想要更新数据库中输入的条目.我有代码并显示错误:

这是我必须使用表单编辑更新来更新数据库中条目的代码：

<?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..我尝试过isset，然后也为分配的变量显示相同的错误。

Notice: Undefined index: name in C:xampphtdocsSignupupdate.php on line 3
Notice: Undefined index: age in C:xampphtdocsSignupupdate.php on line 4
Notice: Undefined index: email in C:xampphtdocsSignupupdate.php on line 5
Notice: Undefined index: contact in C:xampphtdocsSignupupdate.php on line 6
Notice: Undefined index: address in C:xampphtdocsSignupupdate.php on line 7
Notice: Undefined variable: id in C:xampphtdocsSignupupdate.php on line 10
Updated

<?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:";
}
?>
希望这能回答你的问题。如果是这样，请评价我的答案