Error updating database

1

I have the following problem:

I am trying to send data from a form and the problem arises when I make a decision and try to update

Code:

<?php 
include "plantillas/conexion.php";
    $clave = $_POST["clave"];
    $centro = $_POST["centro"];
    $site = $_POST["site"];
    $destino = $_POST["destino"];
    $marca = $_POST["marca"];
    $conector = $_POST["conector"];
    $puerto = $_POST["puerto"];

    $consulta_site = mysqli_query($con, "SELECT * FROM site WHERE centro = '".$centro."' AND nombre = '".$site."'")or die("Error: ". mysqli_error($con));

    while ( $row_site = mysqli_fetch_array($consulta_site)) {
        $id = $row_site["id"];
    } 

        if ($id != "") {

            $update = mysqli_query ($con,"UPDATE site 
                                        SET odf = '$clave'                                          


                                        WHERE id = '$id'") or die("fallo al actualizar los datos: ".mysqli_error($con));

                if ($update) {

                    $insertar = mysqli_query ($con,"INSERT INTO odf VALUES (NULL,'$clave','$centro', '$site', '$marca', '$conector', '$puerto', '$destino')");

                    header('Location: sistema.php');
                }else{
                    echo "Error al Insertar los datos";
                }

        } else{
            echo "El centro de trabajo no esta registrado";
        }


 ?>

and the error that marks me is:

  

Notice: Undefined variable: id in F: \ xampp \ htdocs \ CFE \ odf_site.php on   line 17

I hope and you can help me!

    
asked by Sharly Infinitywars 31.01.2018 в 21:56
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1 answer

1

Add $id=""; before your block while , or change if ($id != "") { for condition if (isset($id)) { on line 17 of your code

    
answered by 31.01.2018 / 22:04
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