Php “解析错误：语法错误，意外的$end in…”但我的语法还好吗？

我似乎不明白为什么我会犯这个错误。语法完全正确，不是吗？很明显，这并不像切换开关那么简单

<? 

为了

谢谢大家

<?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>

---

您的{和}不匹配。把它们排成一行，数一数。

你缺少一个紧括号

固定的：

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

---

另外，您肯定应该使用来防止SQL注入攻击。

您缺少一个右括号：

---

括号不匹配。查看格式稍有不同的代码：

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;
            }
        }

    }

---

您永远不会关闭第一个..

语法显然不正确。这样的随机缩进永远不会让你发现问题。你在哪一行得到错误？我是说，得了吧，那是在代码的最后一行。这一行包括了我的html表单结束标记。恭喜，你赢得了最快的速度。我不会在这样的问题上花费超过一分钟的时间。我说的是你回答的时间，因为在同一时间出现了这么多的答案，说的是同一件事！我还在学习。我还在努力掌握窍门。一旦我了解了安全性的基本知识，就会涉及到安全性。阻止SQL注入攻击的基本经验法则是避开SQL语句中包含的任何用户输入。mysql\u real\u escape\u字符串非常适合这种情况。祝你好运，坚持下去！

        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>

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;
            }
        }

    }