发送表格后发生错误

Question

你好，我的表格有问题，我不知道为什么我仍然没有填写整个表格。我看不出我做错了什么，你能帮助我吗？

<?php 
    if(isset($_POST['submit'])) 
    { 
     $username = mysqli_real_escape_string($_POST['username']); 
     $password1 = mysqli_real_escape_string($_POST['password1']); 
     $password2 = mysqli_real_escape_string($_POST['password2']); 
     $email = mysqli_real_escape_string($_POST['email']); 

     if (!empty($username) && !empty($password1) && !empty($password2) && !empty($email) && ($password1 == $password2)) 
     { 

      //my code 

     } 
     else 
     { 
      echo "u wot m8?!"; 
     } 

<form id="form" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> 
    <input type="text" id="username" name="username" placeholder="Your nickname" maxlength="15" /><br /> 
    <div id="check_av"></div> 
    <input type="password" id="password1" name="password1" placeholder="Your password" /><br /> 
    <input type="password" id="password2" name="password2" placeholder="Retype password" /><br /> 
    <input type="text" id="email" name="email" placeholder="Your email address" /><br /> 
    <input type="submit" id="submit" value="Sign up" name="submit" /> 
</form> 

Answer 1

我想说mysqli_real_escape_string错误的，因为它没有数据库连接，这是它的需要第一个参数（mysqli_real_escape_string (mysqli $link , string $escapestr)）。

$ php -r 'var_dump(mysqli_real_escape_string("something"));' 
Warning: mysqli_real_escape_string() expects exactly 2 parameters, 1 given in Command line code on line 1 

Call Stack: 
    0.0002  130868 1. {main}() Command line code:0 
    0.0002  130924 2. mysqli_real_escape_string() Command line code:1 

NULL 

NULL是empty所以，要么建立数据库连接第一，还是后来逃跑的时候你拥有了它。

道德故事：enable proper error reporting and catch those errors early & quickly