Question

与本书相比，我仔细检查了我的代码，我确信我正确输入了代码。页面加载，如果我将字段留空或者密码和确认密码字段不匹配，则显示错误。但是，如果我正确完成表单并提交，则MySQL数据库不会获取传递给它的信息。 “注册！”并且登录页面也没有显示。我直接通过Ubuntu检查了MySQL数据库，用户表仍为空（空集（0.00秒））。 phpMyAdmin也返回空，“MySQL返回一个空结果集（即零行）。（查询耗时0.0011秒）”，对于users表也是如此。

<!DOCTYPE HTML>                                   
<html lang="en">                                  
    <head>                                         
        <metacharset="UTF-8">                       
        <title>Register Page</title>             
    </head>                                        
    <body>                                         
        <?php
            $page_title = 'Register';
            include('/var/www/html/learn/php/header.html');

            if($_SERVER['REQUEST_METHOD'] == 'POST')
            {

                require('/var/www/connect_db.php');
                $errors = array();

                if(empty($_POST['first_name']))
                {
                    $errors[] = 'Enter your first name.';
                }
                else
                {
                    $fn = mysqli_real_escape_string($dbc, trim($_POST['first_name']));
                }

                if(empty($_POST['last_name']))
                {
                    $errors[] = 'Enter your last name.';
                }
                else
                {
                    $ln = mysqli_real_escape_string($dbc, trim($_POST['last_name']));
                }

                if(empty($_POST['email']))
                {
                    $errors[] = 'Enter your email address.';
                }
                else
                {
                    $e = mysqli_real_escape_string($dbc, trim($_POST['email']));
                }

                if(!empty($_POST['pass1']))
                {
                    if($_POST['pass1'] != $_POST['pass2'])
                    {
                        $errors[] = 'Passwords do not match.';
                    }
                    else
                    {
                        $p = mysqli_real_escape_string($dbc, trim($_POST['pass1']));
                    }
                }
                else
                {
                    $errors[] = 'Enter your password.';
                }

                if(empty($errors))
                {
                    $q = "SELECT user_id FROM users WHERE email='$e'";
                    $r = mysqli_query($dbc, $q);

                    if(mysqli_num_rows($r) != 0)
                    {
                        $errors[] = 'Email address already registered. <a href="login.php">Login</a>';
                    }
                }

                if(empty($errors))
                {
                    $q = "INSERT INTO users
                        (first_name, last_name, email, pass, reg_date)
                        VALUES ('$fn', '$ln', '$e', SHA1('$p'), NOW())";
                    $r = mysqli_query($dbc, $q);

                    if($r)
                    {
                        echo '<h1>Registered!</h1>
                            <p>You are now registered.</p>
                            <p><a href="login.php">Login</a></p>';
                    }

                    mysqli_close($dbc);
                    include('/var/www/html/learn/php/footer.html');
                    exit();
                }
                else
                {
                    echo '<h1>Error!</h1>
                    <p id="err_msg">The following error(s) occurred:<br>';
                    foreach($errors as $msg)
                    {
                        echo "- $msg<br>";
                    }
                    echo 'Please try again.</p>';
                    mysqli_close($dbc);
                }
            }
        ?>

        <h1>Register</h1>
        <form action="register.php" method="POST">
            <p>
                First Name: <input type="text" name="first_name" value="<?php if(isset($_POST['first_name'])) echo $_POST['first_name'];?>">
                Last Name: <input type="text" name="last_name" value="<?php if(isset($_POST['last_name'])) echo $_POST['last_name'];?>">
            </p>
            <p>
                Email Address: <input type="text" name="email" value="<?php if(isset($_POST['email'])) echo $_POST['email'];?>">
            </p>
            <p>
                Password: <input type="password" name="pass1" value="<?php if(isset($_POST['pass1'])) echo $_POST['pass1'];?>">
                Confirm Password: <input type="password" name="pass2" value="<?php if(isset($_POST['pass2'])) echo $_POST['pass2'];?>">
            </p>
            <p>
                <input type="submit" value="Register">
            </p>
        </form>

        <?php include('/var/www/html/learn/php/footer.html');?>
    </body>                                        
</html>

这是我认为我的问题所在，但我尝试了各种替代方案，仍然无法获取数据以将用户插入MySQL数据库：

if(empty($errors))
{
    $q = "INSERT INTO users
        (first_name, last_name, email, pass, reg_date)
        VALUES ('$fn', '$ln', '$e', SHA1('$p'), NOW())";
    $r = mysqli_query($dbc, $q);

    if($r)
    {
        echo '<h1>Registered!</h1>
            <p>You are now registered.</p>
            <p><a href="login.php">Login</a></p>';
    }

    mysqli_close($dbc);
    include('/var/www/html/learn/php/footer.html');
    exit();
}
else
{
    echo '<h1>Error!</h1>
    <p id="err_msg">The following error(s) occurred:<br>';
    foreach($errors as $msg)
    {
        echo "- $msg<br>";
    }
    echo 'Please try again.</p>';
    mysqli_close($dbc);
}

我不确定为什么这本书会让我输入“if（$ r）”，但请告诉我它的含义并显示我搞砸的地方。

Answer 1

你的问题：

您正在复制变量$ q＆amp; $ R。在这里改变

     if(empty($errors))
        {
       // changing the variable name here to $s and $m

            $s = "SELECT user_id FROM users WHERE  email='$e'"; 
            $m = mysqli_query($dbc, $s);

            if(mysqli_num_rows($m) != 0)
            {
                $errors[] = 'Email address already registered. <a href="login.php">Login</a>';
            }
        }

            $q = "INSERT INTO users
                (first_name, last_name, email, pass, reg_date)
                VALUES ('$fn', '$ln', '$e', SHA1('$p'), NOW())";
            $r = mysqli_query($dbc, $q);

同时删除附加的if语句，因为它没有任何用途。另外请使用参数化查询来防止sql注入。希望你明白。

PHP代码没有插入MySQL数据库

1 个答案: