使用Objective-C解析JSON字符串响应时出错

Question

我得到的回应是：

{“response”：{“id”：“R1”，“cmd”：[{“batchSize”：50，“startRow”：0，“name”：“doLogin”，“result”：“OK”，“attributes”：[{“name”：“businessName”，“type”：“String”}，{“name”：“objId”，“type”：“Long”}，{“name”；“type”；“String”}，{“name”：“businessName”，“type”：“String”}，{“name”；“type”；“{“name”；“name”；“objId”，“type”：“type”：“Long”；“name”；“name”；“name”；“{“name”；“type”；“String”；“String”；“firsttype”；“name”；“}，“记录”：[“商业名称”：“帕洛阿尔托鸡蛋-分销商”，“objId”：“200”，“商业类型”：“D”，“名字”：“系统”]，[“商业名称”：“帕洛阿尔托鸡蛋-分销商”，“objId”：“200”，“商业类型”：“D”，“名字”：“系统”]，[“商业名称”：“帕洛阿尔托鸡蛋-分销商”，“objId”：“200”，“商业类型”：“D”，“名字”：“系统”，[“businessName\：\“Palo Alto Egg-分销商”，“objId\”：“200”，“businessType\”：“D”，“firstName\”：“System”]}}

现在的问题是。。。。。。。NSDictionary*json=[NSJSONSerialization JSONObjectWithData:（NSData*）对象选项：针织错误：

NSDictionary *first = [json objectForKey:@"response"];
NSArray *second = [first objectForKey:@"cmd"];
NSArray *attribute_array = [[second objectAtIndex:0] objectForKey:@"result"];
NSLog(@"Resultttttttttt=%@",attribute_array);

//Value of Result
NSString *resultVal = [NSString stringWithFormat:@"%@",attribute_array];

NSArray *record_array = [[second objectAtIndex:0] objectForKey:@"records"];

NSLog(@"Resultttttttttt Nisarg = %@",[[record_array objectAtIndex:0] objectForKey:@"firstName\\"]);

在最后一句话中，当我尝试获取键firstName的值时，它会出现错误，因为结构类似于“firstName\”而不是键“firstName”，因此任何使用“firstName\”解析字符串的建议Key......

Answer 1

为什么要转述

[["businessName\":\"Palo Alto Egg - Distributor","objId\":\"200","businessType\":\"D","firstName\":\"System"],["businessName\":\"Palo Alto Egg - Distributor","objId\":\"200","businessType\":\"D","firstName\":\"System"],["businessName\":\"Palo Alto Egg - Distributor","objId\":\"200","businessType\":\"D","firstName\":\"System"]

它必须是：

[["businessName":"Palo Alto Egg - Distributor","objId":"200","businessType":"D","firstName":"System"],["businessName":"Palo Alto Egg - Distributor","objId":"200","businessType":"D","firstName":"System"],["businessName":"Palo Alto Egg - Distributor","objId":"200","businessType":"D","firstName":"System"]

Answer 2

您的JSON格式良好，但每条记录都是字符串列表，而不是字典，即您有

[
    "businessName\":\"Palo Alto Egg - Distributor",
    "objId\":\"200",
    "businessType\":\"D",
    "firstName\":\"System"
],

而不是

{
    "businessName": "Palo Alto Egg - Distributor",
    "objId": "200",
    "businessType": "D",
    "firstName": "System"
},

因此，[record_array objectAtIndex:0]是一个数组，而不是字典。

如果您可以控制JSON输出，那么应该检查代码并使其返回正确的格式。如果无法更改输出，并且不想对原始JSON数据进行复杂的字符串操作，那么最好的选择是：

NSString *firstName = NULL;

for (NSString *line in record) {
    if ([line hasPrefix:@"firstName"]) {
        firstName = [[line componentsSeparatedByString:@"\":\""] objectAtIndex:1];
    }
}

NSLog(@"%@", firstName);