importcom.google.gson.Gson;publicclassJsonToJsonStringExample{publicstaticvoidmain(String[]args){// 创建Java对象Personperson=newPerson("John",25);// 将Java对象转换为Json字符串Gsongson=newGson();StringjsonString=gson.toJson(person);// 打印结果System.out.println(jsonString);}// 定义Person类static...
Staff staff= JSON.parseObject(jsonString, Staff.class); System.out.println(staff.toString());/*** 对象转化为json字符串*/String jsonStr=JSON.toJSONString(staff); System.out.println(jsonStr); } } 输出结果: Staff{name='Antony', age=12, sex='male', birthday=null} {"age":12,"name":"...
1.问题 在使用fastjson中的JSON.toJSONString方法将对象转换成json字符串的时候,发现有些字段没有了。如: 1 2 3 4 5 6 7 8 publicstaticvoidmain(String[] args) { Map<String,Object>map=newHashMap<>(); map.put("id","1"); map.put("name",null); System.out.println(JSON.toJSONString(map)...
*/String jsonString="{name:'Antony',age:'12',sex:'male',telephone:'88888'}";Staff staff=JSON.parseObject(jsonString,Staff.class);System.out.println(staff.toString());/** * 对象转化为json字符串 */String jsonStr=JSON.toJSONString(staff);System.out.println(jsonStr);}} 输出结果 Staff{nam...
publicclassFastJsonTest{publicstaticvoidmain(String[]args){Personperson=newPerson();person.setBirth(newDate());System.out.println(JSON.toJSONString(person));}publicstaticclassPerson{privateIntegerage=123;privatetransientDatebirth;publicDategetBirth(){returnbirth;}publicvoidsetBirth(Datebirth){this.birth...
JSON.stringify(obj)将JSON转为字符串。JSON.parse(string)将字符串转为JSON格式; var a={"name":"tom","sex":"男","age":"24"}; var b='{"name":"Mike","sex":"女","age":"29"}'; var aToStr=JSON.stringify(a); var bToObj=JSON.parse(b); alert(typeof(aToStr)); //string alert...
public static void main(String[] args) { String str = FileUtil.readCsv("E:\\IdeaProjects\\javaStudy\\src\\test\\data\\capability.json"); System.out.println("读取的文件String:\n"+str); str= JSON.parseObject(str).toJSONString(); ...
在使用JSON.toJSONString 序列化Kafka的 org.apache.kafka.clients.producer.RecordMetadata 这个类的对象时,返回的String为{},但实际的对象字段是有值的。 fastJSON版本 1.2.73
三个属性 @JSONField(format="yyyy-MM-dd HH:mm:ss") private Date createTime @JSONField(serialize = false); private String password; private String email; 使用toJSON输出{"createTime":1483413683714,"email":"eee"} 使用toJSONString输出{"createTime":"2017-...
Applies to: Oracle Integration-OIC - Version 21.2.2.0.0 and laterInformation in this document applies to any platform.GoalHow to convert JSON data to a string by using Integration?SolutionSign In To view full details, sign in with your My Oracle Support account. Register Don't have a ...