示例
';
$id=2;
$name="Rahul";
$salary=80000;
$sql = "update emp4 set name="$name", salary=$salary where id=$id";
if(mysqli_query($conn, $sql)){
echo "Record updated successfully";
}else{
echo "Could not update record: ". mysqli_error($conn);
}
mysqli_close($conn);
?>
执行上面示例代码，得到以下结果 -
Connected successfully
Record updated successfully
