Object result = initMapContainer(OffheapCacheConst.MAP_CONTAINER_FOR_STRING).get(key); if(result == null){ return null; } //反序列化出entityWrapper EntityWrapper entityWrapper = JSON.parseObject(result.toString()); return (T)entityWrapper.getEntity(); } catch (Exception ex) { logger.error(Of...
接下来,我们使用Gson库将JSON数据转为Java对象: importcom.google.gson.Gson;publicclassJsonToJavaExample{publicstaticvoidmain(String[]args){Stringjson="{\"name\": \"Alice\", \"age\": 30}";Gsongson=newGson();Personperson=gson.fromJson(json,Person.class);System.out.println("Name: "+person.g...
Gson gson=newGson();//把json字符串转对象Object object = gson.fromJson(json, Object.class); System.out.println("JsonToObject-->"+object.toString());//把对象转json字符串String objectjson =gson.toJson(object); System.out.println("ObjectToJson-->"+objectjson); }...
2、json转object ObjectMapper objectMapper = new ObjectMapper(); YourClass class = objectMapper.readValue(YourJson, YourClass.class); 如果json中有新增的字段并且是YourClass类中不存在的,则会转换错误 1)需要加上如下语句: ObjectMapper objectMapper = new ObjectMapper(); objectMapper.configure(DeserializationFea...
1、将java对象转换成json字符串: Person p1 = new Person(); p1.setName(“xxx”); p1.setAge(18); String jsonStr = JSONObject.fromObject(p1).toString(); 2、将java集合转换成json字符串: Person p1 = new Person(); p1.setName(“xxx1”); ...
一、提取JsonObject(或Map)中的key-value值 1、获取JsonObject 的key 2.获取JsonObject中的School的key-value值 二、Gosn 三、String字符串分隔 日常进行json格式的转换 一、Fastion 使用阿里的fastjson <dependency><groupId>com.alibaba</groupId><artifactId>fastjson</artifactId><version>1.2.83</version><...
Returns the number value to which the specified name is mapped. JsonObjectgetJsonObject(Stringname) Returns the object value to which the specified name is mapped. JsonStringgetJsonString(Stringname) Returns the string value to which the specified name is mapped. ...
使用JSONObject.toJavaObject(JSON.parseObject("{"typeCharKey":"0"}", Type.class),对象内有Character属性转换异常。 public class Type{ private Character typeCharKey; } 若json里typeCharKey的值是0之类的数字,就会抛出异常com.alibaba.fastjson.JSONException: can not cast to char, value : 0。因为会把...
使用OSS PostObject的callback没有触发回调,但是通过PutObject用同样的callback触发了回调。一般情况下,如果JSON格式有误或者回调失败,都会返回相应的消息,这里需要分别测试Put和Post回调效果: 错误原因 发送请求时callback参数在file下面。 解决方法 调整callback参数与file的位置。
public class JSONObjectSample { public static void main(String[] args) { createJsonByMap(); } private static void createJsonByMap() { Map<String, Object> data = new HashMap<String, Object>(); data.put("name", "zhangsan"); data.put("age", 22); data.put("hobbies", new String[]...