists. Take a look at Managing Hierarchical Data in MySQL.
What you have is an adjacency list. No there isn't a way of recursively grabbing all descendants with a single SQL statement. If possible, just grab them all and map them all in code.
Nested sets can do what you want but I tend to avoid it because the cost of inserting a record is high and it's error-prone.
回答2:
Here is a simple single-query MySql-solution:
SELECT GROUP_CONCAT(Level SEPARATOR ',') FROM (
SELECT @Ids := (
SELECT GROUP_CONCAT(`ID` SEPARATOR ',')
FROM `table_name`
WHERE FIND_IN_SET(`parent_id`, @Ids)
) Level
FROM `table_name`
JOIN (SELECT @Ids := <id>) temp1
) temp2
Just substitute <id> with the parent element's ID.
This will return a string with the IDs of all descendants of the element with ID = <id>, separated by ,. If you would rather have multiple rows returned, with one descendant on each row, you can use something like this:
SELECT *
FROM `table_name`
WHERE FIND_IN_SET(`ID`, (
SELECT GROUP_CONCAT(Level SEPARATOR ',') FROM (
SELECT @Ids := (
SELECT GROUP_CONCAT(`ID` SEPARATOR ',')
FROM `table_name`
WHERE FIND_IN_SET(`parent_id`, @Ids)
) Level
FROM `table_name`
JOIN (SELECT @Ids := <id>) temp1
) temp2
))
Including the root/parent element
The OP asked for the children of an element, which is answered above. In some cases it might be useful to include the root/parent element in the result. Here are my suggested solutions:
Comma-separated string of ids:
SELECT GROUP_CONCAT(Level SEPARATOR ',') FROM (
SELECT <id> Level
UNION
SELECT @Ids := (
SELECT GROUP_CONCAT(`ID` SEPARATOR ',')
FROM `table_name`
WHERE FIND_IN_SET(`parent_id`, @Ids)
) Level
FROM `table_name`
JOIN (SELECT @Ids := <id>) temp1
) temp2
Multiple rows:
SELECT *
FROM `table_name`
WHERE `ID` = <id> OR FIND_IN_SET(`ID`, (
SELECT GROUP_CONCAT(Level SEPARATOR ',') FROM (
SELECT @Ids := (
SELECT GROUP_CONCAT(`ID` SEPARATOR ',')
FROM `table_name`
WHERE FIND_IN_SET(`parent_id`, @Ids)
) Level
FROM `table_name`
JOIN (SELECT @Ids := <id>) temp1
) temp2
))
回答3:
You could probably do it with a stored procedure, if that's an option for you.
Otherwise you can't do it with a single sql-statement.
Ideally you should make the recursive calls to walk the tree from your program
回答4:
create table it should be look like below
DROP TABLE IF EXISTS `parent_child`;
CREATE TABLE `parent_child` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`name` varchar(255) DEFAULT NULL,
`parent_id` int(11) DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=9 DEFAULT CHARSET=latin1;
insert into `parent_child`(`id`,`name`,`parent_id`)
values (1,'cat1',0),(2,'subcat1',1),
(3,'sub-subcat1',2),(4,'sub-subcat2',2),
(5,'cat2',0);
Create function for getting parent child element
DELIMITER $$
USE `yourdatabase`$$
DROP FUNCTION IF EXISTS `GetAllNode1`$$
CREATE DEFINER=`root`@`localhost` FUNCTION `GetAllNode1`(GivenID INT) RETURNS TEXT CHARSET latin1
DETERMINISTIC
BEGIN
DECLARE rv,q,queue,queue_children TEXT;
DECLARE queue_length,front_id,pos INT;
SET rv = '';
SET queue = GivenID;
SET queue_length = 1;
WHILE queue_length > 0 DO
SET front_id = queue;
IF queue_length = 1 THEN
SET queue = '';
ELSE
SET pos = LOCATE(',',queue) + 1;
SET q = SUBSTR(queue,pos);
SET queue = q;
END IF;
SET queue_length = queue_length - 1;
SELECT IFNULL(qc,'') INTO queue_children
FROM (SELECT GROUP_CONCAT(id) AS qc
FROM `parent_child` WHERE `parent_id` = front_id) A ;
IF LENGTH(queue_children) = 0 THEN
IF LENGTH(queue) = 0 THEN
SET queue_length = 0;
END IF;
ELSE
IF LENGTH(rv) = 0 THEN
SET rv = queue_children;
ELSE
SET rv = CONCAT(rv,',',queue_children);
END IF;
IF LENGTH(queue) = 0 THEN
SET queue = queue_children;
ELSE
SET queue = CONCAT(queue,',',queue_children);
END IF;
SET queue_length = LENGTH(queue) - LENGTH(REPLACE(queue,',','')) + 1;
END IF;
END WHILE;
RETURN rv;
END$$
DELIMITER ;
write query for desire output
SELECT GetAllNode1(id) FROM parent_child
or
SELECT GetAllNode1(id) FROM parent_child where id =1 //for specific parent's child element
回答5:
Seeing that the answer is basically no or at least not very easy with a single MYSQL statement, I'll post my php/mysql code to do the hierarchy list..
function createCategorySubArray()
{
$categories = getSQL("SELECT pos_category_id FROM pos_categories");
for($i=0;$i<sizeof($categories);$i++)
{
//here we need to find all sub categories
$pos_category_id = $categories[$i]['pos_category_id'];
$cat_list[$pos_category_id] = recursiveCategory($pos_category_id,array());
}
return $cat_list;
}
function recursiveCategory($pos_category_id, $array)
{
$return = getSql("SELECT pos_category_id FROM pos_categories WHERE parent = $pos_category_id");
for($i=0;$i<sizeof($return);$i++)
{
$sub_cat = $return[$i]['pos_category_id'];
$array[] = $sub_cat;
$array = recursiveCategory($sub_cat, $array);
}
return $array;
}
Then you call it by $cat_array = createCategorySubArray();
I need this to find out which promotions based on product categories are being applied to the sub categories.
回答6:
Your question seems a bit imprecise. Why do you want to have them, and what do you mean by having them, "in a tree" ?
The table you've got IS (the relational way to represent) the tree.
If you want them "in a table" with rows that hold the pairs (ID 4 , ParentID 0), then you need your SQL engine's version of recursive SQL to do this, if that engine supports it.
I wouldn't know about MySQL specifically, but my understanding is that they once planned to implement recursive SQL using the same syntax as Oracle, i.e. with CONNECT BY.
If you look in your manual's table of contents for keywords such as "recursive queries" or "CONNECT BY", I imagine you should be able to find the answer.
(Sorry for not being able to provide a more ready-to-consume answer.)
来源：https://stackoverflow.com/questions/990529/how-to-find-all-ids-of-children-recursively