Question

我需要将json传递给POST参数。输入参数看起来像这样：

{ "type":"facebook",
  "friends":[{ "id": "facebook id", "avatar": "url to avatar image", "first_name" : "first_name", "last_name" :"last_name"}] }

服务器响应错误表明它是一个错误的请求参数

Error - code: 400, message: {"status":"bad request","message":"undefined method `each' for #<String:0x00000005473e48>"}

当我在“朋友”部分准备JSON数据时，这可能是错误的。你能帮我复习吗？

这是我的代码 -

{
NSMutableArray *aFriends = [[NSMutableArray alloc] init];
int nCount = [[self savedSelIndexPath] count];

    for (int i = 0; i < nCount && nCount > 0; i++) 
    {
        NSIndexPath *path = [[self savedSelIndexPath] objectAtIndex:i];
        NSDictionary *data = nil;

            NSString *key = [[self keysArray] objectAtIndex:path.section]; 
            NSArray *nameSection = [[self listContentDict] objectForKey:key];
            data = [nameSection objectAtIndex:[path row]];

        NSDictionary *dictFriends = [NSDictionary dictionaryWithObjectsAndKeys:
                                     [data objectForKey:@"id"], @"id",
                                     [data objectForKey:@"picture"], @"avatar",
                                     [data objectForKey:@"first_name"], @"first_name",
                                     [data objectForKey:@"last_name"], @"last_name",
                                     nil];

        [aFriends addObject:dictFriends];
        dictFriends = nil;
    }

DLog(@"aFriends: %@", aFriends);

NSDictionary *teamProfile = [[Global sharedGlobal] getPlistFile];
NSDictionary *params = [NSDictionary dictionaryWithObjectsAndKeys:
                                @"facebook", @"type",
                                aFriends, @"friends", nil];
DLog(@"params: %@", params);

NSString *sPath = [NSString stringWithFormat:@"/users/%@/friends", [teamProfile valueForKey:@"userId"]];

NSMutableURLRequest *request = [[[Global sharedGlobal] httpClient] requestWithMethod:@"POST" path:sPath parameters:params]; 
DLog(@"request: %@", request);  
….
[aFriends release]; 
    }

Answer 1

如果你想将JSON传递给服务器，你必须像这样设置它：

    NSMutableURLRequest *networkRequest;
    networkRequest = [[NSMutableURLRequest alloc] initWithURL:
                          [NSURL URLWithString:[networkServerAddress stringByAppendingString:@""]]];
    send = [[json objectAtIndex:i] dataUsingEncoding:NSUTF8StringEncoding];
    //Set the headers appropriately
    [networkRequest setHTTPMethod:@"POST"];  
    [networkRequest setValue:@"application/json"    
          forHTTPHeaderField: @"Content-type"];
    [networkRequest setValue:[NSString stringWithFormat:@"%d", [send length]]  
          forHTTPHeaderField:@"Content-length"]; 
    [networkRequest setValue:@"application/json"    
          forHTTPHeaderField:@"Accept"];
    //Set the body to the json encoded string
    [networkRequest setHTTPBody:send];

发送必须是NSData，而不是字典。如果你想这样做，你必须首先解析字典。 JSON有许多优秀的解析器/编写器，我个人喜欢并使用SBJson。使用SBJson，您可以执行以下操作：

SBJsonWriter *writer = [[SBJsonWriter alloc] init];
NSData *dataToSend = [writer dataWithObject:data];

Answer 2

我不确定我是否理解你的要求，但你说你需要将数据编码为JSON，我看不到你的参数的任何JSON编码指令。你只是传递一个NSDictionary * dictFriends ......它是否正确？

见到你！

解析json数据

2 个答案: