Question

使用它作为参考我已经描述了我的Json数据的结构，并且可以根据需要获取信息，直到我得到嵌套记录和数组。 Parsing a complex Json Object using GSON in Java

但是我的JSON数据已多次嵌套。例如;

{
  "meetings": [
    {
      "meetingName": "GaryVon",
      "location": "USA",
      "meetingType": "P",
      "meetingDate": "2016-03-25",
      "weatherCondition": "FINE",
      "showCode": {
        "meetingCode": "A",
        "scheduledType": "R"
      },
      "venueType": "ANI",
      "showPools": [
        {
          "showProduct": "GaryVon",
          "showStatus": "Open",

        }
      ]
    }
  ]
}

我有我的包装器和描述json数据格式的类。新java文件中的每个类。

public class meetingContainer { 
    public List<meetings> meetings;
}

顶级课程

public class meetings { 
   private String meetingName;
   private String location;
   private String meetingType;
   private String meetingDate;
   private String weatherCondition;
   private ShowCode showCode;
   private String venueType;
   private ShowPools[] showPools;

   public String getMeetingName() { return meetingName; }
   public String getLocation() { return location; }
   public String getMeetingType() { return meetingType; }
   public String getMeetingDate() { return meetingDate; }
   public String getWeatherCondition() { return weatherCondition; }
   public ShowCode getShowCode() { return showCode; }
   public String getVenueType() { return venueType; }
   public ShowPools[] getShowPools() { return showPools; }
}

二级课程

public class ShowCode {
   private String meetingCode;
   private String scheduledType;

   public String getMeetingCode() { return meetingCode; }
   public String getScheduledType() { return scheduledType; }
}

二级课程

public class ShowPools {
   private String showProduct;
   private String showStatus;

   public String getShowProduct() { return showProduct; }
   public String getShowStatus() { return showStatus; }
}

然后我尝试解析它并获取正常工作的数据，直到我进入嵌套数组/记录

Gson g = new Gson();
meetingContainer mc = g.fromJson(jsonMeetingsString, meetingContainer.class);
for(meetings m: mc.meetings){
    System.out.println(m.getMeetingName()); //Result = "GaryVon"
    System.out.println(m.getLocation()); //Result = "USA"
    System.out.println(m.getmeetingType()); //Result = "P"
    System.out.println(m.getShowCode());  //Result = "packagename.ShowCode@210366b4"          
}

我的问题是如何声明嵌套数组/记录，然后从不同的类调用这些方法，即调用showcode和showpools中的方法。另一篇文章没有说明如何。很抱歉，如果这是一个简单的答案，因为我是java的新手。

Answer 1

m.getShowCode().getMeetingCode()

这将返回ShowCode类型的引用，以使用getter访问内部值，例如：

private List<ShowPools> showPools;

您应该使用showPools的列表

lst = ['a', 'b', 'c']
pad = len('results:') * ' '  # Number of spaces to insert (2nd, 3rd, ... lines)
for i, x in enumerate(lst):
    if i == 0:
        print('results:', x)
    else:
        print(pad, x)

Answer 2

您提供的JSON字符串无效。它有一个额外的, -

        {
          "showProduct": "GaryVon",
          "showStatus": "Open",
                              ^

回答您在评论中提出的问题：m.getShowCode().getShowProduct()无效，因为 showCode 节点只有两个属性 meetingCode 和 scheduledType

下面的代码列出了JSON的所有值。如果它没有涵盖您的问题，请告诉我

    Gson g = new Gson();
    meetingContainer mc = g.fromJson(jsonMeetingsString,
            meetingContainer.class);
    for (meetings m : mc.meetings) {
        System.out.println("meetingName: " + m.getMeetingName()); 
        System.out.println("location: "+ m.getLocation()); 
        System.out.println("meetingType: "+ m.getMeetingType()); 
        System.out.println("meetingDate: "+ m.getMeetingDate()); 
        System.out.println("weatherConditio: "+ m.getWeatherCondition()); 
        System.out.println("showCode->meetingCode: "+ m.getShowCode().getMeetingCode()); 
        System.out.println("showCode->scheduledType: "+ m.getShowCode().getScheduledType());
        System.out.println("venueType: "+ m.getVenueType()); 
        for(ShowPools showPool : m.getShowPools()){
            System.out.println("showPools->showProduct: "+ showPool.getShowProduct());
            System.out.println("showPools->showStatus: "+ showPool.getShowStatus());
        }
    }

输出：

meetingName: GaryVon
location: USA
meetingType: P
meetingDate: 2016-03-25
weatherConditio: FINE
showCode->meetingCode: A
showCode->scheduledType: R
venueType: ANI
showPools->showProduct: GaryVon
showPools->showStatus: Open

使用Java中的GSON反序列化复杂的Json对象

2 个答案: