Question

我有一个文件与我的HTML表单和另一个PHP文件与下面的代码。单击提交按钮后，我能够成功连接。但是当我在phpMyAdmin控制面板中刷新我的数据库时。没有数据被推送到我的数据库。

HTML表格

<form class="well" id="contactForm" name="sendMsg" novalidate="" action="contact_form.php" method="post">
        <div class="control-group">
            <div class="controls">
                <input class="form-control" name="fullname" id="name" type="text" placeholder="Full Name" required="" data-validation-required-message="Please enter your name" />
            </div>
        </div>
        <div class="control-group">
            <div class="controls">
                <input class="form-control" name="phonenumber" id="phone" type="text" placeholder="Phone Number" required="" data-validation-required-message="Please enter your phone number" />
            </div>
        </div>
        <div class="control-group">
            <div class="controls">
                <input class="form-control" name="emailaddress" id="email-address" type="email" placeholder="Email Address" required="" data-validation-required-message="Please enter your email" />
            </div>
        </div>
        <div class="control-group">
            <div class="controls">
                <textarea rows="6" class="form-control" name="message" id="msg" type="msg" placeholder="Enter detailed question/concern, and we will get back to you." required="" data-validation-required-message="Please enter your question/concern"></textarea>
            </div>
        </div>
        <div class="control-group">
            <div class="controls submit-btn">
                <button class="btn btn-primary" type="submit" value="Submit">Submit</button>
            </div>
        </div>
    </form>

CONTACT_FORM.PHP

<?php

define('DB_NAME', 'xxx');
define('DB_USER', 'xxx');
define('DB_PASSWORD', 'xxx');
define('DB_HOST', 'xxx');

$connection = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD);

if(!$connection){
die('Database connection failed: ' . mysqli_connect_error());
}

$db_selected = mysqli_select_db($connection, DB_NAME);

if(!$db_selected){
die('Can\'t use ' .DB_NAME . ' : ' . mysqli_connect_error());
}

echo 'Connected successfully';

$name = @$_POST ['fullname'];
$phone = @$_POST ['phonenumber'];
$email = @$_POST ['emailaddress'];
$msg = @$_POST ['message'];

$sql = "INSERT INTO contact_form (fullname, phonenumber, emailaddress, message) VALUES ('$name',   '$phone', '$email', '$msg')";

if (!mysqli_query($connection, $sql)){
die('Error: ' . mysqli_connect_error($connection));
}
?>

enter image description here

Answer 1

首先使用那种结构：

if (isset($_POST['fullname'])) {
    $name = $_POST['fullname'];
}

等等其他值。

或者更好的是： -

$name = isset($_POST['fullname']) ? $_POST['fullname'] : '';

这样，字段$name总是被设置为某种东西，所以当你稍后在代码中使用它时你不会出错

Answer 2

很明显，你没有传递列ID的值，以太改变了列AUTO_INCREMENT， enter image description here 第1步：单击更改列ID 第2步：检查A_I（AUTO_INCREMENT）

第3步：点击保存

或者另外一种方法也可以传递ID的值，如下所示：

$name = @$_POST ['fullname'];
$phone = @$_POST ['phonenumber'];
$email = @$_POST ['emailaddress'];
$msg = @$_POST ['message'];
$id = 1 // Set value for ID

$sql = "INSERT INTO contact_form (ID, fullname, phonenumber, emailaddress, message) VALUES ($id, '$name',   '$phone', '$email', '$msg')";

Answer 3

从@变量中删除$_POST符号，您还需要删除空格。我想让您意识到，在插入SQL值之前，如果没有清理/验证这些值，请使用ctype或filter_var。最后，我建议您将排序规则从latin1更改为utf8，除非您对该字符集有特定原因。

在$phone之前的插入值中有一些额外的空格可能导致它失败

VALUES ('$name',   '$phone', '$email', '$msg')

如另一个答案中所述，在尝试使用变量之前，您始终需要验证变量中是否有数据。此外，如果您在`contact_form.php中重定向，那么使用您的失败消息进行退出非常重要，否则您可能永远不会看到它们。

if($_POST['fullname']){
    $name = $_POST['fullname']
}else{
    echo "name not received";
    exit;
}
if($_POST['phonenumber']){
    $phone = $_POST['phonenumber']
}else{
    echo "phone not received";
    exit;
}
if($_POST['emailaddress']){
    $email = $_POST['emailaddress']
}else{
    echo "email not received";
    exit;
}
if($_POST['message']){
    $msg = $_POST['message']
}else{
    echo "message not received";
    exit;
}

$sql = "INSERT INTO contact_form (fullname, phonenumber, emailaddress, message) 
        VALUES ('$name', '$phone', '$email', '$msg')";

if (!mysqli_query($connection, $sql)){
die('Error: ' . mysqli_connect_error($connection));
}

Answer 4

$sql = "INSERT INTO contact_form 
        (fullname, phonenumber, emailaddress, message) 
        VALUES ('$name',   '$phone', '$email', '$msg')";

使用"代替'

表单不会将数据提交到MySQL数据库

4 个答案: