Question

我有一张表包含一些这样的数据。这是问题/答案脚本的示例。 ID是自动递增，如果PID = 0，那么问题是，当回复任何问题时，PID被设置为问题的ID。没有回复的主题。

ID    PID    SUBJECT                CONTENT                   DATE
1      0     First Question         This is my first          09/01/2013
2      0     Second Question        This is second            09/01/2013
3      1                            Yes this is first         09/01/2013
4      2                            I agree this is second    10/01/2013
5      0     Third Question         This is third question    11/01/2013
6      1                            Reply to first            11/01/2013
7      1                            Another reply to first    12/01/2013
8      5                            This is reply of 5th      13/01/2013
9      2                            Last try for second       14/01/2013

我的问题是，

如何选择回复计数的问题？

Ex.
First Question (3)
Second Question (2)
Third Question (1)

如何选择今天已回答的问题或答案？

Ex. For 09/01/2013
First Question (2) ---> 1 question and 1 answer but 2 actions
Second Question (1) ---> just 1 question

Answer 1

选择问题并加入答案：

select q.id, q.subject, count(a.id)
from yourtable q
left join yourtable a on q.id=a.pid
where q.pid=0
group by q.id;

Answer 2

问题1

select PID,count(*) from table
where pid<>0 
group by PID

问题2

select PID,count(*) from table
where pid<>0 and date=current_date()
group by PID

Answer 3

尝试加入第一个任务

SELECT 
    q.id as ID,
    q.pid as PID,
    q.subject as SUBJECT,
    COUNT(lq.id) as Total
FROM questions as q
LEFT JOIN questions as lq ON lq.pid = q.ID
WHERE q.PID = 0
GROUP BY q.id

输出

ID  PID     SUBJECT             TOTAL
1   0       First Question      3
2   0       Second Question     2
5   0       Third Question      1

Demo

编辑：
第二部分。您应该注意，可以有许多其他方法来执行相同的任务。

SELECT 
    q.id as ID,
    q.pid as PID,
    q.subject as SUBJECT,
    (COUNT(lq.id) - 1) as Total,
    q.date
FROM questions as q
LEFT JOIN questions as lq ON lq.pid = q.ID OR lq.id = q.PID
WHERE q.date = DATE(NOW())

输出

ID  PID     SUBJECT             TOTAL   DATE
1   0       First Question      2       January, 09 2012 00:00:00+0000
2   0       Second Question     1       January, 09 2012 00:00:00+0000

Demo

Mysql从2行中选择记录

3 个答案: