Question

我有一个嵌套的Json：

[
    {
        "result":"1",
        "roleId":4
    },
    {
        "projectInfo":[
            {
                "result":true
            },
            {
                "Project":[
                    {
                        "ProjectId":5378,
                        "ProjectName":"ASAG",
                        "CountryId":146,
                        "ProjectGroupId":743,
                        "Description":"Axel Spinger AG"
                    },
                    {
                        "ProjectId":5402,
                        "ProjectName":"BIZ",
                        "CountryId":146,
                        "ProjectGroupId":759,
                        "Description":"Bizerba Win 7 BAU"
                    },
                    {
                        "ProjectId":5404,
                        "ProjectName":"BOM",
                        "CountryId":146,
                        "ProjectGroupId":743,
                        "Description":"Bombardier Transportation ThinApp Migration"
                    },
                    {
                        "ProjectId":5394,
                        "ProjectName":"REDBULL",
                        "CountryId":149,
                        "ProjectGroupId":762,
                        "Description":"Red Bull Mac Packaging"
                    },
                    {
                        "ProjectId":5397,
                        "ProjectName":"VHV",
                        "CountryId":146,
                        "ProjectGroupId":743,
                        "Description":"VHV Win7 Migration"
                    }
                ]
            }
        ]
    }
]

我需要的是将它分成小块以获得某些特定键的价值，例如：How to parse JSON into Objective C - SBJSON

我的代码是：

SBJsonParser* jParser = [[SBJsonParser alloc] init];
NSDictionary* root = [jParser objectWithString:string];
NSDictionary* projectInfo = [root objectForKey:@"projectInfo"];
NSArray* projectList = [projectInfo objectForKey:@"Project"];
for (NSDictionary* project in projectList)
{
    NSString *content = [project objectForKey:@"ProjectId"];
    NSLog(@"%@", content);
}

但是在尝试从根节点获取projectInfo时出现错误。我的代码有什么问题吗？请给我一个例子来分割我的JSON。任何帮助都会很棒。

Answer 1

你JSON包含类似嵌套数组。只需将每个内容吐入字典即可获得结果。

工作代码：

SBJsonParser* jParser = [[SBJsonParser alloc] init];
NSArray* root = [jParser objectWithString:string];

NSDictionary* projectDictionary = [root objectAtIndex:1];
NSArray* projectInfo = [projectDictionary objectForKey:@"projectInfo"];
NSDictionary* projectData = [projectInfo objectAtIndex:1];
NSDictionary *projectList = [projectData objectForKey:@"Project"];
NSLog(@"\n\n Result = %@",projectList
      );
for (NSDictionary* project in projectList)
{
    NSString *content = [project objectForKey:@"ProjectId"];
    NSLog(@"\n Project Id =%@", content);
}

Answer 2

根据您的JSON结构，您的顶级结构是一个数组，而不是字典。

试试这个：

SBJsonParser* jParser = [[SBJsonParser alloc] init];
NSArray* root = [jParser objectWithString:string];
NSDictionary* projectDictionary = [root objectAtIndex:1];
NSArray* projectInfo = [projectDictionary objectForKey:@"projectInfo"];

在iOS中将嵌套的Json解析为NSDictionary

2 个答案: