SQL查询:如何使没有孩子的标签成为父母? [英] sql query: how to make the tags without children to become parents?

问题描述

我有这个标签表，

tag_id  tag_name    parent_id   cat_id
3       Tagname-1   NULL        NULL    
5       tagname-2   NULL        NULL    
6       tagname-3   NULL        NULL
9       tagname-4   NULL        NULL
11      tagname-5   3           NULL
13      tagname-6   3           NULL
15      tagname-8   5           NULL
17      tagname-9   5           NULL
18      tagname-10  NULL        NULL
20      tagname-11  6           NULL
22      tagname-12  9           NULL
24      tagname-13  NULL        NULL
26      tagname-14  NULL        NULL
28      tagname-15  NULL        NULL

我想返回这样的结果

ParentID    ParentName  TotalChildren
3           Tagname-1   2
5           tagname-2   2
6           tagname-3   1
9           tagname-4   1
18          tagname-10  0
24          tagname-13  0
26          tagname-14  0
28          tagname-15  0

所以在这里我到目前为止提出的查询，

So here the query I came out with so far,

SELECT 
a.tag_id as ParentID,
a.tag_name as ParentName,
b.TotalChildren

FROM root_tags a INNER JOIN
(
    SELECT parent_id, COUNT(1) as TotalChildren
    FROM root_tags
    WHERE parent_id <> tag_id
    GROUP BY parent_id
) b 

ON a.tag_id = b.parent_id
ORDER BY ParentID

但是，不幸的是，它只会返回这样的结果，

But, unfortunately, it only return the result like this,

ParentID    ParentName  TotalChildren
3           Tagname-1   2
5           tagname-2   2
6           tagname-3   1
9           tagname-4   1

这意味着它失去了没有孩子的父母.

Which means it is missing the parents without any children.

我如何制作没有孩子的标签也可以成为父母?或者换句话说，如何让没有父母的标签自己成为父母?

How can I make the tags without children to become parents too? or in other words, how to make the tags without parents to become parents themselves?

SELECT 
a.tag_id as ParentID,
a.tag_name as ParentName,
b.TotalChildren

FROM root_tags a LEFT OUTER JOIN
(
    SELECT parent_id, COUNT(1) as TotalChildren
    FROM root_tags
    WHERE parent_id <> tag_id
    GROUP BY parent_id
) b 

ON a.tag_id = b.parent_id
ORDER BY ParentID

以上答案返回

ParentID    ParentName  TotalChildren
3           Tagname-1   2
5           tagname-2   2
6           tagname-3   1
9           tagname-4   1
11          tagname-5   NULL
13          tagname-6   NULL
15          tagname-8   NULL
17          tagname-9   NULL
18          tagname-10  NULL
20          tagname-11  NULL
22          tagname-12  NULL
24          tagname-13  NULL
26          tagname-14  NULL
28          tagname-15  NULL

这是不正确的，因为它会返回所有子项.

which is incorrect as it returns all the children.

推荐答案

首选对子查询的联接:

SELECT parents.tag_id AS ParentID,
       parents.tag_name AS ParentName,
       COUNT(childs.tag_id) AS TotalChildren
FROM root_tags AS parents
    LEFT OUTER JOIN root_tags AS childs
        ON parents.tag_id = childs.parent_id
WHERE parents.parent_id IS NULL
GROUP BY parents.tag_id, parents.tag_name
ORDER BY parents.tag_id

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