mysql - 想要更新数据库中输入的条目。我有代码并显示错误 :

Question

这是我必须使用表单编辑更新来更新数据库中的条目的代码:

<?php

$nm=$_POST['name'];
$ag=$_POST["age"];
$em=$_POST['email'];
$cn=$_POST['contact'];
$ad=$_POST['address'];

require("connection.php");
$sql = "update signup set name='$nm', age='$ag', email='$em', contact='$cn', address='$ad' WHERE id='$id'";
if(mysqli_query($conn,$sql))
{
    echo "Updated";
} else 
{
    echo "Error:";
}
?>

这些是显示的错误.. 我已经尝试过 isset 然后也显示了分配变量的相同错误。

Notice: Undefined index: name in C:\xampp\htdocs\Signup\update.php on line 3

Notice: Undefined index: age in C:\xampp\htdocs\Signup\update.php on line 4

Notice: Undefined index: email in C:\xampp\htdocs\Signup\update.php on line 5

Notice: Undefined index: contact in C:\xampp\htdocs\Signup\update.php on line 6

Notice: Undefined index: address in C:\xampp\htdocs\Signup\update.php on line 7

Notice: Undefined variable: id in C:\xampp\htdocs\Signup\update.php on line 10
Updated

Answer 1

<div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-html lang-html prettyprint-override"><code>    <?php
if(isset($_POST['name'])){
    $nm=$_POST['name'];
}
if(isset($_POST["age"])){
    $ag=$_POST["age"];
}
if(isset($_POST["email"])){
    $em=$_POST['email'];
}
if(isset($_POST["contact"])){
    $cn=$_POST['contact'];
}
if(isset($_POST["address"])){
    $ad=$_POST['address'];
}
    require("connection.php");
    $sql = "update 'signup' set 'name'='{$nm}', age='$ag', 'email'='{$em}', 'contact'='{$cn}', 'address'='{$ad}' WHERE 'signup'.'id'='{$id}'";
    if($conn->query($sql)==="TRUE")
    {
        echo "Updated";
    } 
else 
    {
       echo "Error:";
    }
    ?> </code></pre>
</div>
</div>

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