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Evaluating undefined JS variable returns error


By : APi
Date : October 16 2020, 08:10 PM
Does that help So, this is apparently not allowed in Javascript today? , You can simply do:
code :
typeof fbq || fbq === null // undefined


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templated html returns server error when encountering an undefined variable


By : Vinicius Simionato
Date : March 29 2020, 07:55 AM
may help you . If you run the doGet() method in the editor you should see the error:
code :
<? if (this.a) {

My First Function returns undefined variable ($pageName) error on line #14


By : xlambein
Date : March 29 2020, 07:55 AM
help you fix your problem Restructured Answer
You are being shown an error because your function is looking for data & using the information from your comment I don't know why but stack overflow kills anything i plop inside title tags: This is what is in the title tags:
code :
function TestFunc(){  }
function GetPageName() {
    $fileName = basename($_SERVER['PHP_SELF']);
    $pageName = str_replace('.php','', $fileName);
    return $pageName;
}

session redirect with php returns undefined variable and call to a member function error


By : paul egberts
Date : March 29 2020, 07:55 AM
hope this fix your issue i have the following code which comes from another page on submit with a value, and gets the rest of the values from the database. but im getting 2 errors. , You need to change your code like below:-
code :
<?php
error_reporting(E_ALL);
ini_set('display_errors',1);
session_start(); // must be first line
if(isset($_POST["submit"])){
    $tmpqid = $_POST["mdqid"];
    require_once 'dbconfig.php';
    try {
        $conn = new PDO("mysql:host=$host;dbname=$dbname", $username, $password);
        $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
        $sql = "SELECT * FROM as_questions WHERE Qid= $tmpqid";
        $q = $conn->query($sql);
        $q->setFetchMode(PDO::FETCH_ASSOC);
        // enter your while code in try block
        while ($r = $q->fetch()){
            $tempQid = $r['Qid'];
            $tempMclass = $r['M_class'];
            $rawMclass = $r['M_class'];
            $tempSclass = $r['S_class'];
            $rawSclass = $r['S_class'];
            $tempQuestion = $r['Question'];
            $tempAnswer = $r['Answer'];
            $tempDoc = $r['Doctor'];
            $rawDoc = $r['Doctor'];
            $tempTime = $r['Time'];
            $tempAtime = $r['A_time'];

            $_SESSION['qid'] = $tempQid;
            $_SESSION['mclass'] = $tempMclass;
            $_SESSION['rmclass'] = $rawMclass;
            $_SESSION['sclass'] = $tempSclass;
            $_SESSION['rsclass'] = $tempSclass;
            $_SESSION['question'] = $tempQuestion;
            $_SESSION['answer'] = $tempAnswer;
            $_SESSION['doc'] = $tempDoc;
            $_SESSION['time'] = $tempTime;
            $_SESSION['atime'] = $tempAtimeime;

            header("Location: http://localhost/pro/multipledelete.php");
            exit();
        }
    } catch (PDOException $pe) {
        die("Could not connect to the database $dbname :" . $pe->getMessage());
    }
}

?>

Variable can be echoed but PHP returns undefined error when trying to use it with PDO


By : Mårten
Date : March 29 2020, 07:55 AM
This might help you I am trying to save a variable's value into a db using PDO, however I am getting Undefined variable error. , You have to prepare your statement before you can use it.
code :
<?php
    $sql = "SELECT po_number FROM po ORDER BY po_number DESC LIMIT 1 ;";
    $STH = $pdo ->query($sql);
    $row = $STH->fetch();
    $myVar = $row['po_number'];
?>

<?php
    if(isset($_POST['addProduct'])){
        echo $myVar;  //the variable can be echoed with the expected value here

        $sql1 = "INSERT INTO po_parts (po_number, part_id, quantity) VALUES (?,?,?)";

        // Inserted this row
        $statement1 = $pdo->prepare($sql1);

        $statement1->bindParam(1, $myVar);
        //I am getting error that $myVar is undefined on the previous line

        $statement1->bindParam(2, $_POST['part']);
        $statement1->bindParam(3, $_POST['quantity']);
        if($statement1->execute()){
            $message = "Product was added successfully to the order number $newVar";
        }else{
            $message = "OP's was NOT added successfully.";
        }
    }
?>

PHP - Read more page returns 'undefined variable' error


By : vimal
Date : March 29 2020, 07:55 AM
it fixes the issue I've been working on a 'read more' page for quite a while now. What I basically want is when you click read more, it will redirect to the news item page. , You'd probably have to change your code for your newspage like this:
code :
<?php
    $con = mysqli_connect("localhost", "root", "", "groenevingers");

    if (mysqli_connect_errno()) {
    die("Failed to connect to MySQL: " . mysqli_connect_error());
    }
    if(isset($_GET['nieuwsitem'])) {
        $nieuws = intval($_GET['nieuwsitem']);
        $p=mysqli_query($con, "SELECT id, titel, tekst, datum FROM nieuws WHERE id='$nieuws'");

        if(mysqli_num_rows($p) > 0) {
            if ($row = mysqli_fetch_array($p)){
                echo "<h2>".$row['1']."</h2>";
                echo "<div class='clr'></div>";
                echo "<p>".$row['2']."</p>";
                echo "<p> Geplaatst op: ".$row['3']."</p>";
            }
        } else {
            echo "Item niet gevonden.";
        }
    } else {
        echo "Item niet gevonden.";
    }  
?>
if(isset($_GET['nieuwsitem'])) {
$nieuws = intval($_GET['nieuwsitem']);
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