Question

所以我一直在寻找论坛，但没有找到任何帮助我的东西！这是完整的代码。错误发生在$ result = $ conn-＆gt; query（$ query）; 我是php的新手，所以对我很轻松，哈哈。

<?php
include 'Register.html';
if(isset($_POST['Submit']))
{
    SignUp();
}
global $conn;
$servername = "localhost";
$username1 = "root";
$password1 = "";
$dbname = "fiesta_db";

// Create connection
$conn = new mysqli($servername, $username1, $password1, $dbname);

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
    return $conn;
}

function NewUser()
{
  $username = $_POST['username'];
  $email = $_POST['email'];
  $password = $_POST['password'];


  $register = "INSERT INTO website (username, email, password)
  VALUES ($username, $email, $password)";

  if ($conn->query($register) == TRUE) {
    echo "Successfully registered.";
  }
  else {
    echo "Wrong syntax";
  }
}
function SignUp()
{
  if(!empty($_POST['username']))
  {
    global $conn;
    $query = "SELECT * FROM website WHERE username ='$_POST[username]' AND email ='$_POST[email]'; ";
    $result = $conn->query($query);


    if ($result->num_rows != 0)
    {
      NewUser();
    }
    else {
      echo "Username or Email already exist!";
    }
  }
  $conn->close();
}
?>

希望你们能提供帮助（完全不只是想增加更多细节）

<!DOCTYPE HTML>
<html>
<head>
<title>Sign-Up</title>
</head>
<body id="body-color">
<div id="Sign-Up">
<form action='register_do.php' method='post'
    accept-charset='UTF-8'>
<fieldset >
<legend>Register</legend>
<input type='hidden' name='submitted' id='submitted' value='1'/>
<label for='username' >Username:</label>
<input type='text' name='username' id='username' maxlength="50" />
<label for='email' >Email Address:</label>
<input type='text' name='email' id='email' maxlength="50" />
<label for='password' >Password:</label>
<input type='password' name='password' id='password' maxlength="50" />
<input type='submit' name='Submit' value='Submit' />

</fieldset>
</form>
</div>
</body>
</html>

这是注册页面，我没有任何代码可以显示这就是一切。凸块

Answer 1

在您分配给主代码中的SignUp()之前，您正在调用$conn。移动该代码。

$servername = "localhost";
$username1 = "root";
$password1 = "";
$dbname = "fiesta_db";

// Create connection
$conn = new mysqli($servername, $username1, $password1, $dbname);

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
    return $conn;
}

if(isset($_POST['Submit']))
{
    SignUp();
}

您也不需要主代码中的global $conn;语句。在函数外部分配的变量是自动全局的，您只需要函数内部的语句来允许它访问全局变量。

您在NewUser()中的查询中也缺少引号：

$register = "INSERT INTO website (username, email, password)
    VALUES ('$username', '$email', '$password')";

在SignUp()中，您使用了错误的条件。如果用户名可用，SELECT查询将返回0行，因此应该是：

if ($result->num_rows == 0) {
    NewUser();
} else {
    echo "Username or Email already exist!";
}

致命错误：调用成员函数query（）错误

1 个答案: