PHP网站调试

Question

我有一个问题，当 无法运行此程序， 它显示

Parse error: syntax error, unexpected 'username' (T_STRING) in C:\xampp\htdocs\fypp\index.php on line 9

<?php 
    session_start(); 
    if (isset($_POST['bttLogin'])){ 
     require 'connect.php'; 
     $username = $_POST['username']; 
     $password = $_POST['password']; 
     $result = mysquli_query($con, 'select * from account where username="'.$username.'" and password="'.$password.'") 
     if(mysqli_num_rows($result)==1) { 
      $_SESSION['username'] = $username; 
      header("Location: welcome.php"); 
     } 
     else 
      echo "account is invalid"; 
    } 
    ?> 
    <form method="post"> 
    <table cellpadding="2" cellspacing="2" border="1"> 
     <tr> 
      <td>Username</td> 
      <td><input type="text" name="username"></td> 
     </tr> 
     <tr> 
      <td>Password</td> 
      <td><input type="password"" name="password"></td> 
     </tr> 
     <tr> 
      <td>&nbsp;</td> 
      <td><input type="submit" name="bttLogin" value="Login"></td> 
     </tr> 

谢谢回答:)

Answer 1

首先，正确粘贴代码.. 。您最后缺少</table>和</form> ...

其次，您错过了一个简单的撇号和一个半色调n的行

$result = mysquli_query($con, 'select * from account where username="'.$username.'" and password="'.$password.'")

此外，你应该使用mysqli_query代替mysqlui_query ...错字！

你的代码应该如下：

<?php 
session_start(); 
if (isset($_POST['bttLogin'])){ 
    require 'connect.php'; 
    $username = $_POST['username']; 
    $password = $_POST['password']; 
    $result = mysqli_query($con, 'select * from account where username="'.$username.'" and password="'.$password.'"'); 
    if(mysqli_num_rows($result)==1) { 
     $_SESSION['username'] = $username; 
     header("Location: welcome.php"); 
    } 
    else 
     echo "account is invalid"; 
} 
?> 

<form method="post"> 
<table cellpadding="2" cellspacing="2" border="1"> 
    <tr> 
     <td>Username</td> 
     <td><input type="text" name="username"></td> 
    </tr> 
    <tr> 
     <td>Password</td> 
     <td><input type="password"" name="password"></td> 
    </tr> 
    <tr> 
     <td>&nbsp;</td> 
     <td><input type="submit" name="bttLogin" value="Login"></td> 
    </tr> 
</table> 
</form> 

Answer 2

变化mysquli_query为mysqli_query，因为它不是一个有效的扩展，我认为它的错字错误。

修改的查询：

$result = mysqli_query($con, "SELECT * FROM account 
          WHERE username='" . $username . "' 
          AND password ='". $password ."'"); 

还要注意的是，你错过了终止分号（;）在同一行。