php - android 的 api 不工作

Question

我为我的安卓应用创建了一个注册服务android 端工作正常(我猜)并返回“失败”这是我的 PHP 代码:

$json = file_get_contents('php://input');
$User = json_decode($json);

function ConnectDatabase()
{
    $connect = mysqli_connect("localhost", "root", "", "my_db");
    mysqli_set_charset($connect, "utf8");
    return ConnectDatabase();
}

$username = $User->user;
$password = $User->pass;

function AddUser($user, $pass)
{
    $connect = ConnectDatabase();
    if (!(trim($user) == "" || trim($pass) == ""))
    {
        $check = mysqli_query($connect, "select count(*) from usertable where Username='$user'");
        $row = mysqli_fetch_array($check);
        if ($row[0] > 0)
        {
            return "usernameTaken";

        }
        else
        {
            $SQL = "INSERT INTO usertable(Username,Password) VALUES ('$user','$pass')";
            $query = mysqli_query($connect, $SQL);
            if ($query != "")
            {
                return "ok";
            }
            else
            {
                return "failed";
            }
        }
    }
    else
    {
        return "badUsernameOrPassword";
    }
}

$status = AddUser($username, $password);
echo json_encode(["status"=>$status]);

检查了 uri 及其正确...你们看到任何错误了吗？谢谢。

Answer 1

来自 mysqli_query 的 PHP 文档，来自 php.net :

Returns FALSE on failure. For successful SELECT, SHOW, DESCRIBE or EXPLAIN queries mysqli_query() will return a mysqli_result object. For other successful queries mysqli_query() will return TRUE.

因此您的代码必须是:

if ($query != FALSE){
    return "ok";
}else{
    return "failed";
}