使用 Gson 将 JSON 数组解析为 java.util.List
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Parsing JSON array into java.util.List with Gson
提问by Abel Callejo
I have a JsonObjectnamed "mapping"with the following content:
我有一个具有以下内容的JsonObject命名"mapping"：
{
"client": "127.0.0.1",
"servers": [
"8.8.8.8",
"8.8.4.4",
"156.154.70.1",
"156.154.71.1"
]
}
I know I can get the array "servers"with:
我知道我可以通过以下方式获取数组"servers"：
mapping.get("servers").getAsJsonArray()
And now I want to parse that JsonArrayinto a java.util.List...
现在我想把它解析JsonArray成一个java.util.List......
What is the easiest way to do this?
什么是最简单的方法来做到这一点？
采纳答案by MikO
Definitely the easiest way to do that is using Gson's default parsing function fromJson().
绝对最简单的方法是使用 Gson 的默认解析函数fromJson()。
There is an implementation of this function suitable for when you need to deserialize into any ParameterizedType(e.g., any List), which is fromJson(JsonElement json, Type typeOfT).
还有当你需要反序列化到任何合适的这一功能的实现ParameterizedType（例如，任何List），这是fromJson(JsonElement json, Type typeOfT)。
In your case, you just need to get the Typeof a List<String>and then parse the JSON array into that Type, like this:
在您的情况下，您只需要获取TypeaList<String>然后将 JSON 数组解析为 that Type，如下所示：
import java.lang.reflect.Type;
import com.google.gson.reflect.TypeToken;
JsonElement yourJson = mapping.get("servers");
Type listType = new TypeToken<List<String>>() {}.getType();
List<String> yourList = new Gson().fromJson(yourJson, listType);
In your case yourJsonis a JsonElement, but it could also be a String, any Readeror a JsonReader.
在您的情况下yourJson是 a JsonElement，但它也可以是 a String、 anyReader或 a JsonReader。
You may want to take a look at Gson API documentation.
您可能需要查看Gson API 文档。
回答by Prateek
Below code is using com.google.gson.JsonArray.
I have printed the number of element in list as well as the elements in List
下面的代码正在使用com.google.gson.JsonArray. 我已经打印了列表中的元素数量以及列表中的元素
import java.util.ArrayList;
import com.google.gson.Gson;
import com.google.gson.JsonArray;
import com.google.gson.JsonObject;
import com.google.gson.JsonParser;
public class Test {
static String str = "{ "+
"\"client\":\"127.0.0.1\"," +
"\"servers\":[" +
" \"8.8.8.8\"," +
" \"8.8.4.4\"," +
" \"156.154.70.1\"," +
" \"156.154.71.1\" " +
" ]" +
"}";
public static void main(String[] args) {
// TODO Auto-generated method stub
try {
JsonParser jsonParser = new JsonParser();
JsonObject jo = (JsonObject)jsonParser.parse(str);
JsonArray jsonArr = jo.getAsJsonArray("servers");
//jsonArr.
Gson googleJson = new Gson();
ArrayList jsonObjList = googleJson.fromJson(jsonArr, ArrayList.class);
System.out.println("List size is : "+jsonObjList.size());
System.out.println("List Elements are : "+jsonObjList.toString());
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
OUTPUT
输出
List size is : 4
List Elements are : [8.8.8.8, 8.8.4.4, 156.154.70.1, 156.154.71.1]
回答by Hoang
I read solution from official website of Gson at here
我在这里阅读了 Gson 官方网站上的解决方案
And this code for you:
而这个代码给你：
String json = "{"client":"127.0.0.1","servers":["8.8.8.8","8.8.4.4","156.154.70.1","156.154.71.1"]}";
JsonObject jsonObject = new Gson().fromJson(json, JsonObject.class);
JsonArray jsonArray = jsonObject.getAsJsonArray("servers");
String[] arrName = new Gson().fromJson(jsonArray, String[].class);
List<String> lstName = new ArrayList<>();
lstName = Arrays.asList(arrName);
for (String str : lstName) {
System.out.println(str);
}
Result show on monitor:
结果显示在监视器上：
8.8.8.8
8.8.4.4
156.154.70.1
156.154.71.1
回答by Gene Bo
I was able to get the list mapping to work with just using @SerializedNamefor all fields.. no logic around Typewas necessary.
我能够让列表映射仅@SerializedName用于所有字段.. 不需要任何逻辑Type。
Running the code - in step #4 below- through the debugger, I am able to observe that the List<ContentImage> mGalleryImagesobject populated with the JSON data
运行代码 -在下面的第 4 步中- 通过调试器，我能够观察到List<ContentImage> mGalleryImages用 JSON 数据填充的对象
Here's an example:
下面是一个例子：
1. The JSON
1. JSON
{
"name": "Some House",
"gallery": [
{
"description": "Nice 300sqft. den.jpg",
"photo_url": "image/den.jpg"
},
{
"description": "Floor Plan",
"photo_url": "image/floor_plan.jpg"
}
]
}
2. Java class with the List
2.带有List的Java类
public class FocusArea {
@SerializedName("name")
private String mName;
@SerializedName("gallery")
private List<ContentImage> mGalleryImages;
}
3. Java class for the List items
3. List 项的 Java 类
public class ContentImage {
@SerializedName("description")
private String mDescription;
@SerializedName("photo_url")
private String mPhotoUrl;
// getters/setters ..
}
4. The Java code that processes the JSON
4. 处理 JSON 的 Java 代码
for (String key : focusAreaKeys) {
JsonElement sectionElement = sectionsJsonObject.get(key);
FocusArea focusArea = gson.fromJson(sectionElement, FocusArea.class);
}
回答by Alex Moore-Niemi
Given you start with mapping.get("servers").getAsJsonArray(), if you have access to GuavaStreams, you can do the below one-liner:
鉴于您从 开始mapping.get("servers").getAsJsonArray()，如果您可以访问GuavaStreams，则可以执行以下单行操作：
List<String> servers = Streams.stream(jsonArray.iterator())
.map(je -> je.getAsString())
.collect(Collectors.toList());
Note StreamSupportwon't be able to work on JsonElementtype, so it is insufficient.
NoteStreamSupport将无法处理JsonElement类型，因此它是不够的。
