jsonresponse

How to make json post request?

由 ε祈祈猫儿з 提交于 2019-12-12 05:17:32
问题 I have to make json post request with the following request parameters, { "method":"login", "data":{ "username":"korea", "password":"123456" } } I use the following code to make the request, NSString *jsonRequest = [NSString stringWithFormat:@"{""\"\method:\"\login\"\"\,""\data:\"{\"\"username\":\"%@\",\"password\":\"%@\"}",username,password]; NSLog(@"Request: %@", jsonRequest); NSURL *url = [NSURL URLWithString:@"http://myurl..."]; NSMutableURLRequest *request = [[NSMutableURLRequest alloc]

How to handle Array of objects in response while using StringRequest in android volley

由 社会主义新天地 提交于 2019-12-12 02:14:24
问题 I have android app . in that app I'm posting some string data on server and get some response. Problem is ,I'm receiving the response in jsonstring,but I want this data in json array. althouugh when I'm using JsonArrayRequest ,it didn't allow post method in parameter and then my web service is not worked. So I'm stick with StringRequest and service works ok but complete response returns as a whole string.So I'm unable to display my data my list view . So how to resolve this issue? Here is my

How to use JMSSerializer with symfony 4.2

由 耗尽温柔 提交于 2019-12-11 17:21:28
问题 i am building an Api with symfony 4.2 and want to use jms-serializer to serialize my data in Json format, after installing it with composer require jms/serializer-bundle and when i try to use it this way : ``` demands = $demandRepo->findAll(); return $this->container->get('serializer')->serialize($demands,'json');``` it gives me this errur : Service "serializer" not found, the container inside "App\Controller\DemandController" is a smaller service locator that only knows about the "doctrine",

unit test Spring MissingServletRequestParameterException JSON response

由 会有一股神秘感。 提交于 2019-12-07 10:52:38
问题 I have POST method in a Spring boot rest controller as follows @RequestMapping(value="/post/action/bookmark", method=RequestMethod.POST) public @ResponseBody Map<String, String> bookmarkPost( @RequestParam(value="actionType",required=true) String actionType, @RequestParam(value="postId",required=true) String postId, @CurrentUser User user) throws Exception{ return service.bookmarkPost(postId, actionType, user); } now if I test with missing parameter in Postman I get an 400 http response and a

unit test Spring MissingServletRequestParameterException JSON response

由 家住魔仙堡 提交于 2019-12-05 16:43:15
I have POST method in a Spring boot rest controller as follows @RequestMapping(value="/post/action/bookmark", method=RequestMethod.POST) public @ResponseBody Map<String, String> bookmarkPost( @RequestParam(value="actionType",required=true) String actionType, @RequestParam(value="postId",required=true) String postId, @CurrentUser User user) throws Exception{ return service.bookmarkPost(postId, actionType, user); } now if I test with missing parameter in Postman I get an 400 http response and a JSON body: { "timestamp": "2015-07-20", "status": 400, "error": "Bad Request", "exception": "org

How to use Django's assertJSONEqual to verify response of view returning JsonResponse

由 此生再无相见时 提交于 2019-12-03 09:51:05
I'm using Python 3.4 and Django 1.7. I have a view returning JsonResponse . def add_item_to_collection(request): #(...) return JsonResponse({'status':'success'}) I want to verify if that view returns correct response using unit test: class AddItemToCollectionTest(TestCase): def test_success_when_not_added_before(self): response = self.client.post('/add-item-to-collection') self.assertEqual(response.status_code, 200) self.assertJSONEqual(response.content, {'status': 'success'}) However the assertJSONEqual() line raises an exception: Error Traceback (most recent call last): File "E:\Projects

Use JSONResponse to serialize a QuerySet in Django 1.7?

由 ≯℡__Kan透↙ 提交于 2019-11-30 04:46:23
I saw that now in Django 1.7 I can use the http.JSONResponse object to send JSON to a client. My View is: #Ajax def get_chat(request): usuario = request.GET.get('usuario_consultor', None) usuario_chat = request.GET.get('usuario_chat', None) mensajes = list(MensajeDirecto.objects.filter(Q(usuario_remitente = usuario, usuario_destinatario = usuario_chat) | Q(usuario_remitente = usuario_chat, usuario_destinatario = usuario))) return JsonResponse(mensajes, safe=False) But I get the next error: <MensajeDirecto: Towi CrisTowi> is not JSON serializable` Do you know how to serialize a QuerySet to send

Query result in JSON format (key value pair) on using @Query annotation in Spring Boot, Hibernate

由 爷，独闯天下 提交于 2019-11-28 12:56:40
My Controller @GetMapping(value="/getAllDetails") public List<PriceListEntity> getAllDetails() { return MyRepository.getAllDetails(); } My Repository @Repository public interface MyRepository extends CrudRepository<TestNativeQ, String> { @Query( value="SELECT qplt.name price_list_name, qplab.status_code, qplab.start_date, (SELECT charge_definition_code FROM oalfsaas_repl.QP_CHARGE_DEFINITIONS_B WHERE charge_definition_id=qplab.charge_definition_id ) chargedefinitioncode " + "FROM PriceListEntity qplab, PriceListLineEntity qplt " + " WHERE qplab.price_list_id =qplt.price_list_id ", nativeQuery

Query result in JSON format (key value pair) on using @Query annotation in Spring Boot, Hibernate

由 旧巷老猫 提交于 2019-11-27 07:25:43
问题 My Controller @GetMapping(value="/getAllDetails") public List<PriceListEntity> getAllDetails() { return MyRepository.getAllDetails(); } My Repository @Repository public interface MyRepository extends CrudRepository<TestNativeQ, String> { @Query( value="SELECT qplt.name price_list_name, qplab.status_code, qplab.start_date, (SELECT charge_definition_code FROM oalfsaas_repl.QP_CHARGE_DEFINITIONS_B WHERE charge_definition_id=qplab.charge_definition_id ) chargedefinitioncode " + "FROM
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