插入表单数据到数据库故障

Question

好吧，所以我目前正在做一个系统采取的应用程序和它的工作，直到我加入剩余的问题：

<?php 
include_once('db.php'); 

$user =$_POST['username']; 
$job =$_POST['job']; 
$active =$_POST['active']; 
$why =$_POST['Q1'] 
$le =$_POST['Q2'] 
$skype =$_POST['skype'] 

if(mysqli_query($conn, "INSERT INTO app (username,job,active,why,le,skype) VALUES ('$user','$job','$active','$why','$le','$skype')")) 
echo"successfully inserted"; 
else 
echo "failed"; 
?> 

但是，当我有它这样的我得到这个错误

Parse error: syntax error, unexpected '$le' (T_VARIABLE) in C:\xampp\htdocs\app_insert.php on line 8

请记住我使用xampp与Apache和Mysql，任何人都知道发生了什么？

Answer 1

你错过了分号，;为：

$why =$_POST['Q1'] 
$le =$_POST['Q2'] 
$skype =$_POST['skype'] 

它应该是：

$why =$_POST['Q1']; 
$le =$_POST['Q2']; 
$skype =$_POST['skype']; 

Answer 2

你缺少分号和插入SQL语句似乎不正确。您正在将字符串传递给SQL语句而不是值

<?php 
include_once('db.php'); 

$user =$_POST['username']; 
$job =$_POST['job']; 
$active =$_POST['active']; 
$why =$_POST['Q1']; 
$le =$_POST['Q2']; 
$skype =$_POST['skype'] 

if(mysqli_query($conn, "INSERT INTO app (username,job,active,why,le,skype) VALUES ('{$user}','{$job}','{$active}','{$why}','{$le}','{$skype}')")) 
echo"successfully inserted"; 
else 
echo "failed"; 
?>