在 MySQL/MariaDB 关系表中获取给定父节点的所有子节点(及其子节点)
我有一张这样的桌子:
父、子0 20 82 32 63 43 56 76 99 10
我正在寻找一个查询来选择给定父级的子树,即如果给定父级是6",则输出必须是:{10,9,7,6}
解决方案看看这个.@pv := '6' 中指定的值应设置为要查找其所有后代的父级的 id.
您也可以查看实时
如果您仍有任何问题或疑虑,请告诉我们.
I have a table like this:
parent, child
0 2
0 8
2 3
2 6
3 4
3 5
6 7
6 9
9 10
I'm looking for a query to select the sub-tree of a given parent, i.e. If the given parent is "6", the output must be: {10,9,7,6}
解决方案Chek this. The value which specified in @pv := '6' should be set to the id of the parent that you want to find all the descendants of it.
also you can check live Demo updated
select Parent, concat ( "{" ,Parent,",",GROUP_CONCAT(concat (child )SEPARATOR ','),"}") as Child
from (select * from #TableName
order by parent, child) s,
(select @pv := '6') initialisation
where find_in_set(parent, @pv) > 0
and @pv := concat(@pv, ',', child);
Output : {6,7,9,10}
For display childs with parent into one column use below query :
select parent as child from tchilds where parent = @pv2
union
select Child
from (select * from tchilds
order by parent, child) s,
(select @pv2 := '6') initialisation
where find_in_set(parent, @pv2) > 0
and @pv2 := concat(@pv2,',', child)
Output
let us know if you have still any questions or concerns.
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