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。
最佳答案
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";
}
关于php - android 的 api 不工作,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/51791371/