從JSON響應解析

Question

我收到JSON回應

(
        {
        Response = success;
        UserId = 287;
    }
)

我試圖用下面的代碼解析用戶ID

NSDictionary *dataDictionary = [NSJSONSerialization JSONObjectWithData:receivedData options:0 error:&e];
    NSLog(@"%@", dataDictionary);
    NSString *useridstring=[dataDictionary valueForKey:@"UserId"];

但是正在獲取用戶ID ( 287 ) 。 請建議我如何獲取不帶括號的用戶ID

Answer 1

您需要從JSON響應中獲取整數值。

int userid = [[dataDictionary valueForKey:@"UserId"] intValue];

如果在整數響應中也包含括號，那么接收到JSON響應就存在問題。 要解決此問題，您可以使用-

NSString *useridstring=[dataDictionary valueForKey:@"UserId"];
int userid = [[useridstring substringWithRange:NSMakeRange(1, useridstring.length - 2)] intValue]

編輯-

您得到一個響應數組，為此您必須使用-

int userid = [[dataDictionary valueForKey:@"UserId"] firstObject];

Answer 2

您的回答不是字典，而是數組。 因此，您的代碼應如下所示：

NSArray *dataArray = [NSJSONSerialization JSONObjectWithData:receivedData options:0 error:&e];
NSLog(@"%@", dataArray);
NSDictionary *dataDictionary = dataArray[0]; //Same as objectAtIndex:
NSString *userIDString = dataDictionary[@"UserID"]; //Same as objectForKey:
//If you're still getting parens try this:
//int userIDInt = [dataDictionary[@"UserID"] intValue];