我尝试编写一个 sql 代码将三个表连接在一起,但它总是显示
You have an error in your SQL syntax;
check the manual that corresponds to your MySQL server version
for the right syntax to use near
'leave l JOIN employee e ON l.Emp_ID=e.Emp_ID JOIN department d ON e.Dept_ID= d.D' at line 1
这是我的代码
<?php
include("conn.php");
SESSION_START();
$aid = $_SESSION["eid"];
$check_user=mysql_query("select * from employee where Emp_ID='$aid'");
$row=mysql_fetch_assoc($check_user);
$leave = mysql_query("select * from leave");
$_GET['Leave_ID'] = $leave['Leave_ID'];
$leaveID = $_GET['Leave_ID'];
?>
<?php
$sql = mysql_query("select e.Emp_Fname, e.Emp_ID, e.Emp_Email, e.ContactNo_HP, e.ContactNo_Home, l.Date_Apply, l.Leave_Type, l.Leave_Start, l.Leave_End, l.Leave_Reason FROM leave l JOIN employee e ON l.Emp_ID=e.Emp_ID JOIN department d ON e.Dept_ID= d.Dept_ID where l.Leave_ID = $leaveID");
if($sql == FALSE)
{
die(mysql_error());
}
$rows = mysql_fetch_assoc($sql);
?>