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iOS:在PHP服务文件中使用return而不是echo时出现JSON pars错误

afnetworking-2 echo json ios php

Olcay ErtaÅ Mohammad Razipour · 技术社区 · 11 年前

我有以下内容 HTTP 请求 AFNetworking :

NSMutableDictionary *dictionary = [NSMutableDictionary new];
[dictionary setObject:@"login" forKey:@"request"];
[dictionary setObject:@"q" forKey:@"userName"];
[dictionary setObject:@"q" forKey:@"password"];

DDLogDebug(@"LoginView - login - Dictionary: %@", [dictionary description]);

AFHTTPRequestOperationManager *manager = [AFHTTPRequestOperationManager manager];
[manager setRequestSerializer:[AFJSONRequestSerializer serializer]];
[manager setResponseSerializer:[AFJSONResponseSerializer serializer]];
//[manager.securityPolicy setAllowInvalidCertificates:true];

AFHTTPRequestOperation *operation =
[manager POST:WEB_SERVICE_URL
   parameters:dictionary
      success:^(AFHTTPRequestOperation *operation, id responseObject) {

          dispatch_async(dispatch_get_main_queue(), ^{
              DDLogDebug(@"LoginView - Success Response: %@", responseObject);
              NSString *success = [responseObject objectForKey:@"success"];

              if ([success isEqualToString:@"1"]) {
                  DDLogDebug(@"LoginView - Login successful!");
                  [self performSegueWithIdentifier:kSegueLoginToTimeLine sender:self];
              }
              else {
                  DDLogError(@"LoginView - Login failed!");
                  [HMXCommonMethods showMessage:@"KullanÄ±cÄ± adÄ± ya da Åifre hatalÄ±" withTitle:@"Hata!"];
              }
          });
      }

      failure:^(AFHTTPRequestOperation *operation, NSError *error) {

          dispatch_async(dispatch_get_main_queue(), ^{
              DDLogError(@"LoginView - Error Response: %@", [error localizedDescription]);

              NSInteger statusCode = [operation.response statusCode];

              switch (statusCode) {
                  case 500:
                      [HMXCommonMethods showMessage:@"Sunucu hatasÄ± nedeniyle giriÅ yapÄ±lamadÄ±!" withTitle:@"Hata!"];
                      break;
                  case 408:
                      [HMXCommonMethods showMessage:@"Ä°stek zaman aÅÄ±mÄ±na uÄradÄ±!" withTitle:@"Hata!"];
                      break;
                  case 404:
                      [HMXCommonMethods showMessage:@"Sunucuya eriÅilemedi!" withTitle:@"Hata!"];
                      break;
                  default:
                      [HMXCommonMethods showMessage:@"Sunucu hatasÄ± nedeniyle giriÅ yapÄ±lamadÄ±!" withTitle:@"Hata!"];
                      break;
              }
          });
      }];

[operation start];
DDLogError(@"LoginView - login - request started.");

当我在 PHP 包含以下内容的文件 echo 代码工作正常:

$response = "{\"success\":\"1\",";
$response = $response . "\"id\":\"" . $id . "\",";
$response = $response . "\"companyid\":\"" . $cid . "\",";
$response = $response . "\"companyname\":\"" . $cname . "\",";
$response = $response . "\"name\":\"" . $name . "\",";
$response = $response . "\"surname\":\"" . $surname . "\",";
$response = $response . "\"email\":\"" . $email2 . "\",";
$response = $response . "\"password\":\"" . $password2 . "\"}";
$stmt2->close();

$log = "Company ID = " . $cid . "\n" .
       "Company Name = " . $cname . "\n" .
       "Company Address = " . $caddress . "\n" .
       "Company Phone = " . $cphone . "\n" .
       "Company Web Page = " . $cwebpage . "\n";

$this->logger->debug("GetUser: Company Info:\n" . $log);
$this->logger->debug("GetUser: Response:\n" . $response);

echo $response;

如果我改变了 回响 具有 return 我明白了 JSON pars error 3840 。我也尝试过

echo json_encode($response)
return json_encode($response)

但它不起作用,我也犯了同样的错误。我如何使用 回来 没有 JSON 错误

1 回复 | 直到 11 年前

Jay Blanchard Daan 11 年前

为了让AJAX能够读取PHP的输出,PHP必须响应响应。如果您使用 return AJAX无法读取和使用数据,因为它希望接收几种类似文本的数据类型之一:XML、HTML、脚本、JSON、JSONP或纯文本。A. 回来 没有提供这一点。

此外,您不应该手动形成JSON字符串,因为这可能是一件复杂的事情,可能会出现许多语法错误。利用PHP的JSON函数- json_encode() 和 json_decode() .