Question

所以，我试图通过使用网址中的category_id来通过函数获取论坛标题的名称。

它没有归还标题。是的，我包括functions.php。

链接是：

http://www.dxbridge.com/view_category.php?cid=1

的functions.php：

function getForumsCategoriesName($cid) {

    $query = "SELECT * FROM categories WHERE id='" . $cid . "'";

    try {
        global $db;
        // Execute the query against the database
        $stmt = $db->prepare($query); 
        $stmt->execute();
        $result = $stmt->fetchAll();
        foreach($result as $forums) {
            $forumsID = $forums['id'];
            $forumsTitle = $forums['category_title'];
            $forumsTopicAmount = $forums['topic_amount'];
            $forumsCategoriesName = "<h1>" . $forumsTitle . "</h1>";
            echo $forumsCategories3;
        }
    }
    catch(PDOException $ex) { 
        // Note: On a production website, you should not output $ex->getMessage(). 
        // It may provide an attacker with helpful information about your code.  
        die("Error loading names"); 
    }
}

尝试从函数中获取名称

$cid = $_GET['cid'];
getForumsCategoriesName($cid);

另外，我知道变量正在被设置，它不会通过函数。

Answer 1

你还没有回复/回应任何东西（实际上你回应了一些东西，一个未定义的变量）。绑定值，不要将其直接注入查询字符串：

function getForumsCategoriesName($cid) 
{ 
    $result = array();
    try {
        global $db;

        // Execute the query against the database
        $query = 'SELECT * FROM categories WHERE id = :cid '; // put a named placeholder
        $stmt = $db->prepare($query); 
        $stmt->bindParam(':cid', $cid); // bind the value
        $stmt->execute();
        $result = $stmt->fetchAll(PDO::FETCH_ASSOC);
        return $result; // return the values
        // echo $forumsCategories3; // this doesn't make sense, its undefined.
    }
    catch(PDOException $ex) { 
        // Note: On a production website, you should not output $ex->getMessage(). 
        // It may provide an attacker with helpful information about your code.  
        die("Error loading names"); 
    }
}

然后使用：

$cid = $_GET['cid'];
$result = getForumsCategoriesName($cid);

foreach($result as $forums) {
    $forumsID = $forums['id'];
    $forumsTitle = $forums['category_title'];
    $forumsTopicAmount = $forums['topic_amount'];
    $forumsCategoriesName = "<h1>" . $forumsTitle . "</h1>";

    echo $forumsID . '<br/>'; // echo everybody else

}

函数参数php不起作用

1 个答案: