表单提交的数据未保存到phpmyadmin中

我在apache服务器上的localhost中使用下面的php代码，当我以html格式提交数据时，它没有显示任何错误，而且一切似乎都正常，但数据没有保存在phpmyadmin表中。有人能帮忙吗

<?php
$servername = 'localhost';
$username = 'root';
$password = 'xxxx';
$database = 'newtable';
$con = mysqli_connect("$servername","$username","$password","$database");
if (! $con){
    die('Could not connect: ' . mysqli_error());
}
$sql = "INSERT INTO newtable (firstname, lastname) VALUES  ('$_POST[firstname]', '$_POST[lastname]')";
if (! $sql)
{
    die('Error: ' . mysqli_error());
}
echo "Record Added Successfully!";
mysqli_close($con);
?>

---

您只需编写查询，而忘记执行查询

$sql = "INSERT INTO newtable (firstname, lastname) VALUES  ('".$_POST['firstname']."', '".$_POST['lastname']."')";
$result=mysqli_query($con,$sql);// execute it

if (! $result)
{
die('Error: ' . mysqli_error($con));// need to pass connection as parameter
}

阅读



最好使用bind语句来防止表单sql注入

$sql = "INSERT INTO newtable (firstname, lastname) VALUES  (?, ?)";
$stmt = $con->prepare($sql);

$stmt->bind_param("ss", $firstname, $lastname);
$stmt->execute();

您忘了执行查询，请使用下面的语句

$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];

$sql = "INSERT INTO newtable (firstname, lastname) VALUES  (?, ?)";
$stmt = $con->prepare($sql);

$stmt->bind_param("ss", $firstname, $lastname);
$stmt->execute();

您没有在任何地方执行insert查询语句，因此没有添加数据

替换以下行：

if (! $sql)
{
die('Error: ' . mysqli_error());
}

与

谢谢你们的回答，它成功了

我只需要添加$result=mysqli\u查询（$con，$sql）

这是程序的执行吗？
打字速度比我今天快，打得好！以更安全的形式更正！！请回答下面给出的答案。
if (! $sql)
{
die('Error: ' . mysqli_error());
}

if ($mysqli->query($con, $sql) !== TRUE)
{
die('Error: ' . mysqli_error($con));
}

<?php 
$servername = 'localhost';
$username = 'root'; 
$password = ''; 
$database = 'newtable'; $con = mysqli_connect("$servername","$username","$password","$database");
if (! $con){ die('Could not connect: ' . mysqli_error()); } 
$sql = "INSERT INTO yourTableName (firstname, lastname) VALUES ('".$_POST['firstname']."', '".$_POST['lastname']."')"; 
if (! $sql) { die('Error: ' . mysqli_error()); } echo "Record Added Successfully!"; mysqli_close($con); 
?>