Question

嗨，在下面的代码中我具有登录活动。如果用户名和密码通过json发送请求，如果用户名和密码正确，则服务器以json格式发送json响应

在json中，响应包含关键状态和用户类型。如果状态为成功和用户类型，则根据用户类型重定向到特定页面。

但是现在我的问题是，如果服务器以json格式发送响应时电子邮件和密码不正确。其中包含一个名为status的密钥。

如果状态无效，则希望以烤面包格式显示消息。

return redirect()->back();

Answer 1

Retrofit retrofit = new Retrofit.Builder()
        .baseUrl(API.URL_BASE)
        .addConverterFactory(ScalarsConverterFactory.create())
        .addConverterFactory(GsonConverterFactory.create())
        .build();
API service = retrofit.create(API.class);

    try{
    JSONObject parmobject=new JSONObject ();
    parmobject.put("emailMobile",emailMobile);
    parmobject.put("password",password);
    Call<Login> userCall = service.getUser(parmobject.toString ());
    userCall.enqueue(new Callback<Login> () {
        @Override
        public void onResponse(Call<Login> call, Response<Login> response) {


            if (response != null && response.code ()==200) {
                status = response.body ( ).getStatus ( ).toString ( );
                usertype = response.body ( ).getUsertype ( ).toString ( );


                if (status.equalsIgnoreCase ("success") && usertype.equalsIgnoreCase ("admin")) {
                    // dialog.dismiss();
                    makeText (LoginActivity.this, "Login successfully", Toast.LENGTH_SHORT).show ( );
                    Intent mainIntent;
                    mainIntent = new Intent (LoginActivity.this, NavigationViewActivity.class);
                    startActivity (mainIntent);
                    finish ( );
                }
                else if (status.equalsIgnoreCase ("success") && usertype.equalsIgnoreCase ("operator")) {
                    // dialog.dismiss();
                    makeText (LoginActivity.this, "Login successfully", Toast.LENGTH_SHORT).show ( );
                    Intent mainIntent;
                    mainIntent = new Intent (LoginActivity.this, OperatorHome.class);
                    startActivity (mainIntent);
                    finish ( );
                }

                else if (status.equalsIgnoreCase ("success") && usertype.equalsIgnoreCase ("employee")) {
                    // dialog.dismiss();
                    makeText (LoginActivity.this, "Login successfully", Toast.LENGTH_SHORT).show ( );
                    Intent mainIntent;
                    mainIntent = new Intent (LoginActivity.this, EmployeeHome.class);
                    startActivity (mainIntent);
                    finish ( );
                }
                else if (status.equalsIgnoreCase ("invalid"))
                {
                    makeText (LoginActivity.this, status, Toast.LENGTH_SHORT).show ( );

                }
                else
                {
                    makeText (LoginActivity.this, "Invalid login credentials", Toast.LENGTH_SHORT).show ( );

                }
            }
            else
            {
                makeText (LoginActivity.this, "Invalid login credentials", Toast.LENGTH_SHORT).show ( );
            }


        }
        @Override
        public void onFailure(Call<Login> call, Throwable t) {
            Toast.makeText(LoginActivity.this, "Some error occurred -> ", Toast.LENGTH_LONG).show();;
            progressDialog.dismiss ();
        }
    });
} catch (JSONException e) {
    e.printStackTrace();
}
    return retrofit;

基于服务器要在Android中显示消息的状态响应

1 个答案: