你的SQL語法有錯誤;檢查對應於你的MySQL服務器版本正確的語法使用近'假入JOIN

Question

我嘗試寫一個SQL代碼加入三個表在一起，但它始終顯示手動

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 

這裏是我的代碼

<?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); 
    ?> 

Answer 1

LEAVE是標準SQL關鍵字，可能是在MySQL一樣。嘗試使用反引號：

`LEAVE` 

Answer 2

離開保留MYSQL關鍵字（documentation） 只是逃避你的表名以``