PHP/MySQLi联系人表单无法与$\u POST一起使用

因此，当我将测试值放入php文件中时，我的表单似乎可以工作，如下所示：

$name = 'name';
$email = 'test@gmail.com';
$telephone = '123456789';
$message = 'test message';

但是，当我用php_POST替换测试值时，如下所示：

 $name = $_POST['name']

一切似乎都破裂了，我不知道为什么

在contact.html文件中获得了一个非常简单的contact-us表单设置：

<div class="container-fluid">
    <div class="row">            
        <div class="col-xs-12 col-sm-9 col-sm-offset-1 col-md-8 col-md-offset-2 col-lg-8 col-lg-offset-2">
            <div class="form-header">
                <h8>CONTACT US FORM</h8>
            </div>

            <form method="post" action="php/contact.php">
                <div class="row">
                    <div class="form-group col-xs-4 col-sm-4 col-md-4 col-lg-4">
                        <label>Name:</label>
                        <input type="text" name="name" class="form-control" onchange="capitalise()">
                    </div>
                    <div class="form-group col-xs-4 col-sm-4 col-md-4 col-lg-4">
                        <label>Email Address:</label>
                        <input type="email" class="form-control" name="email">
                    </div>
                    <div class="form-group col-xs-4 col-sm-4 col-md-4 col-lg-4">
                        <label>Phone Number:</label>
                        <input type="tel" class="form-control" name="tel">
                    </div>

                    <div class="clearfix"></div>

                    <div class="form-group col-xs-12 col-sm-12 col-md-12 col-lg-12">
                        <label>Message:</label>
                        <textarea class="form-control" rows="6"></textarea>
                    </div>
                    <div class="form-group col-xs-12 col-sm-12 col-md-12 col-lg-12">
                        <input type="hidden" name="msg" value="contact">
                        <button type="submit" class="btn btn-default">Submit</button>
                    </div>
                </div>
            </form>
        </div>            
    </div>
</div>

联系我们表格
姓名：
电邮地址：
电话号码：
信息：
提交

下面是使用准备好的语句的PHP联系人表单

<?php

$servername = "*******";
$username = "********";
$password = "********";
$dbname = "*********";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);

// Check connection
if ($conn->connect_error){
    die("Connection failed: " . $conn->connect_error);
}

$stmt = $conn -> prepare("INSERT INTO contact (name, email, telephone, message) VALUES (?, ?, ?, ?)");
$stmt->bind_param("ssss", $name, $email, $telephone, $message);

$name = $_POST['name'];
$email = $_POST['email'];
$telephone = $_POST['tel'];
$message = $_POST['msg'];
$stmt->execute();

echo "New records created successfully";

$stmt->close();
$conn->close();

?>

---

在将变量插入绑定之前，您需要声明变量：

<?php
    $servername = "*******";
    $username = "********";
    $password = "********";
    $dbname = "*********";

    // Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);

    // Check connection
    if ($conn->connect_error){
        die("Connection failed: " . $conn->connect_error);
    }

    $stmt = $conn -> prepare("INSERT INTO contact (name, email, telephone, message) VALUES (?, ?, ?, ?)");

    $name = $_POST['name'];
    $email = $_POST['email'];
    $telephone = $_POST['tel'];
    $message = $_POST['msg'];

    $stmt->bind_param("ssss", $name, $email, $telephone, $message);

    $stmt->execute();

    echo "New records created successfully";

    $stmt->close();
    $conn->close();
?>

定义“一切似乎都破裂了”。有错误吗？意外的结果？还有什么吗？尝试将POST数组/变量放在查询上方。使用错误报告并检查查询中的错误；你没那么做。使用条件语句检查isset/empty。在传递给函数之前，未定义变量。您确定没有收到任何错误吗？顺便说一句，

onchange=“capitalize（）”

建议您正在使用JS，这可能是造成此问题的原因，并且不知道是否与此相关。为帮助伙伴干杯！现在工作很好。看到一个被不公正地接受的错误答案总是很有趣的。答案怎么会错呢？它解决了OP的问题。更有趣的是，我想说，看到一个不公正的批评而没有任何批评。