左外连接（MySQL）问题，这个查询有什么问题？答案

【问题标题】：Left Outer Join (MySQL) Issues, what's wrong with this query?左外连接（MySQL）问题，这个查询有什么问题？
【发布时间】：2016-06-16 10:43:50
【问题描述】：

我在 mySQL 中有五个表，我正在研究学生费用模块，但我在查询时遇到了一些问题，所以我无法获得适当的结果，所以请帮助我，如果你能给我一些反馈，我将不胜感激在这个查询上。

1.a )class_details 表创建

CREATE TABLE `class_details` 
(`class_id_pk` int(11) NOT NULL AUTO_INCREMENT
,`class_name`varchar(200) NOT NULL
,`session` varchar(50) DEFAULT NULL
,`class_status` varchar(50) DEFAULT NULL
,PRIMARY KEY (`class_id_pk`)
,UNIQUE KEY `UNIQUE` (`class_name`,`session`)) ENGINE=InnoDB AUTO_INCREMENT=4 DEFAULT CHARSET=latin1;

1.b) class_details 插入

insert  into `class_details`
(`class_id_pk`,`class_name`,`session`,`class_status`) 
VALUES 
(1,'1st','2016-2017',NULL)
,(2,'2nd','2016-2017',NULL)
,(3,'3rd','2016-2017',NULL);

2.a) feedetails 表创建

CREATE TABLE `feedetails` 
(`section_id_fk` int(50) NOT NULL
,`fees` varchar(30) DEFAULT NULL
,PRIMARY KEY (`section_id_fk`)) ENGINE=InnoDB DEFAULT CHARSET=latin1;

2.b) 饲料详情插入

insert  into `feedetails`(`section_id_fk`,`fees`) 
values 
(1,'1000')
,(2,'2000')
,(3,'3000')
,(4,'4000')
,(5,'5000')
,(6,'6000');

3.a) section_details 创建

CREATE TABLE `section_details` 
(`section_id_pk` int(11) NOT NULL AUTO_INCREMENT
,`class_id_fk` int(11) NOT NULL
,`section_name` varchar(50) NOT NULL
,`section_status` varchar(50) DEFAULT NULL
,PRIMARY KEY (`section_id_pk`,`class_id_fk`,`section_name`)
,UNIQUE KEY `UNIQUE` (`class_id_fk`,`section_name`)
,CONSTRAINT `FK_section_details` FOREIGN KEY (`class_id_fk`) REFERENCES `class_details` (`class_id_pk`) ON DELETE CASCADE ON UPDATE CASCADE ) ENGINE=InnoDB AUTO_INCREMENT=4 DEFAULT CHARSET=latin1;

3.b) section_details 插入

insert  into `section_details` (`section_id_pk`,`class_id_fk`,`section_name`,`section_status`) 
values 
(1,1,'A',NULL)
,(2,2,'A',NULL)
,(3,3,'A',NULL);

4.a) 学生费用

CREATE TABLE `student_fee` 
( `sr_no` int(200) NOT NULL AUTO_INCREMENT
,`scholar_no`int(50) NOT NULL
,`paid_amount` int(200) DEFAULT NULL
,`due_amount` int(200) DEFAULT NULL
,`fee_date` date DEFAULT NULL
,`section_id_fk` int(50) DEFAULT NULL
,PRIMARY KEY (`sr_no`)
,KEY `FK_student_fee`(`section_id_fk`)
,CONSTRAINT `FK_student_fee` FOREIGN KEY (`section_id_fk`) REFERENCES `section_details` (`section_id_pk`) ON DELETE CASCADE ON UPDATE CASCADE ) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=latin1;

4.b) student_fee 插入

insert  into student_fee` (`sr_no`,`scholar_no`,`paid_amount`,`due_amount`,`fee_date`,`section_id_fk`) 
values 
(3,5,800,200,'2016-06-16',1)
,(4,29,1000,0,'2016-06-16',1)
,(5,5,200,0,'2016-06-16',1);

5.a) student_details 创建

CREATE TABLE `student_details` 
(`scholar_no` int(30) NOT NULL
,`fname` varchar(30) DEFAULT NULL
,`lname` varchar(30) DEFAULT NULL
,`stu_class` varchar(30) DEFAULT NULL
,`rte` varchar(30) DEFAULT NULL
,`active` varbinary(10) DEFAULT NULL
,PRIMARY KEY (`scholar_no`)) ENGINE=InnoDB DEFAULT CHARSET=latin1;

5.b) student_details 插入

insert  into `student_details` (`scholar_no`,`fname`,`lname`,`stu_class`,`rte`,`active`) 
values 
(5,'KP','PK','1','N','y')
,(29,'Abc','Xyz','1','N','y');

上面我已经提到了用于此查询的所有表详细信息。第一次在 student_details 表 stu_class=1 中，费用为 1000。当我在 student_fee 中插入一些值并使用我的查询时，结果是正确的，但是当我在 student_fee 中向另一个学生插入一些金额时，查询添加了支付的金额对于不正确的同一个学生，我想显示那些实际插入的学生的插入费用..

SELECT 
student_details.scholar_no
,student_details.fname
,student_details.lname
,student_details.stu_class
,feedetails.fees
,class_name
,section_name
,IF(sssf.paid_amount IS NULL,0,sssf.paid_amount) AS paid_amount 
FROM 
    (student_details 
    LEFT OUTER JOIN feedetails 
        ON student_details.stu_class = feedetails.section_id_fk
    ) 
LEFT OUTER JOIN 
    (SELECT 
    scholar
    , SUM(pa) AS paid_amount
    , SUM(pva) AS prev_paid_amount
    ,SUM(da) AS due_amount
    , SUM(dva) AS prev_due_amount
    ,section_id_fk
    ,fee_date
    ,stu_class 
    FROM 
        (SELECT 
        scholar
        ,CASE WHEN section_id_fk = stu_class THEN paid_amount ELSE 0 END AS pa
        ,CASE WHEN section_id_fk != stu_class THEN paid_amount ELSE 0 END AS pva
        ,CASE WHEN section_id_fk = stu_class THEN due_amount ELSE 0 END AS da
        ,CASE WHEN section_id_fk != stu_class THEN due_amount ELSE 0 END AS dva
        ,due_amount
        ,paid_amount
        ,section_id_fk
        ,fee_date
        ,stu_class 
        FROM 
            (SELECT 
            scholar
            ,due_amount
            ,SUM(paid_amount) AS paid_amount
            ,section_id_fk
            ,fee_date 
            FROM 
                (SELECT 
                student_fee.due_amount AS due_amount
                ,student_fee.paid_amount AS paid_amount
                ,student_fee.scholar_no AS scholar
                ,section_id_fk
                ,fee_date 
                FROM student_fee 
                ORDER BY STR_TO_DATE(fee_date,'%Y-%m-%d')DESC
                ) AS kkk 
            GROUP BY kkk.scholar,section_id_fk ORDER BY scholar
        ) AS k 
        LEFT OUTER JOIN student_details sd 
            ON k.scholar = sd.scholar_no
    ) AS lk
) AS sssf 
ON student_details.scholar_no = sssf.scholar 
LEFT OUTER JOIN 
    (SELECT * 
    FROM section_details AS sd 
    LEFT OUTER JOIN class_details cd 
        ON sd.class_id_fk = cd.class_id_pk
    ) AS sc 
ON student_details.stu_class = sc.section_id_pk 
WHERE student_details.active = 'y' AND rte = 'N'

【问题讨论】：

我刚刚格式化了您的代码，以便您可以更轻松地阅读它。告诉我，您在最终查询中发现任何问题吗？
sqlfiddle.com 在这里会有所帮助。
@RichBenner 是的，谢谢！
这不是问题。正确格式化代码始终是一个好主意，这样可以更明显地发现错误。
@RichBenner 当我使用 SQLYog 时，使用 formate_current_query 可以很好地格式化查询，但我不知道在 stackoverflow 中编写任何代码......！

标签： mysql sql database join left-join

【解决方案1】：

复杂的查询。不幸的是有缺陷。

首先，您可以从子查询中删除所有ORDER BY 子句。子查询只是返回无序集，因此它们是否包含ORDER BY 子句没有任何影响（除了可能给 DBMS 带来不必要的工作）。

最里面的子查询（kkk）没有WHERE子句和GROUP BY子句，所以你也可以直接从student_fee中选择。

下一个子查询 (kk) 按 scholar 和 section_id_fk 分组，但您选择 due_amount 和 fee_date 而不进行任何聚合。这为您提供了任意选择的值。不应该是sum(due_amount) 或max(due_amount) 之类的吗？

那么在 sssf 子查询中你没有 GROUP BY 子句。这为您提供了一个结果行。但是您选择未聚合的scholar、section_id_fk、fee_date、stu_class，因此您再次获得任意选择的值，例如所有学者之一。

检查所有聚合。设置ONLY_FULL_GROUP_BY模式可能会有所帮助，以免出错。

【讨论】：

该死，打败我+1 :)

【解决方案2】：

检查下面的 SQL，我有一个快速修复。如果没有解决，请多解释一下你的问题。

SELECT 
student_details.scholar_no
,student_details.fname
,student_details.lname
,student_details.stu_class
,feedetails.fees
,class_name
,section_name
,IF(sssf.paid_amount IS NULL,0,sssf.paid_amount) AS paid_amount 
FROM 
    (student_details 
    LEFT OUTER JOIN feedetails 
        ON student_details.stu_class = feedetails.section_id_fk
    ) 
LEFT OUTER JOIN 
    (SELECT 
    scholar
    , SUM(pa) AS paid_amount
    , SUM(pva) AS prev_paid_amount
    ,SUM(da) AS due_amount
    , SUM(dva) AS prev_due_amount
    ,section_id_fk
    ,fee_date
    ,stu_class 
    FROM 
        (SELECT 
        scholar
        ,CASE WHEN section_id_fk = stu_class THEN paid_amount ELSE 0 END AS pa
        ,CASE WHEN section_id_fk != stu_class THEN paid_amount ELSE 0 END AS pva
        ,CASE WHEN section_id_fk = stu_class THEN due_amount ELSE 0 END AS da
        ,CASE WHEN section_id_fk != stu_class THEN due_amount ELSE 0 END AS dva
        ,due_amount
        ,paid_amount
        ,section_id_fk
        ,fee_date
        ,stu_class 
        FROM 
            (SELECT 
            scholar
            ,due_amount
            ,SUM(paid_amount) AS paid_amount
            ,section_id_fk
            ,fee_date 
            FROM 
                (SELECT 
                student_fee.due_amount AS due_amount
                ,student_fee.paid_amount AS paid_amount
                ,student_fee.scholar_no AS scholar
                ,section_id_fk
                ,fee_date 
                FROM student_fee 
                ORDER BY STR_TO_DATE(fee_date,'%Y-%m-%d')DESC
                ) AS kkk 
            GROUP BY kkk.scholar,section_id_fk ORDER BY scholar
        ) AS k 
        LEFT OUTER JOIN student_details sd 
            ON k.scholar = sd.scholar_no
    ) AS lk
    group by scholar
) AS sssf 
ON student_details.scholar_no = sssf.scholar 
LEFT OUTER JOIN 
    (SELECT * 
    FROM section_details AS sd 
    LEFT OUTER JOIN class_details cd 
        ON sd.class_id_fk = cd.class_id_pk
    ) AS sc 
ON student_details.stu_class = sc.section_id_pk 
WHERE student_details.active = 'y' AND rte = 'N'

【讨论】：

非常感谢@ghkhan，是的，它现在按照要求工作。我可以在这里看到我在这里犯了一个愚蠢的错误#group by Academic is the key..thanks man

【解决方案3】：

我看到了您的查询，但您犯了一点错误，请在查询中的 AS lk 之后使用 GROUP BY scholar 或在查询中查找 AS lk 并替换 AS lk GROUP BY scholar...！

【讨论】：

是的，你检查得很好......！

【解决方案4】：

试试这个

SELECT 
student_details.scholar_no
,student_details.fname
,student_details.lname
,student_details.stu_class
,feedetails.fees
,class_name
,section_name
,IF(sssf.paid_amount IS NULL,0,sssf.paid_amount) AS paid_amount 
FROM 
    (student_details 
    LEFT OUTER JOIN feedetails 
        ON student_details.stu_class = feedetails.section_id_fk
    ) 
LEFT OUTER JOIN 
    (SELECT 
    scholar
    ,pa AS paid_amount
    ,pva AS prev_paid_amount
    ,da AS due_amount
    ,dva AS prev_due_amount
    ,section_id_fk
    ,fee_date
    ,stu_class 
    FROM 
        (SELECT 
        scholar
        ,CASE WHEN section_id_fk = stu_class THEN paid_amount ELSE 0 END AS pa
        ,CASE WHEN section_id_fk != stu_class THEN paid_amount ELSE 0 END AS pva
        ,CASE WHEN section_id_fk = stu_class THEN due_amount ELSE 0 END AS da
        ,CASE WHEN section_id_fk != stu_class THEN due_amount ELSE 0 END AS dva
        ,due_amount
        ,paid_amount
        ,section_id_fk
        ,fee_date
        ,stu_class 
        FROM 
            (SELECT 
            scholar
            ,due_amount
            ,paid_amount AS paid_amount
            ,section_id_fk
            ,fee_date 
            FROM 
                (SELECT 
    student_fee.due_amount AS due_amount,SUM(student_fee.paid_amount) AS paid_amount,student_fee.scholar_no AS scholar,section_id_fk,fee_date 
    FROM student_fee  GROUP BY scholar 
                ORDER BY STR_TO_DATE(fee_date,'%Y-%m-%d')DESC
                ) AS kkk 
             ORDER BY scholar
        ) AS k 
        LEFT OUTER JOIN student_details sd 
            ON k.scholar = sd.scholar_no
    ) AS lk
) AS sssf 
ON student_details.scholar_no = sssf.scholar 
LEFT OUTER JOIN 
    (SELECT * 
    FROM section_details AS sd 
    LEFT OUTER JOIN class_details cd 
        ON sd.class_id_fk = cd.class_id_pk
    ) AS sc 
ON student_details.stu_class = sc.section_id_pk 
WHERE student_details.active = 'y' AND rte = 'N'

【讨论】：

感谢您的努力，但在您的情况下，即使我为已支付金额插入了一些金额，payed_amount 仍显示为 0.. 看看上面的查询是否有效......！