Question

我正在制作一个表格来更新我的MySQL数据库，但由于某种原因，我在浏览器中收到以下错误：

C:\wamp\www\Helpdeskapplicatie\update_hardware.php on line 49

有人能告诉我我做错了吗？

<html>
    <head>
        <link rel="stylesheet" type="text/css" href="css/layout.css"/>
    </head>
    <body>
        <div id="menu">
            <div id="menu_wrapper">
                <ul>
                    <li>Configuratiebeheer<img src="afb/pijltje.png" width="10"/></a>
                        <ul>
                            <li><a href="configuratiebeheer_hardware.php">Lijst hardware</a></li>
                            <li><a href="hardware_toevoegen.php">Hardware toevoegen</a></li>
                            <li><a href="hardware_verwijderen.php">Hardware verwijderen</a></li>
                        </ul>
                    </li>
                </ul>
            </div>
        </div>      
<?php
            $connect=mysql_connect("localhost", "root","");
            mysql_select_db("helpdesk_middenpolder", $connect);
            $id=$_GET['id'];
            $q="SELECT * FROM hardware WHERE hardwareID=$id";


            $r=mysql_query($q);

            echo    "<table border='1'>";
            echo    "<th>merknaam</th><th>producttype</th><th>hardwaretype</th>";
            while   ($x=mysql_fetch_array($r)){
                echo "<tr>";
                echo "<td>";
                echo "<input type='text' value='".$x['merknaam']."'>";
                echo "</td>";
                echo "<td>";
                echo "<input type='text' value='".$x['producttype']."'>";
                echo "</td>";
                echo "<td>";
                echo "<input type='text' value='".$x['hardwaretype']."'>";
                echo "</td>";
                echo "</tr>";
            }
            echo "</table>";

        mysql_close($connect);
?>
<?php
    if(isset($_POST['updatehardware'])){ 
        $query = "UPDATE hardware SET merknaam='".$x['merknaam']."', producttype='".$x['producttype']."', hardwaretype='".$x['hardwaretype']."' WHERE hardwareID='".$id['id']."'";
        mysql_query($query);
    }
?>


    <form method="post">
    <input type="submit" name="updatehardware" value="Hardware updaten">
    </form>
    </body>
</html>

Answer 1

您的更新查询需要像这样更改

hardwareID='".$id."'

而不是

hardwareID='".$id['id']."'

Answer 2

您使用$id['id']代替$_GET['id']。但是从不信任传递GET / POST数据。改为使用：

$id = (int) $_GET['id'];

警告：非法字符串偏移＆＃39; id＆＃39;在

2 个答案: