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mysql join重复行,顺序由问题引起

mysql php

0

Dipak · 技术社区 · 7 年前

我试着取每个班级的名字,这个班级的学生和分配给这个班级的科目。

在这里,问题是由 ORDER BY ,我已分配 订货人 student.id .

这导致 主题行复制 .

如果 获取主题行 正如所料,它们不是重复的,而是 .

表结构

id  |   Rid |   class   |   name
================================
1   |   1   |   1       |   John
2   |   2   |   1       |   Harsita

等级

id  |   title
==============
1   |   One
2   |   two
3   |   three

主题

id  |   title
===============
1   |   science
2   |   math
3   |   english
4   |   social
5   |   hpe

id  |   class   |   subject
===========================
1   |   1       |   1   
2   |   1       |   2
3   |   1       |   3
4   |   1       |   4

要求是

class: One
Science | Math | English | Social | Hpe | 
John | Harsita

如果 订货人

class: One
Science | Math | English | Social | Hpe | Math | English | Social | Hpe | Science | 
John | Harsita |

如果尚未分配给 学生id

class: One
Science | Math | English | Social | Hpe | 
John | Harsita | John | Harsita | John | Harsita | John | Harsita | John | Harsita |

GROUP BY 到 subject.id ,它只显示一个学生(第一行:John)。问题出在哪里?我怎么能不使用 subquery 或 GROUP_CONCAT

//mysql query
    SELECT 
        DISTINCT class_subject.class, 
        subject.title AS sub,
        student.name AS stdt,
        class.title AS class
    FROM 
        student
    INNER JOIN class_subject ON class_subject.class=student.class
    INNER JOIN subject ON subject.id=class_subject.subject
    INNER JOIN class ON class.id=student.class
    WHERE
        student.class=:cid;

//php
    $class='';
    $stdt='';
    $sub='';
    $results = Array();
        while($row=$result->fetch(PDO::FETCH_ASSOC)){
            if ($row['class'] != $class){
                $class = $row['class'];
                echo "Class: ".$class."<br/>";
            }
            if ($row['sub'] != $sub){
                $sub = $row['sub'];
                echo $sub." | ";
            }
            $results[]=$row;
        }
        echo "<br/>";
        foreach ($results as $row) {
            if ($row['stdt'] != $stdt) {
                $stdt = $row['stdt'];
                echo $stdt." | ";
            }
        }

2 回复 | 直到 7 年前

1

ScaisEdge 7 年前

查看预期结果sampel似乎需要name和subject的聚合结果

SELECT 
    group_concat( subject.title) AS sub,
    group_concat(student.name) AS stdt,
    class.title AS class
FROM 
    student
INNER JOIN class_subject ON class_subject.class=student.class
INNER JOIN subject ON subject.id=class_subject.subject
INNER JOIN class ON class.id=student.class
WHERE student.class=:cid
group by class.title

使用aggregation函数和group可以在同一行上显示聚合结果

2

1

Nick SamSmith1986 7 年前

JOIN

SELECT 
    c.title AS class,
    GROUP_CONCAT(DISTINCT s.title ORDER BY s.title SEPARATOR ' | ') AS subjects,
    GROUP_CONCAT(DISTINCT t.name ORDER BY t.name SEPARATOR ' | ') AS students
FROM class c
JOIN class_subject cs ON cs.class=c.id
JOIN subject s ON s.id=cs.subject
JOIN student t ON c.id=t.class
WHERE c.id=1
GROUP BY class

class   subjects                            students
One     english | math | science | social   Harsita | John

请注意,您可以指定 ORDER BY 内部 GROUP_CONCAT