方法一:jsckson的ObjectMapper类
String jsonString = “”;//json类型的字符串ObjectMapper mapper = new ObjectMapper();MyClass myClass = mapper.readValue(jsonString, MyClass.class);
但是用此方法在类匹配属性的时候,如果字符串里的某一个属性和类的某一个属性没有对应上,可能会报错:
com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: Unrecognized field
此时有两种解决方案:
解决方法1:ObjectMapper 对象添加
mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
ObjectMapper mapper = new ObjectMapper();mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);MyClass myClass = mapper.readValue(string, MyClass.class);
解决方法2:在需要转化的对象的类中添加注解,注解信息如下:
@JsonIgnoreProperties(ignoreUnknown = true)public class MyClass{}
方法二:fastjson的JSONObject类
转对象:
String jsonString = “”;//json类型的字符串Myclass myclass = JSONObject.parseObject(jsonString,Myclass.class);
转数组:
String jsonString = “”;//json类型的字符串JSONArray jsonArray = JSONArray.parseArray(jsonString);for(int i=0; i<jsonArray.size(); i++) {JSONObject object = jsonArray.getJSONObject(i);SmartProejctEquipmentMap myclass = JSONObject.parseObject(object.toJSONString() , SmartProejctEquipmentMap.class);// 将string类型直接封装成对象}
方法三:net.sf.json的JSONObject类
String jsonString = “”;//json类型的字符串JSONObject jsonobject = JSONObject.fromObject(jsonString);Myclass myclass = (MYclass)JSONObject.toBean(jsonobject,Myclass.class);