Question

任何人都可以帮我修改以下SQL查询。

select pid from activities_2009_1 where purpose IN(1,2);

在此查询中，我有目的地使用IN运算符。这给了我所有有目的= 1或目的2的pids但是，我想选择所有目的= 1和目的2的pid，

PID         A   P   S       D 
612747617   0   1   0       27300 
612747617   1   2   28740   29280 
612747617   2   1   59700   60 
612747617   3   4   61200   120 
612747617   4   1   62700   37300 
612747631   0   1   0       36000 
612747631   1   4   36600   2100 
612747631   2   4   43200   56800

vt_activities_2009_1表中的示例数据如上所示。

其中A表示anum，P表示目的等。

我无法在目的= 1和目的= 2的情况下使用。

我将非常感谢你的帮助。

Answer 1

怎么样：

    select pid 
      from your_table
     where purpose=1
  intersect
    select pid 
      from your_table
     where purpose=2;

我认为应该这样做。

实际上只是想到了这一点，它将在一次通过表格中完成：

select pid
  from (
        select pid
          from my table
         where purpose in(1,2)
       )
group by pid
having count(*) = 2;

这应该是一次通过表格。

Answer 2

SELECT FIPS, count(pid) from
(SELECT distinct sa.* 
from(
    SELECT p.pid, concat(lpad(h.state, 2, '0'), lpad(h.county,3,'0')) FIPS 
    FROM PERSON_2009_1 p,HOUSEHOLD_2009_1 h
    WHERE p.schg = 4
    AND p.agep BETWEEN 10 AND 25
    and p.hid = h.hid) sa
where sa.pid in (select pid from activities_2009_1 x where exists(select 1 from activities_2009_1 y where x.pid = y.pid and purpose =1) 
and exists(select 1 from activities_2009_1 y where x.pid = y.pid and purpose =2)
)group by FIPS order by FIPS;

Answer 3

SELECT FIPS, count(pid) from(SELECT distinct sa.* from(SELECT p.pid, concat(lpad(h.state, 2, '0'), lpad(h.county,3,'0')) FIPS
                       FROM PERSON_2009_1 p,HOUSEHOLD_2009_1 h
                             WHERE p.schg = 4 AND p.agep BETWEEN 10 AND 25 and p.hid = h.hid)
                             sa  where sa.pid in ((select pid from activities_2009_1 where purpose = 1) as a inner join (select pid from activities_2009_1 where purpose = 2) as b on (a.pid = b.pid))
                             )group by FIPS order by FIPS;

我正在使用内连接（交叉）。不是最有效的，但应该有效。

以下是我第一次想到使用AND：

SELECT FIPS, count(pid) from(SELECT distinct sa.* from(SELECT p.pid, concat(lpad(h.state, 2, '0'), lpad(h.county,3,'0')) FIPS
                   FROM PERSON_2009_1 p,HOUSEHOLD_2009_1 h
                         WHERE p.schg = 4 AND p.agep BETWEEN 10 AND 25 and p.hid = h.hid)
                         sa  where sa.pid in ((select pid from activities_2009_1 where purpose = 1) ANDas sa.pida ininner join (select pid from activities_2009_1 where purpose = 2) as b on (a.pid = b.pid))
                         )group by FIPS order by FIPS;

Answer 4

我想你可能想要这个：

SELECT FIPS, count(pid) from
( SELECT distinct sa.* from
  (SELECT p.pid, concat(lpad(h.state, 2, '0'), lpad(h.county,3,'0')) FIPS
     FROM PERSON_2009_1 p,HOUSEHOLD_2009_1 h
    WHERE p.schg = 4
      AND p.agep BETWEEN 10 AND 25 
      and p.hid = h.hid
  ) sa
  where sa.pid in (select pid from activities_2009_1 where purpose = 1)
    AND sa.pid in (select pid from activities_2009_1 where purpose = 2)
)
group by FIPS order by FIPS;

或者：

SELECT FIPS, count(pid) from
( SELECT distinct sa.* from
  (SELECT p.pid, concat(lpad(h.state, 2, '0'), lpad(h.county,3,'0')) FIPS
     FROM PERSON_2009_1 p,HOUSEHOLD_2009_1 h
    WHERE p.schg = 4
      AND p.agep BETWEEN 10 AND 25 
      and p.hid = h.hid
  ) sa
  where 2 =  (select count(distinct a.purpose)
                from activities_2009_1 a 
               where a.pid = sa.pid
                 and a.purpose in (1,2))
)
group by FIPS order by FIPS;

Answer 5

这对你来说很有用......

SELECT PID
     , A
     , P
     , S
     , D
  FROM ACTIVITIES_2009_1
 WHERE PID IN (SELECT PID
                 FROM (  SELECT PID, COUNT (*)
                           FROM ACTIVITIES_2009_1
                          WHERE P = 1
                             OR P = 2
                       GROUP BY PID
                         HAVING COUNT (*) > 1))

Answer 6

尝试为表a和b创建作业。 a将验证目的是否存在= 1，b将验证目的是否存在= 2。如果缺少任何一个目的，inner join将删除该行。

select a.pid 
from activities_2009_1 a
  inner join activities_2009_1 b    
       on a.pid=b.pid
where a.purpose=1 and b.purpose=2;

Answer 7

我们如何谈论select的含义，而不是给出你眼前问题的答案。

简单形式select <columns> from <table> where <conditions>几乎是自我解释。选择（返回）的是表中条件为真的行。为了实现你想要的条件（“对于同一个pid有一个p = 1和p = 2”），你首先必须确保两个p列在同一行。让我们再说一遍。为了实现你想要的条件（“对于同一个pid有一个p = 1和p = 2”），你首先必须确保两个p列在同一行。

这就是联接可以为您做的事情：将两个表中的行对齐以创建一个包含两个值的新表。在这种情况下，pid上的连接（对齐），以及对left.p = 1和right.p = 2的检查实现了这一点。我确定在其中一个答案中有一个例子（l和r指的是加入的两个表格（“左”和“右”），在这种情况下是同一个表两次。

使用count()=2的示例非常聪明且（可能）有效但依赖于每个pid + p组合唯一的假设（或约束）。

至于你的（连接）查询的效率，我建议你在关闭sql语法之后开始关心它（以及索引，规范化/非规范化数据库设计和约束）。 “高效运行”查询所获得的时间可能不会超过开发人员效率低下/由于代码不清晰而导致的时间损失。硬件比人工软件更容易扩展。

AND运算符跨多行

7 个答案: