我有 3 张桌子
1.employee_info,2.部门,3.职位
喜欢图片
employee_info_table
department_table
position_table
我连接了 2 个表employee_info表和部门表,它成功了
$sql = "SELECT t.*, d.Name AS d_name
FROM employee_info t
LEFT JOIN department d ON t.department_id = d.id
ORDER BY id ASC";
结果结果,但是当我连接 3 个表时出现错误。这是我的代码:
$sql = "SELECT t.*, p.* d.Name AS d_name
FROM employee_info t
JOIN department d ON t.department_id = d.id
JOIN position p ON p.id = t.position_id;
ORDER BY Name ASC";
HTML 代码显示表
<table>
<?php
$result = mysqli_query( $connect, $sql );
while ( $row = mysqli_fetch_assoc( $result ) ) { ?>
<tr>
<td><?php echo $row['Id'] ?></td>
<td><?php echo $row["Name"]; ?></td>
<td><?php echo $row["Sex"]; ?></td>
<td><?php echo $row["Age"]; ?></td>
<td><?php echo $row['email'] ?></td>
<td><?php echo $row['d_name'] ?></td>
</tr>
<?php } ?>
</table>
你的 SELECT 中有语法错误。p.*
后缺少逗号,position_id后还有一个额外的分号:
$sql = "SELECT t.*, p.*, d.Name AS d_name
FROM employee_info t
JOIN department d ON t.department_id = d.id
JOIN position p ON p.id = t.position_id
ORDER BY Name ASC";