Question

我想从mysql db中获取记录。我想从记录中获取第二个最大日期。但我失败了

这是我的代码

    <?php
include ("connection.php");
$q_opinion="SELECT r.client_id,c.id,t.id,a.id,o.id,c.name as opinion, r.notification_date, t.title as ttitle,a.title as atitle,o.title as otitle, l.title as ltitle, s.title as stitle, pr.opinion_id, pc.id, pr.client_id as pr_client, pc.address, pc.liaison_one, city.id, pc.head_office_id, city.city, pc.title as cname
FROM og_ratings r 
    LEFT join
(
  select max(notification_date) notification_date,
    client_id
  from og_ratings
  WHERE notification_date NOT IN (select max(notification_date) FROM og_ratings )
   ) r2
  on r.notification_date = r2.notification_date
  and r.client_id = r2.client_id
LEFT JOIN og_companies c
ON r.client_id = c.id
LEFT JOIN og_rating_types t
ON r.rating_type_id = t.id
LEFT JOIN og_actions a
ON r.pacra_action = a.id
LEFT JOIN og_outlooks o
ON r.pacra_outlook = o.id
LEFT JOIN og_lterms l
ON r.pacra_lterm = l.id
LEFT JOIN og_sterms s
ON r.pacra_sterm = s.id
LEFT JOIN pacra_client_opinion_relations pr
ON pr.opinion_id = c.id
LEFT JOIN pacra_clients pc
ON pc.id = pr.client_id
LEFT JOIN city
ON city.id = pc.head_office_id
WHERE r.client_id  IN (SELECT opinion_id FROM pacra_client_opinion_relations WHERE client_id = 50)

";
$result = mysql_query($q_opinion) or die;
$rating = array();
while($row = mysql_fetch_assoc($result))
{
  $rating[] = $row['client_id'];
  $action[] = $row['atitle'];
  $opinion[] = $row['opinion'];
  $date[] = $row['notification_date'];
  $lrating[] = $row['ltitle'];
  $srating[] = $row['stitle'];
}
for ($i=0; $i<count($rating); $i++) {
    if ($rating[$i] == "")continue;
     ?>
    <table border="1">
    <tr>
          <td><?= $rating[$i] ?> </td>
           <td><?= $date[$i] ?> </td>
          <td><?= $opinion[$i] ?> </td>
         <td><?= $action[$i] ?> </td>
          <td><?= $lrating[$i] ?> </td>
           <td><?= $srating[$i] ?> </td>
    </tr>
    </table>
<?php   
}
?>

这是此代码的输出

在输出图像中，您可以看到它从db获取所有记录。但我想只获取具有第二个最大日期的数据。

我怎么做？

Answer 1

阅读您的查询并不好玩，但我认为问题在于：

LEFT JOIN (
  SELECT max(notification_date) notification_date, client_id
  FROM og_ratings
  WHERE notification_date NOT IN (
    SELECT max(notification_date)
    FROM og_ratings
)

如果您想要GROUP BY client_id所需的每个客户的最长日期：

SELECT client_id, max(notification_date) notification_date
FROM og_ratings
GROUP BY client_id

如果您想要第二个最大值，那么选项很少，我使用的这个更容易理解，但它不一定是性能最高的：

SELECT client_id, max(notification_date) notification_date
FROM og_ratings
WHERE
  (client_id, notification_date) NOT IN (
    SELECT client_id, max(notification_date)
    FROM og_ratings GROUP BY client_id
  )
GROUP BY client_id

第三个问题，你正在使用LEFT JOIN，这意味着你将从og_ratings返回所有值，无论它们是否是第二个最大值。在此上下文中使用INNER JOIN：

SELECT
  r.client_id,
  c.id,
  t.id,
  ..etc...
FROM
  og_ratings r INNER JOIN (
    SELECT client_id, max(notification_date) notification_2nd_date
    FROM og_ratings
    WHERE
      (client_id, notification_date) NOT IN (
        SELECT client_id, max(notification_date)
        FROM og_ratings GROUP BY client_id
      )
    GROUP BY client_id
   ) r2
  ON r.notification_date = r2.notification_2nd_date
     AND r.client_id = r2.client_id
  LEFT JOIN og_companies c ON r.client_id = c.id
  LEFT JOIN og_rating_types t ON r.rating_type_id = t.id
  LEFT JOIN og_actions a ON r.pacra_action = a.id
  LEFT JOIN og_outlooks o ON r.pacra_outlook = o.id
  LEFT JOIN og_lterms l ON r.pacra_lterm = l.id
  LEFT JOIN og_sterms s ON r.pacra_sterm = s.id
  LEFT JOIN pacra_client_opinion_relations pr ON pr.opinion_id = c.id
  LEFT JOIN pacra_clients pc ON pc.id = pr.client_id
  LEFT JOIN city ON city.id = pc.head_office_id
WHERE
  r.client_id IN (
    SELECT opinion_id FROM pacra_client_opinion_relations
    WHERE client_id = 50
  )

Answer 2

请注意，架构设置QueryA和QueryB都只是用于可视化。

然而，QueryC是您尝试使用数据的地方。

这不是一个简单的order by和limit @Musa做的原因很简单：你可以拥有许多第二个最大日期的行，而不是一个。这就是为什么使用@grp和@preate的变量来定位第二组。

架构设置

-- drop table specimenA;
create table specimenA
(   mypk int auto_increment primary key,
    id int not null,    -- note, not autoinc or pk
    theDate date not null,
    title varchar(255) not null,
    otherThing varchar(20) not null
    -- etc
    -- not other indexes whatsoever
);
-- truncate table specimenA
insert specimenA (id,theDate,title,otherThing) values
(170,'2007-09-19','whatever','whatever'),
(170,'2008-09-12','whatever','whatever'),
(170,'2010-01-15','whatever','whatever'),
(170,'2011-02-03','whatever','whatever'),
(170,'2012-06-26','whatever','whatever'),
(170,'2013-03-05','whatever','whatever'),
(170,'2014-06-25','whatever','whatever'),
(170,'2015-06-09','whatever','whatever'),

(917,'2009-10-14','whatever','whatever'),
(917,'2008-12-31','whatever','whatever'),

(109,'2010-04-26','whatever','whatever'),
(109,'2011-03-02','whatever','whatever'),
(109,'2012-06-25','whatever','whatever'),
(109,'2013-01-04','whatever','whatever'),
(109,'2014-03-28','whatever','whatever'),
(109,'2015-03-18','whatever','whatever'),

(1057,'2014-03-28','whatever','whatever'),
(1057,'2014-11-21','whatever','whatever'),
(1057,'2015-08-13','whatever','whatever');

QueryA

set @rn:=0,@grp:=0,@prevdate:=''; 
select id,theDate,title,otherThing,
@rn:=@rn+1 as rownum, 
@grp:=if(@prevdate=theDate,@grp,@grp+1) as descGrp, 
@prevdate:=theDate as unused 
from specimenA 
order by theDate DESC -- **** Note this
-- DESC means greatest first, as is most-recent first for dates
+------+------------+----------+------------+--------+---------+------------+
| id   | theDate    | title    | otherThing | rownum | descGrp | unused     |
+------+------------+----------+------------+--------+---------+------------+
| 1057 | 2015-08-13 | whatever | whatever   |      1 |       1 | 2015-08-13 |
|  170 | 2015-06-09 | whatever | whatever   |      2 |       2 | 2015-06-09 |
|  109 | 2015-03-18 | whatever | whatever   |      3 |       3 | 2015-03-18 |
| 1057 | 2014-11-21 | whatever | whatever   |      4 |       4 | 2014-11-21 |
|  170 | 2014-06-25 | whatever | whatever   |      5 |       5 | 2014-06-25 |
| 1057 | 2014-03-28 | whatever | whatever   |      6 |       6 | 2014-03-28 |
|  109 | 2014-03-28 | whatever | whatever   |      7 |       6 | 2014-03-28 |
|  170 | 2013-03-05 | whatever | whatever   |      8 |       7 | 2013-03-05 |
|  109 | 2013-01-04 | whatever | whatever   |      9 |       8 | 2013-01-04 |
|  170 | 2012-06-26 | whatever | whatever   |     10 |       9 | 2012-06-26 |
|  109 | 2012-06-25 | whatever | whatever   |     11 |      10 | 2012-06-25 |
|  109 | 2011-03-02 | whatever | whatever   |     12 |      11 | 2011-03-02 |
|  170 | 2011-02-03 | whatever | whatever   |     13 |      12 | 2011-02-03 |
|  109 | 2010-04-26 | whatever | whatever   |     14 |      13 | 2010-04-26 |
|  170 | 2010-01-15 | whatever | whatever   |     15 |      14 | 2010-01-15 |
|  917 | 2009-10-14 | whatever | whatever   |     16 |      15 | 2009-10-14 |
|  917 | 2008-12-31 | whatever | whatever   |     17 |      16 | 2008-12-31 |
|  170 | 2008-09-12 | whatever | whatever   |     18 |      17 | 2008-09-12 |
|  170 | 2007-09-19 | whatever | whatever   |     19 |      18 | 2007-09-19 |
+------+------------+----------+------------+--------+---------+------------+

并且，使用上面的select语句，并将其作为别名inR作为派生表，嵌套：

QueryB

set @rn:=0,@grp:=0,@prevdate:=''; 
select id,theDate,title,otherthing 
from 
(   select id,theDate,title,otherThing,
    @rn:=@rn+1 as rownum, 
    @grp:=if(@prevdate=theDate,@grp,@grp+1) as descGrp, 
    @prevdate:=theDate as unused 
    from specimenA 
    order by theDate DESC  -- **** Note this
) inR 
where descGrp=2;
+-----+------------+----------+------------+
| id  | theDate    | title    | otherthing |
+-----+------------+----------+------------+
| 170 | 2015-06-09 | whatever | whatever   |
+-----+------------+----------+------------+

还有你的第二个最伟大的约会。意思是第二个最近的日期。

所以，采取你原来的选择陈述，相同的概念。显示上述内容的动机很简单：InR只是Derived Table，不会比您的select语句更多或更少，它将成为派生表。

QueryC

set @rn:=0,@grp:=0,@prevdate:=''; 

select client_id, cid, tid, aid, oid, opinion, notification_date,
ttitle, atitle, otitle, ltitle, stitle,
opinion_id, pcid, pr_client, address, liaison_one, 
cityid, head_office_id, city, cname
from
(   SELECT r.client_id as client_id,c.id as cid,t.id as tid,a.id as aid,o.id as oid,c.name as opinion, r.notification_date, 
    t.title as ttitle,a.title as atitle,o.title as otitle, l.title as ltitle, s.title as stitle, 
    pr.opinion_id, pc.id as pcid, pr.client_id as pr_client, pc.address, pc.liaison_one, 
    city.id as cityid, pc.head_office_id, city.city, pc.title as cname,
    @rn:=@rn+1 as rownum, 
    @grp:=if(@prevdate=r.notification_date,@grp,@grp+1) as descGrp, 
    @prevdate:=r.notification_date as unused 

    FROM og_ratings r 
        LEFT join
    (
      select max(notification_date) notification_date,
        client_id
      from og_ratings
      WHERE notification_date NOT IN (select max(notification_date) FROM og_ratings )
       ) r2
      on r.notification_date = r2.notification_date
      and r.client_id = r2.client_id
    LEFT JOIN og_companies c
    ON r.client_id = c.id
    LEFT JOIN og_rating_types t
    ON r.rating_type_id = t.id
    LEFT JOIN og_actions a
    ON r.pacra_action = a.id
    LEFT JOIN og_outlooks o
    ON r.pacra_outlook = o.id
    LEFT JOIN og_lterms l
    ON r.pacra_lterm = l.id
    LEFT JOIN og_sterms s
    ON r.pacra_sterm = s.id
    LEFT JOIN pacra_client_opinion_relations pr
    ON pr.opinion_id = c.id
    LEFT JOIN pacra_clients pc
    ON pc.id = pr.client_id
    LEFT JOIN city
    ON city.id = pc.head_office_id
    WHERE r.client_id  IN (SELECT opinion_id FROM pacra_client_opinion_relations WHERE client_id = 50)
    order by r.notification_date DESC
) inR
where descGrp=2

您使用的是QueryC。因为这是两个陈述

变量的初始化
和大型查询字符串

...你需要按顺序运行查询，每个按顺序运行一次，或者使用PHP multi-query并在一次调用中组合两者。上一句话中的那个链接是针对mysqli的，这个概念在此显示，并据此进行修改。

Answer 3

您可以使用下面的查询来获取表格中的第二个最大日期或您的要求： -

select joiningDate from t_member order by joiningDate desc limit 1,1

Answer 4

您可以使用ORDER BY DATE_COLUMN DESC LIMIT 1 OFFSET 1。我在查询结尾添加了此代码：

SELECT 
    r.client_id,
    c.id,
    t.id,
    a.id,
    o.id,
    c.name as opinion, 
    r.notification_date, 
    t.title as ttitle,
    a.title as atitle,
    o.title as otitle, 
    l.title as ltitle, 
    s.title as stitle, 
    pr.opinion_id, 
    pc.id, 
    pr.client_id as pr_client, 
    pc.address, 
    pc.liaison_one, 
    city.id, 
    pc.head_office_id, 
    city.city, 
    pc.title as cname
FROM og_ratings r 
    LEFT join (
        select max(notification_date) notification_date,
            client_id
        from og_ratings
        WHERE notification_date NOT IN (select max(notification_date) FROM og_ratings )
   ) r2
  on r.notification_date = r2.notification_date
  and r.client_id = r2.client_id
LEFT JOIN og_companies c 
    ON r.client_id = c.id
LEFT JOIN og_rating_types t
    ON r.rating_type_id = t.id
LEFT JOIN og_actions a
    ON r.pacra_action = a.id
LEFT JOIN og_outlooks o
    ON r.pacra_outlook = o.id
LEFT JOIN og_lterms l
    ON r.pacra_lterm = l.id
LEFT JOIN og_sterms s
    ON r.pacra_sterm = s.id
LEFT JOIN pacra_client_opinion_relations pr
    ON pr.opinion_id = c.id
LEFT JOIN pacra_clients pc
    ON pc.id = pr.client_id
LEFT JOIN city
    ON city.id = pc.head_office_id
WHERE 
    r.client_id  IN (
        SELECT opinion_id FROM pacra_client_opinion_relations WHERE client_id = 50
    )
ORDER BY r.notification_date DESC # Add this line
LIMIT 1 offset 1                  # and this line

如何在MYSQL中获得第二个MAXIMUM DATE

4 个答案:

架构设置

QueryA

QueryB

QueryC