Question

有人可以提供java代码来创建一个json对象，如下所示

 {"main":[
       ["one","two","three","four","five"],
       ["one","two","three","four","five"],
       ["one","two","three","four","five"],
       ["one","two","three","four","five"],
       ["one","two","three","four","five"]
]}

我尝试了类似

的内容

        Gson gson = new Gson();
        JsonArray array = new JsonArray();

        array.add(new JsonPrimitive("One"));
        array.add(new JsonPrimitive("two"));
        array.add(new JsonPrimitive("three"));
        array.add(new JsonPrimitive("four"));
        array.add(new JsonPrimitive("five"));

        JsonObject jsonObject = new JsonObject();
        jsonObject.add("main", array);

即使我在循环

，我也会得到如下结果

{"main":["one","two","three","four","five"]}

就像一个单一的对象。但我期待结果如

{"main":[
       ["one","two","three","four","five"],
       ["one","two","three","four","five"],
       ["one","two","three","four","five"],
       ["one","two","three","four","five"],
       ["one","two","three","four","five"]
]}

非常感谢提前。

Answer 1

尝试使用此代码创建json

Gson gson = new Gson();
JsonArray array = new JsonArray();
JsonArray child = new JsonArray();

child.add(new JsonPrimitive("One"));
child.add(new JsonPrimitive("two"));
child.add(new JsonPrimitive("three"));
child.add(new JsonPrimitive("four"));
child.add(new JsonPrimitive("five"));

for(int i=0;i<5;i++)
array.add(child);


JsonObject jsonObject = new JsonObject();
jsonObject.add("main", array);

System.out.println(jsonObject);

Answer 2

假设您正在使用Gson，这不是它的设计使用方式。虽然支持这种方式，但不建议使用，因为您可以使用任何json库来执行此操作（SimpleJson）。

相反，Gson能够直接序列化我们熟悉的java对象，因此您应该将json对象表示为java对象。 JsonObject映射到Map。 JsonArray映射到List或数组。 JsonPrimitives映射到它们各自的java原始类型（boolean，double，string，null）

// generate the object
Map<List<List<String>>> object = new HashMap<>();
List<List<String>> main = new ArrayList<>();
List<String> counts = Arrays.asList("one", "two", "three", "four", "five");
for (int i = 0; i < 5; i++) {
    main.add(counts);
}
object.put("main", main);

// serialize it
String json = new Gson().toJson(object);

// deserializing it requires a typetoken or separate class representing the map object.
Map<List<List<String>>> desObj = new Gson().fromJson(json, new TypeToken<Map<List<List<String>>>>(){}.getType());

Answer 3

看来"main"包含一个数组数组，所以您只需要将array五次添加到另一个新数组（例如将其称为mainArray）然后添加mainArray的{{1}}：

创建一个新的空数组：jsonObject
将mainArray五次添加到array
将mainArray添加到mainArray

Answer 4

您可以尝试将表示json数据的字符串转换为JsonObject：

import com.google.gson.JsonObject;
import com.google.gson.JsonParser;


public class JsonQuestion {

    public static void main(String[] args) {

        String myJSON = "{\"main\":[\n"
                + "   [\"one\",\"two\",\"three\",\"four\",\"five\"],\n"
                + "   [\"one\",\"two\",\"three\",\"four\",\"five\"],\n"
                + "   [\"one\",\"two\",\"three\",\"four\",\"five\"],\n"
                + "   [\"one\",\"two\",\"three\",\"four\",\"five\"],\n"
                + "   [\"one\",\"two\",\"three\",\"four\",\"five\"]\n"
                + "]}";

        JsonParser jsonParser = new JsonParser();
        JsonObject jsonObject = (JsonObject) jsonParser.parse(myJSON);
        System.out.println("jsonObject: " + jsonObject.toString());

    }
}

Answer 5

<style>
.modal {
    display: none;
    position: fixed;
    z-index: 1;
    padding-top: 100px;
    left: 0;
    top: 0;
    width: 100%;
    height: 100%;
    overflow: auto;
    background-color: rgb(0,0,0);
    background-color: rgba(0,0,0,0.4);
}
.modal-content {
    background-color: #fefefe;
    margin: auto;
    padding: 20px;
    border: 1px solid #888;
    width: 80%;
}
.close {
    color: #aaaaaa;
    float: right;
    font-size: 28px;
    font-weight: bold;
}
.close:hover,
.close:focus {
    color: #000;
    text-decoration: none;
    cursor: pointer;
}
</style>

<script>
  var modal = document.getElementById('myModal');
  var btn = document.getElementById("myBtn");
  var span = document.getElementsByClassName("close")[0];
  btn.onclick = function() {
      modal.style.display = "block";
  }
  span.onclick = function() {
      modal.style.display = "none";
  }
  window.onclick = function(event) {
      if (event.target == modal) {
          modal.style.display = "none";
      }
  }
</script>



 <?php      
  $sql = "select * from enquires2";
  $result = mysqli_query($link,$sql);
  while ($row = mysqli_fetch_array($result)) 
  {
  ?>
  <tr>
    <td>
      <span><img src='gridview/view.png' alt='View' id='myBtn<?php echo $row['id']; ?>'></span>
    </td>
  </tr>
  <div id='myModal' class='modal'>
    <div class='modal-content'>
      <span class='close'>&times;</span>
      <p>Some text in the Modal..</p>
    </div>
  </div>
<?php
  }
?>

Answer 6

您可以使用以下方法制作json：

 private void createJsonData(){
        final String[] units = {"One","Two","Three","Four",
                "Five"};
        try {
        JSONObject jsonObject = new JSONObject();
        JSONArray  jsonArray = new JSONArray();

            JSONArray  jsonArray1 = new JSONArray();
            for (int j = 0; j < 5 ; j++) {
                jsonArray1.put(units[j])
            }
            jsonArray.put(jsonArray);

            jsonObject.put("main",jsonArray);
        } catch (JSONException e) {
            e.printStackTrace();
        }
    }

Answer 7

使用此方法

public static void newjson() {
        JSONObject json =new JSONObject();
        List<List<String>> listoflist  = new ArrayList<List<String>>();
        List<String> list=new ArrayList<String>();
        list.add("one");
        list.add("one");
        list.add("one");


        listoflist.add(list);
        listoflist.add(list);
        try {
            json.put("main",listoflist);
        } catch (JSONException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        System.out.println(json);
    }

在java中创建一个json数据对象

7 个答案: