Question

如何从一次JSONString.stringify简单数组解析JSONString，该数组现在“显得扁平化”，并将其值转换回Java List或Java Array？（使用Jersey 1.x＆amp; Java）？数组最初在字符串化之前以[1,2,3]开始。

items =（3）[“To”，“8357”，“30028”] - ＆gt;通过休息呼叫发送的JSON.stringify（项目）

Chrome Dev Tools在休息电话后的请求有效负载：项=％5B％22％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0 ％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2 ％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0To％22％2C％228357％22％2C％2230028％22％5D

/*inside (Jersey) Rest Resource
@POST
@Path("/...")
@Consumes(MediaType.APPLICATION_JSON)
@Produces(MediaType.APPLICATION_JSON)
public Response receive(@Context SecurityContext securityContext, @Context 
HttpServletRequest srequest, String jsonString) throws URISyntaxException, 
JSONException ...

/ * eclipse监视jsonString里面（Jersey）Rest Resource 项=％5B％22％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0 ％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2 ％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0To％22％2C％228357％22％2C％2230028％22％5D * / [在此输入图像说明] [2]

注意：没有名称值。没有实体。

只有一个非常简单的ID字符串，因为这就是我所需要的。（Jersey 1.X或JAX-RS 1.X支持吗？）

JSONArray jSONArray = new JSONArray(java.util.Arrays.asList(jsonString));

Eclipse jSONArray表达式：jSONArray
  --myArrayList
  ---- elementData中
  ------ [0]“项目=％5B％22％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2 ％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0 ％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0％C2％A0To％22％2C％228357％22％2C％2230028％22 ％5D“

----------值
  ------------ [0..99]
  --------------- [0]我
  --------------- 1 t
  --------------- [2] e
  --------------- [3] m
  --------------- [4] s
  --------------- [5] =
  --------------- [6]％
  --------------- [7] 5
  --------------- [8] B ....

Answer 1

我无法完全理解你原来的问题，但其中一个方法是：

import java.io.File;
import java.io.FileNotFoundException;
import java.io.IOException;
import javax.swing.JFileChooser;
import org.json.simple.JSONObject;
import org.json.simple.parser.JSONParser;

public class JsonParser {

    JFileChooser chooser = new JFileChooser();
    File f;

    static String fn = "";
    static String js1 = "{\"name\": \"LALA\", \"email\": \"tst@tst.com\"}";
    String name = "name";
    String email = "email";
    String fName = "firsName";
    String city = "city";
    // ... other needed fields
    User u1 = null;

    public JsonParser() {
        parseFile();
        System.out.println("\n" + u1.toShortString());
    }

    private String openFchooser() throws FileNotFoundException, IOException, InterruptedException, Exception {
        int returnVal = chooser.showOpenDialog(null);
        if (returnVal == JFileChooser.APPROVE_OPTION) {
            f = chooser.getSelectedFile();
        }
        return f.getAbsolutePath();

    }

    // To parse JSON files with data
    //===========================================
    public void parseFile() {
        JSONParser parser = new JSONParser();

        try {
            // To parse obj 1
            Object obj1 = parser.parse(js1);
            System.out.println("User 1: " + obj1.toString());
            System.out.println();

            JSONObject jobj1 = (JSONObject) obj1;
            String from_name = jobj1.get(name).toString();
            String from_email = jobj1.get(email).toString();
//            String from_fName = jobj1.get(fName).toString();
//            String from_city = jobj1.get(city).toString();
            u1 = new User(from_name, from_email, null, null);
//            System.out.println(u1.toString() + "\n");

        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    public static void main(String[] args) {
        new JsonParser();
    }

    class User {

        String name = null;
        String email = null;
        String fName = null;
        String city = null;

        public User(String n, String e, String f, String c) {
            this.name = n;
            this.email = e;
            this.fName = f;
            this.city = c;
        }

        public String getFirsName() {
            return this.name;
        }

        public String setFirsName(String s) {
            return this.name = s;
        }

        public String getEmail() {
            return this.email;
        }

        public String setEmail(String s) {
            return this.email = s;
        }

        public String toString() {
            return "{\"name\":" + this.name + ", "
                    + "\"email\":" + this.email + ", "
                    + "\"firsName\":" + this.fName + ", "
                    + "\"city\":" + this.city + "\"}";
        }

        public String toShortString() {
            return "{\"name\": \"" + this.name + "\", "
                    + "\"email\": \"" + this.email + "\"}";
        }
    };
}

<强>输出：

User 1: {"name":"LALA","email":"tst@tst.com"}


{"name": "LALA", "email": "tst@tst.com"}

Answer 2

谢谢你们。我设法找到另一种发送和放大器接收：

现在首先发送数组而不进行字符串化收到： public Response archiveSelectedApplicants（@Context SecurityContext securityContext，@ Context HttpServletRequest srequest， @FormParam（＆＃34; items []＆＃34;）List items）抛出URISyntaxException

如何解析JSONString并将其值转换为数组？

2 个答案: