Question

我有一个字符串数据，我想将它添加到json并检索它以将其添加到excel。但它没有发生请帮忙这是我的代码......

String table = “[{” ID “：1，” 名称 “：” XXX “ ”位置“： ”XX“} { ”ID“：2 ”名称为“： ”YYY“， ”位置“： ”YY“} {” ID “：3”，名称为 “：” ZZZ “ ”位置“： ”ZZ“}]”;

 JSONObject jObject  = new JSONObject(table); // json       String
 projectname = jObject.getString("name"); // get the name from data.
 System.out.println(projectname);

Answer 1

您可以使用JSONObject和JsonArray，如下所示

JSONObject mainObj = new JSONObject();
    JSONArray somearr = new JSONArray();
                        while (iter.hasnext()) {
                             ClassName someObject=iter.next();
                            JSONObject jsonobj = new JSONObject();
                            jsonobj.put("id", id_from_someObject);;
                            jsonobj.put("name", name_from_someObject);
                            jsonobj.put("name", location_from_someObject);;

                        somearr.put(seatObj);

                        }

mainObj.put("response",somearr);

return mainObj.toString();

Answer 2

简单有效的方式我会告诉你使用json.org和jackson-mapper-asl首先创建一个类，假设它的名字是Person

public class Person {
    private int id;
    private String name;
    private String location;
    //getters and setters
    @Override
    public String toString() {
        return "Person{" +
                "id=" + id +
                ", name='" + name + '\'' +
                ", location='" + location + '\'' +
                '}';
    }
}

创建另一个类假设mainclass

public static void main(String[] args) throws IOException, JSONException {
    String jsonString ="[{\"id\":1,\"name\":\"xxx\",\"location\":\"xx\"},
                                 {\"id\":2,\"name\":\"yyy\",\"location\":\"yy\"},
                                 {\"id\":3,\"name\":\"zzz\",\"location\":\"zz\"}]";
    ObjectMapper mapper = new ObjectMapper();
    JSONArray jsonArray = new JSONArray(jsonString);
    List<Person> listFromJsonArray = new ArrayList<Person>();
    for(int i =0 ;i<jsonArray.length();i++){
        String firstObjectAsString = jsonArray.get(i).toString();
        Person person = mapper.readValue(mapper.readTree(firstObjectAsString), 
                                                                   Person.class);
        listFromJsonArray.add(person);
    }
    System.out.println(listFromJsonArray);

}

现在使用列表获取单个Person对象，使用getter获取各个值并执行任何操作

Answer 3

嗯，在定义表字符串时确实有错误。我猜猜你需要

String table = "[{'id':1,'name':'xxx','location':'xx'}{'id':2,'name':'yyy','location':'yy'}{'id':3,'name':'zzz','location':'zz'}]";
JSONObject jObject = new JSONObject(table);

而不是

String table = [{"id":1,"name":"xxx","location":"xx"}{"id":2,"name":"yyy","location":"yy"}{"id":3,"name":"zzz","location":"zz"}];
JSONObject jObject = new JSONObject(table);

Answer 4

您可以尝试使用以下代码来解析json对象

String table = "{"keyElement" : {"id":1,"name":"xxx","location":"xx"}{"id":2,"name":"yyy","location":"yy"}{"id":3,"name":"zzz","location":"zz"}}";
JSONObject jObject  = new JSONObject(table);
JSONArray array = jObject.getJSONArray("keyElement");    
for(int i = 0 ; i < array.length() ; i++)
{
    System.out.println(array.getJSONObject(i).getString("id"));
    System.out.println(array.getJSONObject(i).getString("name"));
    System.out.println(array.getJSONObject(i).getString("location"));
}

现在你可以了解......

使用java将字符串转换为json对象

4 个答案: