Undefined variable in PHP file

Question

I am getting error Undefined variable: con at line:

$filter_Result = mysqli_query($con,$query);

in this PHP file:

<?php require_once('../Connections/conexion.php'); 

if(isset($_POST['search']))
{
    echo "estoy filtrando";
    $valueToSearch = $_POST['valueToSearch'];
    // search in all table columns
    // using concat mysql function
    $query = "SELECT * FROM `tblusuario` WHERE CONCAT(`idUsuario`, `strEmail, `strNombre`) LIKE '%".$valueToSearch."%'";
    $search_result = filterTable($query);

    $totalRows_DatosConsulta = mysqli_num_rows($search_result);

}
 else {

     echo "estoy sin filtrar";

     $query = "SELECT * FROM `tblusuario`";
     $search_result = filterTable( $query );

     $totalRows_DatosConsulta = mysqli_num_rows( $search_result );
}


function filterTable($query)
{
    $filter_Result = mysqli_query($con,$query);
    return $filter_Result;
}
?>

Here is file conexion.php, where variable $con is defined:

$hostname_con = "p:localhost";
$database_con = "ondemand";
$username_con = "root";
$password_con = "";
$con = mysqli_connect($hostname_con, $username_con, $password_con, $database_con);
mysqli_set_charset($con, 'utf8');

I have checked if file conexion.php is found correctly, and it is loaded.

I am not able to find the reason for the error.

Hari Lamichhane · Accepted Answer · 2017-03-25T10:45:19.673

1

You need to use it like this

function filterTable($query)
{
    global $con;
    $filter_Result = mysqli_query($con,$query);
    return $filter_Result;
}

or you can pass the $con as argument while calling the function.

   function filterTable($con, $query)
    {
        $filter_Result = mysqli_query($con,$query);
        return $filter_Result;
    }

It is due to scope of variable in PHP.

edited Mar 25 '17 at 10:45

answered Mar 25 '17 at 10:34

Hari Lamichhane

520
5
11

Using a global is the NASTY way of doing it – RiggsFolly Mar 25 '17 at 10:43
@RiggsFolly Added alternate solution too! – Hari Lamichhane Mar 25 '17 at 10:45
Thank yoy @HariLamichhane. – mvasco Mar 25 '17 at 10:50
Its my pleasure :) Remember to read more about variable scopes so that it will help you later. – Hari Lamichhane Mar 25 '17 at 10:52

score 1 · Answer 2 · answered Mar 25 '17 at 10:35

1

function filterTable($con, $query)
{
    $filter_Result = mysqli_query($con,$query);
    return $filter_Result;
}

answered Mar 25 '17 at 10:35

Tkachenko

25
4

Undefined variable in PHP file

2 Answers2