Question

我的子表包含以下列： idSubs，LastName，FirstName，Phone，Email，Type

我的subProjects表包含以下列：idSubs，idProjects，PaidOutAmount

我正在尝试获取PaidOutAmount中显示的最小金额和最大金额。我还希望承包商姓氏这些金额都支付给了。

我运行以下内容：

Select min(PaidOutAmount) least, subs.LastName as PaidTo,   
max(PaidOutAmount) max, subs.LastName as PaidTo
from subprojects, subs 
where subprojects.idSubs = subs.idSubs

结果表是：

least   PaidTo   max      PaidTo
126.00  Torrey  5650.55   Torrey

这就是我想要的，除了第一个＆＃34; PaidTo＆＃34;专栏是＆＃34;托里＆＃34;它应该是＆＃34; Bassler＆＃34;。在我的subProjects表中，最小金额是126.00，但idSubs是6.如果我去我的子表，6实际上是指具有Bassler姓氏的承包商。我不知道自己做错了什么。

根据Jorge的回答，我尝试了以下方法：

Select subs.Lastname as PaidTo, PaidOutAmount as Value from subprojects,
subs where subprojects.idSubs = subs.idSubs and PaidOutAmount =    
min(PaidOutAmount) or PaidOutAmount = max(PaidOutAmount)

在stackoverflow上的其他地方找到max对于where子句无效并且应该使用子查询后，我将其更改为：

Select subs.Lastname as PaidTo, PaidOutAmount as Value from subprojects,
subs where subprojects.idSubs = subs.idSubs and (select PaidOutAmount =  
min(PaidOutAmount)or PaidOutAmount = max(PaidOutAmount))

这将生成一个包含所有子和所有PaidOutAmounts的表。换句话说，它不像我正在尝试那样只提取最小值和最大值。经过实验，我发现上述任何配置或变体只会产生相同的结果，就像我只是使用以下内容一样：

Select subs.Lastname as PaidTo, PaidOutAmount as Value from subprojects,
subs

可能是什么问题？

对于密集的抱歉，我只是没有看到问题。根据你的建议，我去了SQlFiddle。我修改了你建议的查询：

SELECT
 subs.LastName AS PaidTo, subprojects.PaidOutAmount as Value
FROM subprojects AS S
  INNER JOIN (
        SELECT
              idSubs
            , MIN(PaidOutAmount) MinPaidOutAmount
            , MAX(PaidOutAmount) MaxPaidOutAmount
        FROM subprojects
        GROUP BY
              idSubs
  ) AS
  subs ON S.idSubs = subs.idSubs
 WHERE S.PaidOutAmount =  subs.MinPaidOutAmount
  OR S.PaidOutAmount = subs.MaxPaidOutAmount

我在SqlFiddle中添加的相关表格如下：

CREATE TABLE subs
(idSubs int(11) NOT NULL ,
LastName VARCHAR(20),
FirstName VARCHAR(15)  ,
Phone VARCHAR(10)  ,
Email VARCHAR(45)  ,
Type VARCHAR(15),
PRIMARY KEY (`idSubs`));

CREATE TABLE  subprojects (
idSubs int(11) NOT NULL ,
idProjects INT(11) NOT NULL,
PaidOutAmount DECIMAL(7,2) NULL DEFAULT NULL ,
PRIMARY KEY (`idSubs`, `idProjects`));

INSERT INTO SubProjects VALUES (1, 200, 755.23);
INSERT INTO SubProjects VALUES (6, 200, 126.00);
INSERT INTO SubProjects VALUES (8, 200, 2523.53);
INSERT INTO SubProjects VALUES (1, 300, 5650.55);
INSERT INTO SubProjects VALUES (6, 300, 750.19);
INSERT INTO SubProjects VALUES (7, 300, 2500.00);
INSERT INTO SubProjects VALUES (8, 100, 374.14);



INSERT INTO Subs VALUES    
(1,'Torrey','Brian','5635435877','torrey@torrey.com', 'Carpenter');
INSERT INTO Subs  VALUES (2, 'McDonough', 'Mark', '5635997115',   
'McDonough@gmail.com', 'Carpenter');
INSERT INTO Subs VALUES (3, 'Basler', 'Bruce', '5635904457',  
'BB@yahoo.com', 'Carpenter');
INSERT INTO Subs  VALUES (4, 'ThePainter', 'Tyler', '5635800330',  
'paintpros@gmail.com', 'Painter');
INSERT INTO Subs VALUES (5, 'ThePainter2', 'Adam', '5635642819',  
'adampnt@gmail.com', 'Painter');
INSERT INTO Subs VALUES (6, 'A&G', 'Dean', '5635902919',  
'dean12@agelectric.com', 'Electrician');
INSERT INTO Subs VALUES (7, 'AllSeasons', 'Dean', '5635431296',  
'service@allseasons.com', 'Heating/Cooling');
INSERT INTO Subs VALUES (8, 'Gregory', 'Tom', '5634950997',   
'tgregory@floorshowcompanies.com', 'Flooring');
INSERT INTO Subs VALUES (9, 'Jaeger', 'Kevin', '5635836677',   
'kevin@jaegerplmb.com', 'Plumbing');

Answer 1

要过滤您需要的聚合：

一个。分组数据和
湾使用HAVING条款

但是您的查询需要其他东西 - 子查询，类似这样：

SELECT
      s.*
FROM subprojects AS S
      INNER JOIN (
            SELECT
                  idSubs
                , MIN(PaidOutAmount) MinPaidOutAmount
                , MAX(PaidOutAmount) MaxPaidOutAmount
            FROM subprojects
            GROUP BY
                  idSubs
      ) AS
      subs ON S.idSubs = subs.idSubs
WHERE S.PaidOutAmount =  subs.MinPaidOutAmount
      OR S.PaidOutAmount = subs.MaxPaidOutAmount

Answer 2

好吧，这个令人费解的混乱是我必须要做的，但实际上它可以获得我想要的信息。如果有人想告诉我如何做得更好，那就太棒了。

select
(select min(PaidOutAmount)
 from  subprojects) as min,
 (Select Lastname
 from subs, subprojects 
 where subs.idSubs = subprojects.idSubs
 order by PaidOutAmount asc
 limit 1) as PaidTo,
 (Select max(PaidOutAmount) from subprojects) as Max,
(Select Lastname
from subs, subprojects 
where subs.idSubs = subprojects.idSubs
order by PaidOutAmount desc
limit 1) as PaidTo

它产生了这个：

min     PaidTo    Max      PaidTo
126.00   A&G    5650.55     Torrey

这也有效。

SELECT LastName,PaidOutAmount       
FROM subs, subprojects
WHERE PaidOutAmount = (SELECT Max(PaidOutAmount)
    FROM subprojects ) and subs.idSubs = subprojects.idSubs 

 or (PaidOutAmount = (SELECT Min(PaidOutAmount)
    FROM subprojects ) and subs.idSubs = subprojects.idSubs )

LastName  PaidOutAmount
Torrey     5650.55
A&G        126.00

Where子句不从表中提取正确的数据

2 个答案: