Mysql从2行中选择记录

我有一个表格，包括一些类似的数据。这是问答脚本的示例。ID是自动递增的，如果PID=0，则为问题，当回答任何问题时，则PID设置为问题的ID。回答时没有主题

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,

我如何选择具有回复计数的问题

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

我如何选择今天回答的问题或答案

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

关于问题1

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

从表中选择PID、计数（*）
pid0在哪里
按PID分组

关于问题2

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

从表中选择PID、计数（*）
其中，pid0和日期=当前_日期（）
按PID分组

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;

---

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

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

编辑：
第二部分。你应该注意到做同样的任务还有很多其他的方法

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
1   0       First Question      3
2   0       Second Question     2
5   0       Third Question      1

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

---

谢谢你的回复，那么今天的问题我该怎么做呢？@Seyhan：在where子句中添加

和q.date=curdate（）

（假设你的date列实际上是date类型），这并不是我想要的