【发布时间】:2021-08-24 05:03:48
【问题描述】:
我从 mysql 调用一些数据并创建自己的 api 以在 android 应用程序中显示信息。我在 Youtube 上尝试了这个视频,但我意识到这个人假设我们的链接中有“页面”处于活动状态,例如:google.com/page=1。如何激活它,以便我可以使用 Android 传递任何页码和任何数量的项目。同样,问题不在 Android 上。
我在 Youtube 上找到了这个:
$postData = new WP_Query( array(
//if the user have passed a page number, it will load this page number
'posts_per_page' => isset($data['posts']) ? $data['posts'] : 15,
'paged' => isset($data['page']) ? $data['page'] : 1
));
这是我的功能:
function get_all_coin()
{
$servername = "";
$username = "";
$password = "";
$dbname = "";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$query = "SELECT * FROM coins LIMIT 100";
$result = mysqli_query($conn, $query);
$json = array();
if (mysqli_num_rows($result) > 0) {
// output data of each rsow
while ($row = mysqli_fetch_assoc($result)) {
$json['id'] = (int) $row['id'];
$json['name'] = $row['name'];
// ...(More rows)
$json_arr[] = $json;
}
} else {
echo "0 results";
}
$json_ar = json_encode(['data' => $json_arr]);
header('Content-type:application/json;charset=utf-8');
$json_arrd['data'] = $json_arr;
return $json_arrd;
}
【问题讨论】:
-
警告:您对SQL Injections 持开放态度,应该使用参数化的prepared statements,而不是手动构建查询。它们由PDO 或MySQLi 提供。永远不要相信任何形式的输入!即使您的查询仅由受信任的用户执行,you are still in risk of corrupting your data。 Escaping is not enough!
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