Question

我对PHP和mySQL非常熟悉。但是为了改变，我想使用sqlite作为数据库，我只是无法正确连接。我一直在关注this网站上的2.代码示例，所以我的代码如下所示：

<html>
<head>
    <title>
        DB bearbeiten
    </title>
</head>
<body>

    <form action="insert.php" method="post">
        <input type="text" name="name" placeholder="Name">
        <input type="text" name="eName" placeholder="E-Number">
        <input type="text" name="causes" placeholder="Wirkung">
        <input type="number" name="danger" placeholder="Gefahreinstufung">
        <button type="submit" name="submit">Eintragen</button>
    </form>

    <?php
    echo '1';
        if(isset($_POST['submit'])){
            echo '2';
            $table = "zusatzstoffe";
            $filename = "sqlite:/data/Adrian/zusatzstoffe.db";
            $db = new PDO($filename) or die("cannot open DB");

            //$db = SQLite3::_construct();
            settype($danger, "integer");
            $name = $_POST['name'];
            $eName = $_POST['eName'];
            $causes = $_POST['causes'];
            $danger = $_POST['danger'];
            $db->exec("INSERT INTO $table (name, number, causes, danger) VALUES ($name, $eName, $causes, $danger)");
        }

    ?>
</body>
</html>

这条线不起作用，我不知道原因：$db = new PDO(filename, '','') or die("can not open DB");：

无法与sqlite DB建立连接

0 个答案: