Question

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

我想要返回这样的结果，

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

所以这是我到目前为止提出的查询，

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

但不幸的是，它只返回这样的结果，

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

这意味着没有孩子的父母会失踪。

如何让没有孩子的标签成为父母呢？或换句话说，如何让没有父母的标签自己成为父母？

修改

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

上面的答案返回，

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

这是错误的，因为它返回所有孩子。

Answer 1

你快到了..只需要让连接成为外部连接：

已编辑：

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
WHERE b.TotalChildren is not null
ORDER BY ParentID

Answer 2

我在这里并不完全理解您的要求，但我会说，如果您将内部联接更改为左联接并将b.TotalChildren更改为IF(b.TotalChildren is null, 0, b.TotalChildren)，那么您将获得结果集欲望。

Answer 3

SELECT 
a.tag_id as ParentID,
a.tag_name as ParentName,
count(b.child) as TotalChildren
FROM root_tags as a INNER JOIN
(
  SELECT parent_id as child FROM root_tags
  WHERE parent_id is not NULL
)  as b
ON a.tag_id = b.child
where a.parent_id is NULL
ORDER BY ParentID

Answer 4

首选联接子查询：

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

sql查询：如何让没有孩子的标签成为父母？

4 个答案: