PHP和MySQL注册表

我正在学习php和mysql，并试图创建一个带有注册和登录页面的表单，但在获取注册表单将数据写入数据库时遇到了困难。我没有收到与数据库连接有关的错误，但是当我尝试发布数据时，我的表仍然是空的。我发现了错误

“注册时出现了一些问题。请稍后再试

非常感谢您的帮助。

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

 echo '<h2>Register </h2>';
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="Username" />
        Password: <input type="password" name="Password">
        Confirm Password: <input type="password" name="Confirm">

        <input type="submit" value="Submit" />
     </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($_POST['Username']))
    {
        //the user name exists
        //if(!ctype_alnum($_POST['Username']))
        if($_POST['Username'] == ['Username'])  
        {
            $errors[] = 'The username is already in use.';
        }

        }
    else
    {
        $errors[] = 'The username field must not be empty.';
    }

    if(isset($_POST['Password']))
    {
        if($_POST['Password'] != $_POST['Confirm'])
        {
            $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 errors, so save it
        //notice the use of mysql_real_escape_string, keep everything safe!
        //also notice the sha1 function which hashes the password
        $sql = "INSERT INTO
                    Users(Username, Password)
                VALUES('" . mysql_real_escape_string($_POST['Username']) . "',
                       '" . md5($_POST['Password']) . "',
                        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
        {
         header ("location: index.htm"); //redirects to Index Page


        }
    }
}

?> 

---

试试这个。我在

Users

表字段中添加了回勾号

  $username = mysql_real_escape_string($_POST['Username']);
  $password = md5($_POST['Password']);


  $sql= "INSERT INTO  Users(`Username`,`Password`) VALUES('$username',   '$password')";

您现在正在插入（）和0。2额外值

---

注意，调试SQL查询的第一步是首先在MySQL中运行它。因此，首先尝试在MySQL中使用

Username

和

Password

的伪值运行SQL语句，看看它是否有效

很简单：你没有检查错误，上帝只知道

connect.php

中的内容。你告诉脚本只在用户中插入（用户名、密码）但是您现在也在解析另外两个值（）和0），然后您就有了这段“我不知道是什么…”

if（$\u POST['Username']==['Username']）

。即使是这样也不安全！请已在PHP7中删除。了解有关使用PDO的语句，并考虑如何使用代码来检查用户名是否存在。使用mysqli或php更好吗？我建议您从一开始就使用mysqli而不是mysql。要检查用户是否存在，请使用mysql语句

选择COUNT（user_id）FROM Users WHERE

username

='$username'

如果结果大于0，则用户存在，否则如果结果为0，则无用户退出。实际上，正确的逻辑是，如果有一个用户使用该用户名，那么结果应该是1，而不是更多。要更改tpo mysqli，您应该从连接中将代码更改为mysqli