Question

我正在尽最大努力简化我的问题，希望我的解释可以阐明其含义。

情况是这样的：

我有两个表，分别是：tbl_instructors，其中包含诸如教师ID和姓名之类的教师信息；以及tbl_advisory，其中包含由特定教师处理的该类的所有记录。类的Course，Year和Section。所以我的桌子是这样的：

tbl_instructor：

id  | name
----+--------------
001 | Jhon Doe
002 | Kaitlyn Moore

tbl_advisory：

instructor_id  | course        | year      |  section
---------------+---------------+-----------+-----------
001            | BSINT         | 2nd Year  | A
001            | BSINT         | 2nd Year  | C
001            | BSINT         | 2nd Year  | B
002            | BSBA          | 1st Year  | A
002            | BSBA          | 1st Year  | D

现在，我试图在我的PHP查询中使用INNER JOIN选择讲师ID和姓名及其相关的咨询信息：

<?php
$getAdvisory = "SELECT 
  ti.id AS id,
  ti.name AS name,
  ta.course AS ccourse,
  ta.year AS cyear,
  ta.section AS csection
FROM tbl_instructor as ti
INNER JOIN tbl_advisory as ta
ON ti.id = ta.instructor_id";
OpenConn()->query($getAdvisory);

if ($getAdvisory->num_rows > 0) {
    while ($row = $getAdvisory->fetch_assoc()) {
     $id = $row['id'];
     $name = $row['name'];
    }
    ?>
    <td><?php echo $id; ?></td>
    <td><?php echo $name; ?></td>
    <td><button>View Advisory</button></td>
    <?php
 }
}
?>

哪个给了我html / php表中的结果集：

id   | name          |  action
-----+---------------+-----------
001  | John Doe      | View Advisory
001  | John Doe      | View Advisory
001  | John Doe      | View Advisory
002  | Kaitlyn Moore | View Advisory
002  | Kaitlyn Moore | View Advisory

我的问题是我希望我的html表按教师ID进行区分，但仍返回每位教师的所有相关咨询信息，因为当我单击{{{1 }}按钮旁边的每个讲师姓名，以便在讲师删除或添加新咨询后我可以更新表格。

我该如何实现？预先感谢。

Answer 1

您需要DISTINCT子句

 SELECT DISTINCT 
  ti.id AS id,
  ti.name AS name,
  ta.course AS ccourse,
  ta.year AS cyear,
  ta.section AS csection
FROM tbl_instructor as ti
INNER JOIN tbl_advisory as ta ON ti.id = ta.instructor_id

通过这种方式，您应该获得

id   | name          |  action
-----+---------------+-----------
001  | John Doe      | View Advisory
002  | Kaitlyn Moore | View Advisory

如果值在不同的行上，则可以使用（伪）聚合函数

SELECT ti.id AS id,
ti.name AS name,
min(ta.course) AS ccourse,
min(ta.year) AS cyear,
min(ta.section)  AS csection
FROM tbl_instructor as ti
INNER JOIN tbl_advisory as ta ON ti.id = ta.instructor_id
GROUP BY id, name

Answer 2

请检查此查询：

SELECT    ti.*, GROUP_CONCAT( CONCAT( ta.course, ' ' , ta.year,' ', ta.section ) separator ' - ') as ColumnName 
FROM tbl_instructor as ti
INNER JOIN tbl_advisory as ta ON ti.id = ta.instructor_id 
GROUP BY ti.idid

输出：

      | name          |  action
 -----+---------------+-----------
 001  | John Doe      | BSINT 2nd Year a - BSINT 2nd Year b - BSINT 2nd Year c
 002  | Kaitlyn Moore | BSBA 1st Year d - BSBA 1st Year a

如何从一个表中选择数据并从另一表中获取所有相关记录？

2 个答案: