我不理解的phpmyAdmin错误

Question

今天，我正在使用有关使用phpmyAdmin注册论坛的教程 。 直到到达这行代码（应该在SQL中输入）之前，一切都非常顺利。

1. INSERT INTO
2.        users(user_name, user_pass, user_email ,user_date, user_level)
3. VALUES('" . mysql_real_escape_string($_POST['user_name']) . "',
4.        '" . sha1($_POST['user_pass']) . "',
5.        '" . mysql_real_escape_string($_POST['user_email']) . "',
6.        NOW(),   
7.        0);

所以我在phpmyAdmin中输入了这个错误：

1064-您的SQL语法有误； 检查与您的MariaDB服务器版本相对应的手册，以在'user_name']附近使用正确的语法）。 “'，'”。 sha1（$ _ POST ['user_pass']）。 “'，'”。 我的行3

这是我用于signup.php的PHP

<?php
//signup.php
include 'connect.php';
include 'header.php';

echo '<h3>Sign up</h3>';

if($_SERVER['REQUEST_METHOD'] != 'POST')
{
    /*the form hasn't been posted yet, display it
      note that the action="" will cause the form to post to the same page it is on */
    echo '<form method="post" action="">
        Username: <input type="text" name="user_name" />
        Password: <input type="password" name="user_pass">
        Password again: <input type="password" name="user_pass_check">
        E-mail: <input type="email" name="user_email">
        <input type="submit" value="Add category" />
     </form>';
}
else
{
    /* so, the form has been posted, we'll process the data in three steps:
        1.  Check the data
        2.  Let the user refill the wrong fields (if necessary)
        3.  Save the data
    */
    $errors = array(); /* declare the array for later use */

    if(isset($_POST['user_name']))
    {
        //the user name exists
        if(!ctype_alnum($_POST['user_name']))
        {
            $errors[] = 'The username can only contain letters and digits.';
        }
        if(strlen($_POST['user_name']) > 30)
        {
            $errors[] = 'The username cannot be longer than 30 characters.';
        }
    }
    else
    {
        $errors[] = 'The username field must not be empty.';
    }


    if(isset($_POST['user_pass']))
    {
        if($_POST['user_pass'] != $_POST['user_pass_check'])
        {
            $errors[] = 'The two passwords did not match.';
        }
    }
    else
    {
        $errors[] = 'The password field cannot be empty.';
    }

    if(!empty($errors)) /*check for an empty array, if there are errors, they're in this array (note the ! operator)*/
    {
        echo 'Uh-oh.. a couple of fields are not filled in correctly..';
        echo '<ul>';
        foreach($errors as $key => $value) /* walk through the array so all the errors get displayed */
        {
            echo '<li>' . $value . '</li>'; /* this generates a nice error list */
        }
        echo '</ul>';
    }
    else
    {
        //the form has been posted without, so save it
        //notice the use of mysql_real_escape_string, keep everything safe!
        //also notice the sha1 function which hashes the password
        $sql = "INSERT INTO
                    users(user_name, user_pass, user_email ,user_date, user_level)
                VALUES('" . mysql_real_escape_string($_POST['user_name']) . "',
                       '" . sha1($_POST['user_pass']) . "',
                       '" . mysql_real_escape_string($_POST['user_email']) . "',
                        NOW(),
                        0)";

        $result = mysql_query($sql);
        if(!$result)
        {
            //something went wrong, display the error
            echo 'Something went wrong while registering. Please try again later.';
            //echo mysql_error(); //debugging purposes, uncomment when needed
        }
        else
        {
            echo 'Successfully registered. You can now <a href="signin.php">sign in</a> and start posting! :-)';
        }
    }
}

include 'footer.php';
?>

我不知道它在说什么。 请给我另一种方法！

Answer 1

您不能将PHP粘贴到phpMyAdmin中，phpMyAdmin仅适用于SQL。

您可以在phpMyAdmin下使用硬编码数据测试SQL，例如：

INSERT INTO
    users(user_name, user_pass, user_email ,user_date, user_level)
VALUES('admin',
    sha1('password'),
   'admin@example.com',
   NOW(),   
   0);

如果添加echo语句以显示$sql变量的内容，则可以使用该值并将其粘贴到phpMyAdmin中。

另外，请勿使用mysql扩展名-PHP 5.x中已弃用了该扩展名，并将其从PHP 7中删除。请改用mysqli或PDO。

Answer 2

SQL是拼写错误，输出已调试

$sql = "INSERT INTO 
    users(user_name, user_pass, user_email ,user_date, user_level) 
    VALUES('" . mysql_real_escape_string($_POST['user_name']) . "', 
    '" . sha1($_POST['user_pass']) . "', 
    '" . mysql_real_escape_string($_POST['user_email']) . "', 
    NOW(), 
    0)";