I have a column id
, a column parent
and a column path
that is a materialized path.
It looks like
1 | \N | 1
2 | 1 | 1/2
3 | 2 | 1/2/3
4 | 3 | 1/2/3/4
5 | 3 | 1/2/3/5
6 | 2 | 1/2/6
7 | 6 | 1/2/6/7
8 | 2 | 1/2/8
9 | 1 | 1/9
10 | 9 | 1/9/10
11 | 10 | 1/9/10/11 开发者_JAVA百科
12 | 11 | 1/9/10/11/12
13 | 11 | 1/9/10/11/13
14 | 11 | 1/9/10/11/14
15 | 14 | 1/9/10/11/14/15
16 | 14 | 1/9/10/11/14/16
17 | 14 | 1/9/10/11/14/17
18 | 10 | 1/9/10/18
19 | \N | 19
20 | 19 | 19\20
21 | 19 | 19\21
I need to do some queries based off this table.
The queries I need to do are
Select all children of
id
9
SELECT * FROM `tester` WHERE 'path' LIKE '%/9/%';
Would work fine, Until you replace the ID with 1 or 19 as there is no /
at the beginning.
SELECT * FROM `tester` WHERE 'path' LIKE '%1/%';
would select all rows where a number ends in 1, so, 1, 11, 21, 31, 211 etc
SELECT * FROM `tester` WHERE 'path' LIKE '1/%';
would work correctly for either rows 1 or 19
So SELECT * FROM
testerWHERE 'path' LIKE '1/%' OR 'path' LIKE '%/1/%'
;
Select Direct children of 9 but not sub-children For this
Select * from
testerwhere 'parent' = 9
; will work fine.
select an aggregate count of 9's children, x levels deep.
So I want to end up with either one row of level1, level2, level3, ... levelx
or x rows, representing the different levels,
Let us pretend x is 3 for this example
The rows from this example would be 9, 8, 6
(the 4th level if we requested it would be 3)
Any Ideas?
Edit
#select count of children of specific node(5) down to a maximum of three levels, do no include the parent
SELECT COUNT(child.id) children,
LENGTH(REPLACE(child.path, parent.path, '')) - LENGTH(REPLACE(REPLACE(child.path, parent.path, ''), '/', '')) AS LEVEL
FROM `tester` child JOIN `tester` parent ON child.path LIKE CONCAT(parent.path,'%')
WHERE parent.id =5
GROUP BY LEVEL HAVING LEVEL <= 3 AND LEVEL > 0;
**select 9's children's id's down to x levels, with the level relative to 9,
So again for this example we will use 3 as x.
We are looking to get back
10 | 1
11 | 2
18 | 2
12 | 3
13 | 3
14 | 3
Again I am at a complete loss as to how to do this.
Edit:
#select all information, and relative level from parent of children of specific node(5) down to a maximum of three levels, do no include the parent
SELECT child.*,
LENGTH(REPLACE(child.path, parent.path, '')) - LENGTH(REPLACE(REPLACE(child.path, parent.path, ''), '/', '')) AS LEVEL
FROM `tester` child JOIN `tester` parent ON child.path LIKE CONCAT(parent.path,'%')
WHERE parent.id =9
GROUP BY id HAVING LEVEL <= 3 AND LEVEL > 0;
Just to give you a heads up, these solutions are based on string comparisons, are not optimized & cannot use indexes. you should consider normalizing your tables differently. (See Managing Hierarchical Data in MySQL)
Regarding some of the questions:
Select all children of id 9:
Since the Path
column does not include the leading & trailing slashes, you need to concatenate them to the path:
SELECT *
FROM tester
WHERE CONCAT('/', path, '/') LIKE '%/9/%';
select an aggregate count of 9's children, x levels deep:
We need to group by the number of slashes in the path, minus the number of slashes in the parent path:
SELECT (LENGTH(c.Path) - LENGTH(REPLACE(c.Path, '/', '')))
- (LENGTH(p.Path) - LENGTH(REPLACE(p.Path, '/', ''))) AS Level,
COUNT(*)
FROM tester c
JOIN tester p ON c.Parent = p.ID
WHERE CONCAT('/', path, '/') LIKE '%/9/%';
GROUP BY 1
For simplicity i used the query above to show all the levels, If you want to limit x levels deep, use the WHERE
predicate from the query below.
select 9's children's id's down to x levels, with the level relative to 9:
We search the Path
column up to a x number of levels, while taking the parents level into consideration:
SELECT c.*
FROM tester c
JOIN tester p ON c.Parent = p.ID
WHERE CONCAT(
'/',
SUBSTRING_INDEX(
Path,
'/',
(LENGTH(p.Path) - LENGTH(REPLACE(p.Path, '/', ''))) + 4
),
'/') LIKE '%/9/%'
The steps we are taking:
- We need to find out how deep the parent is, we can find that by counting the slashes in the parent's path. (
LENGTH(p.Path) - LENGTH(REPLACE(p.Path, '/', ''))
) - We need to add 1 to that number, since a path with 1 slash is 2 levels deep.
- We add the x number of desired levels.
- Grab the path column up to the level total, (Use the
SUBSTRING_INDEX
function). - Add the leading and trailing slash.
- Search the final string for 9.
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