Question

我似乎无法弄清楚为什么我会收到此错误。语法完全没问题，不是吗？它显然不像切换

那么简单

<?

代表

<?php

谢谢你们！

<?php

if (isset($_POST['username'])&&isset($_POST['password'])) {
$username = $_POST['username'];
$password = $_POST['password'];
$password_hash = md5($password);

if(!empty($username)&&!empty($password)) {

    $query = "SELECT `id` FROM `devand_users` WHERE `username`='$username' AND `password`='$password_hash'";
        if ($query_run = mysql_query($query)) {
            $query_num_rows = mysql_num_rows($query_run);

        if ($query_num_rows==0) {
            echo 'Invalid username or password combination.';
        }   else if ($query_num_rows==1) { 
            echo $password_hash;
            }
    }

}

?>

<form action="<?php echo $current_file; ?>" method="POST">
Username: <input type="text" name="username"> 
Password: <input type="password" name="password">
<input type="submit" value="Login">
</form>

Answer 1

您的{和}不匹配。排列它们，计算它们。

Answer 2

括号不匹配。看到您的代码格式有点不同：

if (isset($_POST['username'])&&isset($_POST['password']))
{
    $username = $_POST['username'];
    $password = $_POST['password'];
    $password_hash = md5($password);

    if(!empty($username)&&!empty($password))
    {
        $query = "SELECT `id` FROM `devand_users` WHERE `username`='$username' AND `password`='$password_hash'";
        if ($query_run = mysql_query($query))
        {
            $query_num_rows = mysql_num_rows($query_run);

            if ($query_num_rows==0)
            {
                echo 'Invalid username or password combination.';
            }   else if ($query_num_rows==1)
            { 
                echo $password_hash;
            }
        }

    }

你永远不会关闭第一个......

Answer 3

你错过了一个近距离。

修正：

if (isset($_POST['username'])&&isset($_POST['password'])) {
  $username = mysql_real_escape_string($_POST['username']);
  $password = $_POST['password'];
  $password_hash = md5($password);

  if(!empty($username)&&!empty($password)) {
    $query = "SELECT `id` FROM `devand_users` WHERE `username`='$username' AND `password`='$password_hash'";

    if ($query_run = mysql_query($query)) {
      $query_num_rows = mysql_num_rows($query_run);

      if ($query_num_rows==0) {
        echo 'Invalid username or password combination.';
      } elseif ($query_num_rows==1) { 
        echo $password_hash;
      }
    }// the missing close bracket?
  }
}

这就是为什么你总是缩进你的代码!!!

此外，您绝对应该使用mysql_real_escape_string()来防止SQL注入攻击。

Answer 4

你错过了一个结束括号：

        if(!empty($username)&&!empty($password)) {

            $query = "SELECT `id` FROM `devand_users` WHERE `username`='$username' AND `password`='$password_hash'";
            if ($query_run = mysql_query($query)) {
                $query_num_rows = mysql_num_rows($query_run);

                if ($query_num_rows==0) {
                    echo 'Invalid username or password combination.';
                }   else if ($query_num_rows==1) { 
                        echo $password_hash;
                    }
            }
        }
    }      <- Missing This
?>

<form action="<?php echo $current_file; ?>" method="POST">
    Username: <input type="text" name="username"> 
    Password: <input type="password" name="password">
    <input type="submit" value="Login">
</form>

“解析错误：语法错误，意外的$ end in ..”我的语法很好但是？

4 个答案: