在NSURLSession中返回错误

尝试返回responseString时出现错误

（'NSData * __ strong，NSURLResponse * __ strong，NSError * __ strong）types的参数发送'NSString *（^）（NSData * __ strong，NSURLResponse * __ strong，NSError * __ strong）'的不兼容块指针types

我从哪里调用它从AViewController类

NSString *responseString=[self callAPI]; 

下面是我在BViewModel类中的代码：

 -(NSString* )callAPI { NSURLSessionConfiguration *configuration = [NSURLSessionConfiguration defaultSessionConfiguration]; NSURLSession *session = [NSURLSession sessionWithConfiguration:configuration delegate:self delegateQueue:nil]; NSURL *url = [NSURL URLWithString:@"http://appersgroup.com/talkawalk/login.php?email=[e-mail]&password=123456&latitude=52.486245&longitude=13.327496&device_token=show"]; NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url cachePolicy:NSURLRequestUseProtocolCachePolicy timeoutInterval:60.0]; [request addValue:@"application/json" forHTTPHeaderField:@"Content-Type"]; [request addValue:@"application/json" forHTTPHeaderField:@"Accept"]; [request setHTTPMethod:@"POST"]; /* NSDictionary *mapData = [[NSDictionary alloc] initWithObjectsAndKeys: @"TEST IOS", @"name", @"IOS TYPE", @"typemap", nil]; NSData *postData = [NSJSONSerialization dataWithJSONObject:mapData options:0 error:&error]; [request setHTTPBody:postData]; */ NSURLSessionDataTask *postDataTask = [session dataTaskWithRequest:request completionHandler:^(NSData *data, NSURLResponse *response, NSError *error) { NSString* responseString = [[NSString alloc] initWithData:data encoding:NSUTF8StringEncoding]; if([responseString rangeOfString:@"nil"].location != NSNotFound) { NSString * newResponse = [[NSString alloc] initWithData:data encoding:NSASCIIStringEncoding]; responseString = newResponse; } NSLog(@"%@",responseString); NSLog(@"response %@",response); NSLog(@"error %@",error); return responseString;//adding this line give me error ? how to return value }]; [postDataTask resume]; } 

Answer 1

你可以像这样使用这个方法：

 -(void)callAPIWithCompletionHandler : (void (^) (NSString * strResponse)) completionHandler { NSURLSessionConfiguration *configuration = [NSURLSessionConfiguration defaultSessionConfiguration]; NSURLSession *session = [NSURLSession sessionWithConfiguration:configuration delegate:self delegateQueue:nil]; NSURL *url = [NSURL URLWithString:@"http://appersgroup.com/talkawalk/login.php?email=[e-mail]&password=123456&latitude=52.486245&longitude=13.327496&device_token=show"]; NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url cachePolicy:NSURLRequestUseProtocolCachePolicy timeoutInterval:60.0]; [request addValue:@"application/json" forHTTPHeaderField:@"Content-Type"]; [request addValue:@"application/json" forHTTPHeaderField:@"Accept"]; [request setHTTPMethod:@"POST"]; /* NSDictionary *mapData = [[NSDictionary alloc] initWithObjectsAndKeys: @"TEST IOS", @"name", @"IOS TYPE", @"typemap", nil]; NSData *postData = [NSJSONSerialization dataWithJSONObject:mapData options:0 error:&error]; [request setHTTPBody:postData]; */ NSURLSessionDataTask *postDataTask = [session dataTaskWithRequest:request completionHandler:^(NSData *data, NSURLResponse *response, NSError *error) { NSString* responseString = [[NSString alloc] initWithData:data encoding:NSUTF8StringEncoding]; if([responseString rangeOfString:@"nil"].location != NSNotFound) { NSString * newResponse = [[NSString alloc] initWithData:data encoding:NSASCIIStringEncoding]; responseString = newResponse; } NSLog(@"%@",responseString); NSLog(@"response %@",response); NSLog(@"error %@",error); completionHandler(responseString); }]; [postDataTask resume]; } 

调用这个方法并返回响应如下：

 [self callAPIWithCompletionHandler:^(NSString *strResponse) { NSString *responseString = strResponse; }]; 

假设这个方法在ViewControllerA实现，并且你想从ViewControllerB调用它。

然后在ViewControllerB导入ViewControllerA ，并在ViewControllerB制作ViewControllerA实例。

 ViewControllerA *vcA = [[ViewControllerA alloc] init]; [vcA callAPIWithCompletionHandler:^(NSString *strResponse) { NSString *responseString = strResponse; }]; 

Answer 2

你正试图从一个不应该返回任何值的块返回一个string。 这就是编译器给你错误的原因。 该块将被asynchronous执行，所以你在做什么没有意义。 你真的需要熟悉块如何工作，特别是当它们asynchronous执行时。

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