PHP Undefined Variable Error (But it is defined?)

Question

I am not sure where my error is but it says Sort is undefined and also an error about Invalid query: Erreur de syntaxe près de 'GROUP BY Mall ORDER BY Counter DESC' à la ligne 1

Here is an excerpt of my code about regarding the display of stores.

if(isset($_GET['formSubmit'])) 
    {
        $varStats = $_GET['formStats'];
        $errorMessage = "";

            switch($varStats)
            {

            case "ByAMEX": $sort = "AMEX = 1"; break;
            case "ByBDOTerminal": $sort = "BDOTerminal = 'YES'"; break;
            case "ByResto": $sort = "Resto = 'YES'"; break;
            case "ByCashOnly": $sort = "Cash = 'YES'"; break;
            case "ByPOSNotSeen": $sort = "POSNotSeen = 'YES'"; break;
            }

            $conn = db_connect();
            sortMall($conn, $sort);
            db_disconnect($conn);

            exit();
    }

Here is an excerpt where I initialized the Sort function:

function sortMall($conn, $sort)
{
$table = "test";
$column1 = "StoreName";
$column2 = "Mall";


$query1 = "Select $column2, Count($column1) AS Counter from $table where $sort GROUP BY $column2 ORDER BY Counter DESC"; 

$result = mysqli_query($conn, $query1);
    if ($result == FALSE)
    {
        echo "Invalid query: " . $conn->error;

        echo "<br/>";
        return;
    }

I actually have another function that almost does the same with the same syntax and all, it works. I don't know why this does not.

where is the warning which tells you about the undefined variable? — B001ᛦ, Jul 27 '16 at 11:06
define variable to null at starting of the script. like $variable=null; — Vamsi Abbineni, Jul 27 '16 at 11:10
It says that the error is on the line sortMall($conn, $sort); — AndyMarty, Jul 27 '16 at 11:11
Thank you guys. Made the Undefined Variable issue disappear. I am now left with the Invalid query: Erreur de syntaxe près de '% GROUP BY Mall ORDER BY Counter DESC' à la ligne 1 error. I don't get it because this syntax works if I run it on SQL — AndyMarty, Jul 27 '16 at 11:17
Could you add the resulting query you get as a text here as well? — Armando Bracho, Jul 27 '16 at 11:18

score 0 · Answer 1 · answered Jul 27 '16 at 11:08

0

$varStats may not be in the case you defined, add a default case

If so, $sort isn't defined that's your problem

answered Jul 27 '16 at 11:08

jeanj

2,106
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22

score 0 · Answer 2 · answered Jul 27 '16 at 11:13

Try this

if(isset($_GET['formSubmit'])) 
    {
        $varStats = isset($_GET['formStats']) ? $_GET['formStats'] : '';
        $errorMessage = "";

            switch($varStats)
            {

            case "ByAMEX": $sort = "AMEX = 1"; break;
            case "ByBDOTerminal": $sort = "BDOTerminal = 'YES'"; break;
            case "ByResto": $sort = "Resto = 'YES'"; break;
            case "ByCashOnly": $sort = "Cash = 'YES'"; break;
            case "ByPOSNotSeen": $sort = "POSNotSeen = 'YES'"; break;
            default: $sort = 'yourvalue';// default case
            }

            $conn = db_connect();
            sortMall($conn, $sort);
            db_disconnect($conn);

            exit();
    }

may be your $sort not getting value from switch

score 0 · Answer 3 · answered Jul 27 '16 at 11:17

Try this,

 if(isset($_GET['formSubmit'])) 
{
    $varStats = isset($_GET['formStats']) ? $_GET['formStats'] : '';
    $errorMessage = "";
    $sort = "1=1";   

        switch($varStats)
        {

        case "ByAMEX": $sort = "AMEX = 1"; break;
        case "ByBDOTerminal": $sort = "BDOTerminal = 'YES'"; break;
        case "ByResto": $sort = "Resto = 'YES'"; break;
        case "ByCashOnly": $sort = "Cash = 'YES'"; break;
        case "ByPOSNotSeen": $sort = "POSNotSeen = 'YES'"; break;
        default: $sort = 'yourvalue';// default case
        }

        $conn = db_connect();
        sortMall($conn, $sort);
        db_disconnect($conn);

        exit();
}

PHP Undefined Variable Error (But it is defined?)

3 Answers3