1.使用Gson类中的toJson()方法 Gson gson = new Gson(); String listToJsonString = gson.toJson(list); 2.使用JSONArray...json=JSONArray.fromobject(list);在调用json.toString()方法转换成字符串 JSONArray jsa = JSONArray.fromObject(list...); String result = jsa.toString(); 版权声明:...
public static void main(String[] args){ String strArr = "[{\"0\":\"zhangsan\",\"1\":\"lisi\",\"2\":\"wangwu\",\"3\":\"maliu\"}," + "{\"00\":\"zhangsan\",\"11\":\"lisi\",\"22\":\"wangwu\",\"33\":\"maliu\"}]"; //第一种方式 List<Map<String,String>...
JSONArray myJsonArray = JSONArray.fromObject(jsonMessage); 七.String转数组 String string = "a,b,c"; String [] stringArr= string.split(","); //注意分隔符是需要转译 如果是"abc"这种字符串,就直接 String string = "abc" ; char [] stringArr = string.toCharArray(); //注意返回值是char数...
String jsonString = “[[1,"zhangsan","male",18,"Beijing"],[2,"lisi","female",18,"Shanghai"]]”List<List<Object>> list = JSON.parseObject([jsonString], new TypeReference<List<List<Object>>>() {});3.对象转json JSON.toJSONString(user);4.List转json JSON.toJSONString(users);JSONOb...
public static <T> T json2Bean(String jsonStr, Class<T> objClass) { return JSON.parseObject(jsonStr, objClass); } } 1. 2. 3. 4. 5. 6. 7. 8. GsonUtil.java public class GsonUtil { private static Gson gson = new GsonBuilder().create(); ...
String test = "[{\"vendorId\":1, \"checkList\":[{\"imageId\":1,\"algorithmType\":\"person\", \"maxCapacity\":50, \"deviceIds\":\"xxxx,yyyy\"}]}]"; //如果是字符串对象,可以强转成Map<String, Object> List<Map<String, Object>> vendors = (List<Map<String, Object>>) JSON....
{"entry":"3","count":1}]我尝试使用编码ArrayList<String> listOne = new ArrayList<String>();if (arrOne != null) { int len = arrOne.size(); for (int i=0;i<len;i++){ listOne.add(arrOne.get(i).toString()); } } System.out.println(arrOne);System.out.println("\nlistOne:"+...
代码语言:javascript 复制 //先将JsonArray转为StringString newIds=JSONObject.getJSONArray("newIds").toString();//再将String转为List<String>List<String>list=JSONObject.parseArray(newIds,String.class); 或者 代码语言:javascript 复制 JSONArray newIds=linkParam.getJSONArray("newIds");//直接利用Json...
String value = object.getString(Api.VALUE); item = new MoreInfo(key, value); items.add(item); } return items; } 这样只能处理list里面只有一组数据的情况。如果循环封装成json,得到的格式就是: [{"name":"name0","age":0}][{"name":"name1","age":5}][{"name":"name2","age":10}]...
String name; int age; } 1.使用FastJson FastJson 是阿里巴巴的开源JSON解析库,它可以解析 JSON 格式的字符串,支持将 Java Bean 序列化为 JSON 字符串,也可以从 JSON 字符串反序列化到 JavaBean 1.1 Maven <dependency> <groupId>com.alibaba</groupId> ...