初学者上路，请赐教！java读写本地json文件

菜鸟上路，请各位高手赐教！java读写本地json文件
11． programmers.json文件中包含如下内容：
{ "programmers": [ { "firstName": "Brett", "lastName":"McLaughlin", "email": "brett@newInstance.com" }, { "firstName": "Jason", "lastName":"Hunter", "email": "jason@servlets.com" }, { "firstName": "Elliotte", "lastName":"Harold", "email": "elharo@macfaq.com" } ],"authors": [ { "firstName": "Isaac", "lastName": "Asimov", "genre": "science fiction" }, { "firstName": "Tad", "lastName": "Williams", "genre": "fantasy" }, { "firstName": "Frank", "lastName": "Peretti", "genre": "christian fiction" } ],"musicians": [ { "firstName": "Eric", "lastName": "Clapton", "instrument": "guitar" }, { "firstName": "Sergei", "lastName": "Rachmaninoff", "instrument": "piano" } ]}
从文件获取programmers、authors 和 musicians的信息，做以下处理：
1.在控制台输出Jason,Isaac,Sergei的相关信息（firstName,lastName等）
2.将分析得到的programmers、authors 和 musicians的信息，以json格式输出到people.json文件中
3.将programmers.json的内容转换为XML格式,根元素为data.
求该题的完整代码！新手刚入门，还请各位高手多多指教，不胜感激！

[解决办法]
就是用file读取文件内容（应该是按行读取）

再用JSONObject解析就行了。
[解决办法]

Java code

    private ObjectMapper mapper;    private List<AppConfig> appConfig;    public AppConfigController()    {        String propsFile = AppConfigController.class.getResource("/").toString().replace("file:", "") + "appConfig.cfg";        this.mapper = new ObjectMapper();        this.mapper.setAnnotationIntrospector(new NopAnnotationIntrospector());        try        {            propsFile = java.net.URLDecoder.decode(propsFile,"utf-8");                        logger.info("获取到配置文件路径{}", new File(propsFile).getAbsolutePath());            this.appConfig = mapper.readValue(new File(propsFile), new TypeReference<List<AppConfig>>()            {            });        } catch (Exception e)        {            e.printStackTrace();            logger.warn(e.getMessage() + "配置文件路径：{}", propsFile);        }    }
[解决办法]
解决了？那就结贴吧
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哦 非常感谢