Question

我是mysqli / php世界的新手，我试图通过直接潜入练习来学习，但我在开始时陷入困境。我做了一个简单的登录/传递表单，我想将数据存储到数据库，我不知道我在哪里犯了错误，但我一直没有得到任何数据。你能指点我吗？谢谢

                           登录：

            </tr>  

          <tr>
                <td>   
                <label class="contact" >Password :</label></td>
                <td>
                  <input class="input" type="password" name="Password" id="Password" /></td>
          </tr>




           <tr> 
           <td> 
            <input class="input" type="submit"  value="submit" /></td>
          </tr>




</table> 

</form>require_once('co.php');
         $username = "root";
         $password = "";
          $dbname = "test";

          // Create connection
          $conn = mysqli_connect("localhost", $username, $password,                    $dbname);
        // Check connection
         if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
                }
                   class insert {

public function insert () {

           if (isset($_POST['submit'] )){
           $Login=$_POST['Login'];
            $Password=$_POST['Password'];

          $sql="INSERT INTO test (Login, Pass)VALUES ('$Login', '$Password')";}
            mysqli_query($conn,$sql);
         if (mysqli_query($conn, $sql)) {
      echo "New record created successfully";
           } else {
                 echo "Error: " . $sql . "<br>" . mysqli_error($conn);
                 }
                }}



              include('inscription.php');





                ?>

Answer 1

试试这个：

<强> PHP

<?php
if (isset($_POST['submit'], $_POST['Login'], $_POST['Password'] )){
    $username = "root";
    $password = "";
    $dbname = "test";
    $conn = mysqli_connect("localhost", $username, $password,  $dbname);
    if (!$conn) {
        die("Connection failed: " . mysqli_connect_error());
    }

    $Login=mysqli_real_escape_string($_POST['Login'], $conn);
    $Password=mysqli_real_escape_string($_POST['Password'], $conn);
    $sql="INSERT INTO test (Login, Pass)VALUES ('$Login', '$Password')";
    mysqli_query($conn,$sql);
    if (mysqli_query($conn, $sql)) {
        echo "New record created successfully";
    } else {
        echo "Error: " . $sql . "<br>" . mysqli_error($conn);
    }
}else{
   echo 'missing fields '.var_dump($_POST);
}
?>

<强> FORM

＆＃13;

    <form action ="" method="POST">
    <table> 
    <tr>
        <td>   
            <label for="login" class="contact" >Login :</label></td>
        <td>
            <input id="login" class="input" type="text" name="Login" id="Login" />
        </td>
    </tr>
    <tr>
        <td>   
            <label for="password" class="contact" >Password :</label></td>
        <td>
            <input id="password" class="input" type="password" name="Password" id="Password" />
        </td>
    </tr>
    <tr> 
        <td> 
            <input class="input" type="submit"  value="submit" name="submit"/>
        </td>
    </tr>
    </table> 
    </form>

＆＃13;

登录/密码初学者的第一次尝试

1 个答案: