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Fatal error: Uncaught Error: Call to a member function query() on null in

Dawid

By Dawid
in PHP Coding Help

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Dawid Newbie

Dawid

Posted
Posted (edited)

Someone can help with this problem? :(

Error:

"Fatal error: Uncaught Error: Call to a member function query() on null in /profiles/d/d4/d4v/d4vez/listazakupow.cba.pl/pages/config.php:16 Stack trace: #0 /profiles/d/d4/d4v/d4vez/listazakupow.cba.pl/index.php(18): listaZakupow->usun('6') #1 {main} thrown in /profiles/d/d4/d4v/d4vez/listazakupow.cba.pl/pages/config.php on line 16"

 

index.php

<!doctype html>

<head>

    <meta charset="utf-8" />

    <title>Lista Zakupów</title>

</head>

<body>

    <?php

        require_once 'pages/config.php';

 

        $lz = new listaZakupow;

 

        if(isset($_GET['akcja'])){

            switch($_GET['akcja']){

                case 'dodaj':

                    $lz->dodaj($_POST['towar']);

                    break;

                case 'usun':

                    $lz->usun($_GET['id']);

                    break;

                case 'zmien':

                    $lz->zmienStan($_GET['id']);

                    break;

            }

        }

 

        $sql = 'SELECT * FROM zakupy';

        $result = mysqli_query($link,$sql);

 

        if(mysqli_num_rows($result) == 0){

            echo 'Nie masz jeszcze dodanego żadnego przedmiotu do kupienia';

        } else{

            echo 'Twoja lista zakupów:';

            while($row = mysqli_fetch_assoc($result)){

 

                $towarid = $row['id'];

                $towar = $row['towar'];

                $stan = $row['stan'];

 

                echo '<ul>';

                    if($stan=='T'){

                        $kupione = 'x';

                    } else{

                        $kupione = ' ';

                    }

                    echo '<li>'.$towar.' <a href="index.php?akcja=zmien&id='.$towarid.'">['.$kupione.']</a>

                    <a href="index.php?akcja=usun&id='.$towarid.'">usuń</a></li>';

                echo '</ul>';

            }

        }

    ?>

    <form action="index.php?akcja=dodaj" method="post">

        Towar: <input type="text" name="towar" />

        <input type="submit" value="Dodaj" />

</body>

</html>

 

config.php

<?php

    $link = new mysqli('localhost,'name','password','database');

        if ($link -> connect_errno) {

            echo 'Nie udało się połączyć z MySQL: ' . $mysqli -> connect_error;

            exit();

        }

   

    class listaZakupow{

 

        function dodaj($nazwa){

            $link->query('INSERT INTO zakupy VALUES(null,\''.mysqli_real_escape_string($nazwa,$link).'\',\'N\');');

            header("location: index.php");

        }

 

        function usun($id){

            $link->query('DELETE FROM zakupy WHERE id='.intval($id).' limit 1');

            header("location: index.php");

        }

 

        function zmienStan($id){

            $akt = mysqli_fetch_assoc($link->query('SELECT stan FROM zakupy WHERE id='.intval($id).' limit 1'));

            if($akt['stan']=='T'){

                $nw = 'N';

            } else{

                $nw = 'T';

            }

            mysqli_query('UPDATE zakupy SET stan=\''.$nw.'\' where id='.intval($id).' limit 1');

            header("location: index.php");

        }

 

    }

?>

Edited by Dawid
requinix Prolific Member

requinix

Posted
Posted 
function dodaj($nazwa){
	$link->query('INSERT INTO zakupy VALUES(null,\''.mysqli_real_escape_string($nazwa,$link).'\',\'N\');');

Variables like $link defined outside of functions are not available inside of functions.

Pass $link as an argument to the function, like you're already doing with $nazwa.

Dawid Newbie

Dawid

Posted
  • Author
Posted (edited)
43 minutes ago, requinix said: 
function dodaj($nazwa){
	$link->query('INSERT INTO zakupy VALUES(null,\''.mysqli_real_escape_string($nazwa,$link).'\',\'N\');');

Variables like $link defined outside of functions are not available inside of functions.

Pass $link as an argument to the function, like you're already doing with $nazwa.

Could you show me how? Because I guess I didn't understand 😕

Edited by Dawid
requinix Prolific Member

requinix

Posted
Posted

Spend some time learning about PHP so that you can understand what sort of changes you will be making. 

// this variable is defined outside of foo()
$outside = 123;

function foo() {
	// you cannot use $outside here
	echo $outside; // warning: undefined variable
}
foo();

// if you want to use a variable inside a function then you must pass it as a function argument

function bar($inside) {
	// you can use $inside here and it will have the same value as $outside
	echo $inside; // 123
}
bar($outside);

 

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