Question

我正在尝试制作一份注册表格。单击提交时，此表单需要将数据添加到数据库。它没有给出任何错误或消息。这是我的代码，希望有人能帮助我。

<body>
 <form action="" method"post" class="form">
    <div class="form-group1">
    <label for="" >Username :</label>
    <input type="text" class="form-control" name="username" id="tb-username" placeholder="Username">
    </div>
    <div class="form-group2">
    <label for="" >Password :</label>
    <input type="password" class="form-control" name="password" id="tb-password" placeholder="Password">
    </div>
    <button type="submit" class="btn btn-primary" id="btn-submit" name="register">Submit</button>
    </form>

    <?php

    if(isset($_POST['register']))
    {
    if(isset($_POST['username'], $_POST['password']))
    {
        $username = $_POST['username'];
        $password = $_POST['password'];

        $link = mysqli_connect("localhost", "root", " ", "vbproject");

        if(mysqli_connect_errno())
        {
            printf("Connect failed: %s\n", mysqli_connect_errno());
            exit();
        }

        $sql = "INSERT INTO users (username, password)
                VALUES('$username', '$password')";

        mysqli_close($link);
    }
    }
     ?>

</body>

Answer 1

你这里有错误

<form action="" method"post" class="form">

用

替换它

<form action="" method="post" class="form">

您错过了导致=请求的GET符号，并且您正在以POST方式访问

Answer 2

你也需要：

 $sql = "INSERT INTO users (username, password)
        VALUES('$username', '$password')";
 $result = $link->query($sql);
 if($result === true){
  echo "Insert Success";
 }else{
  echo "Insert failed";
  }

Answer 3

请在表单标记中将method"post"更改为method="post"。

 <body>

    <form action="" method="post" class="form">
    <div class="form-group1">
    <label for="" >Username :</label>
    <input type="text" class="form-control" name="username" id="tb-username" placeholder="Username">
    </div>
    <div class="form-group2">
    <label for="" >Password :</label>
    <input type="password" class="form-control" name="password" id="tb-password" placeholder="Password">
    </div>
    <button type="submit" class="btn btn-primary" id="btn-submit" name="register">Submit</button>
    </form>

    <?php

    if(isset($_POST['register']))
    {
    if(isset($_POST['username'], $_POST['password']))
    {
        $username = $_POST['username'];
        $password = $_POST['password'];

        $link = mysqli_connect("localhost", "root", " ", "vbproject");

        if(mysqli_connect_errno())
        {
            printf("Connect failed: %s\n", mysqli_connect_errno());
            exit();
        }

        $sql = "INSERT INTO users (username, password)
                VALUES('$username', '$password')";
        if(mysqli_query($link, $sql)){
         echo 'Inserted';
        }else{
          echo 'Not Inserted';
        }
        mysqli_close($link);
    }
    }
     ?>

    </body>

在代码中添加以下行：

if(mysqli_query($link, $sql)){
         echo 'Inserted';
        }else{
          echo 'Not Inserted';
        }

Answer 4

您没有运行mysql查询。请找到以下解决方案：

$sql = "INSERT INTO users (username, password)
            VALUES('$username', '$password')";

mysqli_query($con, $sql); // this Line Missing

mysqli_close($link);

Answer 5

这就是流程的方式

<?php
$con=mysqli_connect("localhost", "root", " ", "vbproject");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

// Perform queries 
 $sql = "INSERT INTO users (username, password)
            VALUES('$username', '$password')";
mysqli_query($con, $sql);

mysqli_close($con);
?>

Answer 6

在关闭连接之前添加此行，它将执行您正在执行的查询

mysqli_query($link,$sql);

Answer 7

只是观察你的代码

<?php
if(isset($_POST['register']))
{

但$ _POST ['register'}尚未分配任何值，if语句将失败建议你在这里添加值=“提交”

<button type="submit" class="btn btn-primary" id="btn-submit" name="register">Submit</button>

注册表单PHP SQL

7 个答案: