Question

如何创建下面给出的json？

<?php
$cid= (string)$_GET['cid'];//I passed this from another page using get method
echo $cid; //My code works up to this point
$record = mysql_query("select * from questions where QType = '$cid'");
$array = array();
while($row = mysql_fetch_assoc($record))
        {
            $array[] = $row;
        }
    for($var = 0; $var<count($array);$var++)
        {
        echo $array[$var]['Question'].'<br>';

        }           
            ?>

Answer 1

这取决于您的JSON库，但是对于org.json.simple或net.sf.json来说，就像这样：

JSONObject objectInArray = new JSONObject();
objectInArray.put("id", "some value");
objectInArray.put("name", "some value");
JSONArray jsonArray = new JSONArray();
jsonArray.add(objectInArray);
JSONObject data = new JSONObject();
data.put("data", jsonArray);

Answer 2

使用Google的gson-2.x.jar 使像这样的实体类

public class MyEntity{
    @SerializedName("data")
    private List<Data> m_listData;

   //Setter getters
   public List<Data> getData() {
        return m_listData;
    }

    public void setData(List<Data> data) {
        this.m_listData= data;
    }
}
class Data{
    @SerializedName("id")
    private String id;

    @SerializedName("name")
    private String name;

    public String getId() {
        return id;
    }

    public void setId(String id) {
        this.id= id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name= name;
    }
}

//and then convert you json string to java object
public class Test{
    public static void main(String[] args){

       String jsonString="your JSON string"
       Gson obj_gson = new Gson();
       MyEntity obj_myEntity=obj_gson.fromJson(jsonString, MyEntity.class);
  }
}

如何在java中创建下面给出的json？

2 个答案: