Question

这是表格：

<form>
            <div class="form-group ">
                                    <label for="exampleInputFirstName">First Name</label>
                                    <input type="text" class="form-control" id="exampleInputFirstName" name="FirstName" placeholder="First Name">
                              </div>
                              <div class="form-group">
                                    <label for="exampleInputLastName">Last Name</label>
                                    <input type="text" class="form-control" id="exampleInputLastName" name="LastName" placeholder="Last Name">
                              </div>
                              <div class="form-group">
                                    <label for="exampleInputEmail1">Email address</label>
                                    <input type="email" class="form-control" id="exampleInputEmail1" name="Email" aria-describedby="emailHelp" placeholder="Enter Email">
                                    <small id="emailHelp" class="form-text text-muted">We'll never share your email with anyone else.</small>
                              </div>
                              <div class="form-group">
                                    <label for="exampleInputPassword">Password</label>
                                    <input type="password" class="form-control" id="exampleInputPassword" name="Password" placeholder="Password">
                              </div>
                              <div class="form-group">
                                    <label for="exampleInputPassword1">Confirm Password</label>
                                    <input type="password" class="form-control" id="exampleInputPassword1" name="confirmPassword" placeholder="Confirm Password">
                              </div>
                              <button type="submit" name="submit" class="btn btn-primary">Submit</button>
                        </form>

这是php：

<?php 
        $link = mysqli_connect("shareddb-i.hosting.stackcp.net", "LoginCredentials-3337b6db", "subrat410", "LoginCredentials-3337b6db");

        if (mysqli_connect_error()) {

            die ("There was an error connecting to the database");

        } 
    if(isset($_POST['submit'])) {
        $FirstName = $_POST['FirstName'];
        $LastName = $_POST['LastName'];
        $Email = $_POST['Email'];
        $Password = $_POST['Password'];
        $confirmPassword = $_POST['confirmPassword'];
    }
        $query ="INSERT INTO `Login-Credentials`( `FirstName`, `LastName`, `Email`, `Password`, `ConfirmPassword`) VALUES (' $FirstName ',' $LastName ',' $Email ',' $Password ',' $confirmPassword ')";

        // $query = "UPDATE `LoginCredentials` SET password = 'uedjUFH7^%' WHERE email = 'robpercival80@gmail.com' LIMIT 1";

        mysqli_query($link, $query);
?>

每次我填写表格并提交时，都会在数据库中创建另一个空行。任何帮助将不胜感激。我已经实现了php.ini文件，它没有显示任何类型的错误。另外，在查看页面源代码时，我在声明$ FirstName = $ _ POST（'FirstName'）的php代码中看到未定义索引错误。谢谢

Answer 1

Lovelace是正确的，也许您的$ _POST变量仍未设置。您的表单标签没有方法或操作属性

<form actions="<?php echo htmlspecialchars($_SERVER['PHP_SELF']); ?>" method="POST">
    //the rest of your code is fine

</form>

Answer 2

“ <form>元素定义了如何发送数据。其所有属性旨在让您配置当用户单击“提交”按钮时要发送的请求。最重要的两个属性是action和method。” see more。

您的HTML代码可能类似于：

<form actions=”[target url]” method=“POST”>
        <div class="form-group ">
             <label for="exampleInputFirstName">First Name</label>
             <input type="text" class="form-control" id="exampleInputFirstName" name="FirstName" placeholder="First Name">
        </div>
        <div class="form-group">
             <label for="exampleInputLastName">Last Name</label>
             <input type="text" class="form-control" id="exampleInputLastName" name="LastName" placeholder="Last Name">
        </div>
       <div class="form-group">
             <label for="exampleInputEmail1">Email address</label>
             <input type="email" class="form-control" id="exampleInputEmail1" name="Email" aria-describedby="emailHelp" placeholder="Enter Email">
             <small id="emailHelp" class="form-text text-muted">We'll never share your email with anyone else.</small>
       </div>
       <div class="form-group">
            <label for="exampleInputPassword">Password</label>
            <input type="password" class="form-control" id="exampleInputPassword" name="Password" placeholder="Password">
       </div>
       <div class="form-group">
            <label for="exampleInputPassword1">Confirm Password</label>
            <input type="password" class="form-control" id="exampleInputPassword1" name="confirmPassword" placeholder="Confirm Password">
       </div>
      <button type="submit" name="submit" class="btn btn-primary">Submit</button>
 </form>

Answer 3

您的表单应定义一个操作和一个方法，它可能是使用GET而不是POST：

<form action="your/php/file.php" method="post">

请参阅此处以了解表单默认为GET请求的信息（如果使用GET请求路由，PHP $ _GET变量将非常有用）：What is the default form HTTP method?

在您的php中，通常使用$ _REQUEST [“ propertyName”]比使用$ _POST [“ propertyName”]或$ _GET [“ propertyName”]总是更好，因为它可以使用任一方法类型滚动

此外，您可能应该将查询放入if语句中，以防止出现空行。

最后，您可以尝试以下php行以查看发生了什么事情：

var_dump($_REQUEST);
var_dump($_GET);
var_dump($_POST);

让我知道您的发现，我们将继续帮助您解决问题

提交完全填写的表单在mysql中创建空行

3 个答案: