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Home/ Questions/Q 6629397
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Editorial Team
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Editorial Team
Asked: May 25, 20262026-05-25T22:15:15+00:00 2026-05-25T22:15:15+00:00

I am trying to declare a function parameter inside my array, but I’m having

  • 0

I am trying to declare a function parameter inside my array, but I’m having trouble getting it to work. I’ve trimmed it down for simplicity purposes, and I have something like:

function taken_value($value, $table, $row, $desc) {
    $value = trim($value); 
    $response = array();

    if (!$value) { 
        $response = array(
            'ok' => false, 
            'msg' => "This can not be blank."
        );
    } else if (mysql_num_rows(
                   mysql_query(
                       "SELECT * FROM $table WHERE $row = '$value'"))) {
        $response = array(
            'ok' => false,
            'msg' => $desc." is already taken."
        );  
    } else {
        $response = array(
            'ok' => true, 
            'msg' => ""
        ); 
    }
    echo json_encode($response); 
}

Notice the function parameter $desc trying to be used in the array here:

'msg' => $desc." is already taken.");

The whole function works fine EXCEPT when I try to add the $desc to the array results.

How could this be done?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-25T22:15:15+00:00Added an answer on May 25, 2026 at 10:15 pm

    Do you have an open resource handle to your database? You are not passing one to the query function.

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