Question

我正在处理一个致命的错误，我在其他地方看到了很多解决方案，但我似乎无法找到适合我代码的正确解决方案...

所以这是出错的代码：

$link = mysqli_connect($server, $gebruiker, $wachtwoord, $database);

if($link === false) {
    die("ERROR: Kon geen verbinding maken. " . mysqli_connect_error());
}
if(isset($_REQUEST['term'])) {
    $sql = "SELECT * FROM producten WHERE naam LIKE ?";

    if($stmt = mysqli_prepare($link, $sql)) {

        mysqli_stmt_bind_param($stmt, "s", $param_term);

        $param_term = '%' . $_REQUEST['term'] . '%';

        if(mysqli_stmt_execute($stmt)) {
            $result = mysqli_stmt_get_result($stmt);
            if(mysqli_num_rows($result) > 0) {
                while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
                    echo "<p><a href=toevoegen.php?id=" . $row['id'] . " style=\"text-decoration: none\">" . $row["naam"] . "</a></p>";
                }
            } else {
                echo "<p>Geen producten gevonden</p>";
            }

        } else {
            echo "ERROR: Kon $sql niet uitvoeren. " . mysqli_error($link);
        }
    }
    mysqli_stmt_close($stmt);
}

mysqli_close($link);

这是让一切混乱的部分：

$result = mysqli_stmt_get_result($stmt);

谁能帮助我朝着正确的方向前进，谢谢！

Answer 1

我修复了它，解决方案：

if(isset($_REQUEST['term'])) {
    $conn = mysqli_connect($server, $gebruiker, $wachtwoord, $database); //Connection to my database
    $query = "SELECT id, naam FROM producten WHERE naam LIKE ?";
    $statement = $conn->prepare($query);
    $term = "%".$_REQUEST['term']."%";
    $statement->bind_param("s", $term);
    $statement->execute();
    $statement->store_result();
    if($statement->num_rows() == 0) {
        echo "<p>Geen producten gevonden ".$_REQUEST['term']."</p>";
        $statement->close();
        $conn->close();
    } else {
        $statement->bind_result($id,$naam);
        while ($statement->fetch()) {
            echo "<p><a href=toevoegen.php?id=" . $id ." style=\"text-decoration: none\">" . $naam . "</a></p>";
        };
        $statement->close();
        $conn->close();
    };};

致命错误：未捕获错误：调用未定义函数mysqli_stmt_get_result（）

1 个答案: