Question

我需要写一个这样的查询，请帮忙吗？

select id, parent, faq, 
(select count(*) from faq_table where parent = 
(select id from faq_questions) group by id) 
 as reply from faq_table

此表存储问题和答案（如常见问题解答），答案在父列中获取问题ID的值。理想情况下，我还想添加名为par的第二列，其中所有问题的值都为1.

像：

id  | parent | faq
19  |    0   | name a president of the US
20  |   19   | Bush
21  |   19   | Obama
22  |    0   | Can Canada win the WC match against the Lankan's today
23  |   22   | Yes because Cheema is going to make a double today

在该查询的结果表中，我必须得到：

id  | parent | faq  |  reply | par
19  |   0    | name a president of the US | 2 | 1
20  |  19    | Bush  | 0 | 0
21  |  19    | Obama | 0 | 0
22  |   0    | Can Canada win the WC match against the Lankan's today | 1 | 1
23  |  22    | Yes because Cheema is going to make a double today | 0 | 0

Answer 1

这仅适用于单个层次结构级别：

SELECT t.id, t.parent, t.faq, IFNULL(g.Cnt, 0) as reply,
  g.Cnt IS NOT NULL AS par
FROM faq_table t LEFT JOIN (
    SELECT parent, COUNT(*) Cnt
    FROM faq_table
    WHERE parent > 0
    GROUP BY parent
  ) g ON t.id = g.parent

否则，您可以使用子查询：

SELECT t.id, t.parent, t.faq,
  (SELECT COUNT(*) FROM faq_table f WHERE f.parent = t.id) as reply,
  (SELECT COUNT(*) FROM faq_table f WHERE f.parent = t.id) > 0 As par
FROM faq_table t

Answer 2

使用您的查询和此事实

答案在父母

要求

第二列名为par，其中所有问题的值都为1.

变得微不足道，似乎与您的数据一致。

select id, parent, faq, 
(select count(*) from faq_table where parent = 
(select id from faq_questions) group by id) 
 as reply,
if(parent=0,1,0) as par
from faq_table

基本上当parent = 0时，它是父级。但由于注释声明只有一个表，因此上面必须是伪 - 正确的查询应该是

select id, parent, faq, 
   (select count(*) from faq_table t where t.parent = faq_table.id) as reply,
   case when parent = 0 then 1 else 0 end as par
from faq_table

Answer 3

此版本也有效;

SELECT q.id, q.parent, q.faq, count(a.id) AS reply, IF(ISNULL(a.id),0,1) as par
FROM faq_table q
LEFT JOIN faq_table a
  ON q.id = a.parent
  AND a.id IS NOT NULL
GROUP BY q.id

Answer 4

试一试

SELECT
    f.id,
    f.parent,
    f.faq,
    (SELECT COUNT(*) FROM faq_table f2 WHERE f2.id = f.parent) AS reply
FROM faq f

Mysql复杂加入，有人吗？

4 个答案: