JSON 响应中 NSDictionary 中的对象数组 - objective c

array of objects inside NSDictionary in JSON response - objective c

我试图从 NSDictionary 的 JSON 响应中获取对象列表,然后将其存储在 NSMutableArray 中,但它一直只给我数组中的一个对象,可以请帮我提供示例代码,

这是 json 响应:

{
  "errorCd": "00",
  "desc": "Success",
  "ref": 83,
  "statusCode": "1",
  "extraData": [
    {
      "key": "bal",
      "value": "80"
    },
    {
      "key": "txs",
      "value": "[{\"id\":2268099999,\"amnt\":100.0,\"curr\":\"JOD\",\"sender\":\"PSPCI\",\"receiver\":\"00962799999992\",\"date\":\"Feb 1, 2016 4:03:25 PM\",\"status\":1,\"type\":5,\"fees\":0.0,\"reference\":40},{\"id\":2357099999,\"amnt\":20.0,\"curr\":\"JOD\",\"sender\":\"00962799999992\",\"receiver\":\"PSPCI\",\"date\":\"Feb 2, 2016 12:52:35 PM\",\"status\":1,\"type\":6,\"fees\":0.0,\"reference\":68}]"
    }
  ]
}

我正在尝试获取键 txs 值中的列表。

这就是我想做的事情:

-(NSDictionary *)getListOfExtraData:(NSMutableArray *)extras{
    NSDictionary *array = [[NSDictionary alloc] init];
    for (NSDictionary *dictionary in extras) {
        if([[dictionary valueForKey:@"key"] isEqualToString:@"txs"])
            array = [dictionary objectForKey:@"value"];
    }
    if (array != nil && array.count > 0)
        return array;
    return nil;
}

JSON 数据无效,因为 value"

中有一个对象

你应该像这样发送你的JSON

{

  "errorCd": "00",
  "desc": "Success",
  "ref": 83,
  "statusCode": "1",
  "extraData": [
    {
      "key": "bal",
      "value": "80"
    },
    {
      "key": "txs",
      "value": [
        {
          "id": 2268099999,
          "amnt": 100.0,
          "curr": "JOD",
          "sender": "PSPCI",
          "receiver": "00962799999992",
          "date": "Feb 1, 2016 4:03:25 PM",
          "status": 1,
          "type": 5,
          "fees": 0.0,
          "reference": 40
        },
        {
          "id": 2357099999,
          "amnt": 20.0,
          "curr": "JOD",
          "sender": "00962799999992",
          "receiver": "PSPCI",
          "date": "Feb 2, 2016 12:52:35 PM",
          "status": 1,
          "type": 6,
          "fees": 0.0,
          "reference": 68
        }
      ]
    }
  ]
}
After the json is in correct format,i.e.


{
  "errorCd": "00",
  "desc": "Success",
  "ref": 83,
  "statusCode": "1",
  "extraData": [
    {
      "key": "bal",
      "value": "80"
    },
    {
      "key": "txs",
      "value": [
        {
          "id": 2268099999,
          "amnt": 100.0,
          "curr": "JOD",
          "sender": "PSPCI",
          "receiver": "00962799999992",
          "date": "Feb 1, 2016 4:03:25 PM",
          "status": 1,
          "type": 5,
          "fees": 0.0,
          "reference": 40
        },
        {
          "id": 2357099999,
          "amnt": 20.0,
          "curr": "JOD",
          "sender": "00962799999992",
          "receiver": "PSPCI",
          "date": "Feb 2, 2016 12:52:35 PM",
          "status": 1,
          "type": 6,
          "fees": 0.0,
          "reference": 68
        }
      ]
    }
  ]
}  

要存储数组,您需要将变量 array 声明为 NSArray 或 NSMutableArray 而不是 NSDictionary

好吧,我会说 JSON 很不寻常;看起来它可以是字符串化数字 ("80") 或 JSON 数组。

您需要进一步解析内部数组,使用:

NSData *jsonResponse = ...;    // Response from server
NSError *error = nil;
NSDictionary *topLevelDict = [NSJSONSerialization JSONObjectWithData:jsonResponse
                                                             options:0
                                                               error:&error];
if (topLevelDict) {
    NSArray *extraData = topLevelDict[@"extraData"];
    for (NSDictionary *innerDict in extraData) {
        if ([innerDict[@"key"] isEqualToString:@"txs"]) {
            NSString *valueStr = innerDict[@"value"];
            NSArray *innerArray = [NSJSONSerialization JSONObjectWithData:[valueStr dataUsingEncoding:NSUTF8StringEncoding]
                                                                  options:0
                                                                    error:&error];
            if (innerArray) {
                // FINALLY!  WE HAVE IT!
            } else {
                // Report error
            }
            break;
        }
    }
} else {
    // report error
}