java - 使用 gson 将 JSON 转换为 java 对象

Question

我正在使用 GSON 库来处理来自 Web 服务的 JSON，但我无法让它工作，我总是得到一个空值。我已经查看了类似的问题，例如 Converting Json to Java such as Simple Json to Java convertion using GSON .但我仍然缺少一些东西

JSON

{"A":"val1","B":"val2","C":"val3","D":"val4","E":"val5","F":"val6","G":"val7"}
         SiteWrapper m = gson.fromJson(json, SiteWrapper.class);

Java 类

SiteWrapper m = gson.fromJson(json, SiteWrapper.class);
System.out.println(m.getMenu());

static class Site {
    static String A;
    static String B;
    static String C;
    static String D;
    static String E;
    static String F;
    static String G;

    public String toString() {
        return String.format(A,B,C,D,E,F,G);}

    public static String getA() {
        return A;
    }
    public static String getB() {
        return B;
    } 
... all the way to getG

    public void setA(String A) {
        Site.A = A;
    }
    public void setB(String B) {
        Site.B = B;
    }
... all the way to setB

和我的包装器

class SiteWrapper {
    private Site site;
    public Site getMenu() { return site; }
    public void setMenu(Site site) { this.site = site; }
}

无论我做什么我都会得到一个空值，有什么想法吗？

Answer 1

因为它是一个静态内部类。如docs指出和评论:

As well, if a field is marked as "static" then by default it will be excluded. If you want to include some transient fields...

你可能想试试

 Gson gson = new GsonBuilder()
    .excludeFieldsWithModifier()
    .create();

此外，由于它是一个内部类，您可能需要更改 JSON 如果可以:

 {
   "site":{
      "A":"val1",
      "B":"val2",
      "C":"val3",
      "D":"val4",
      "E":"val5",
      "F":"val6",
      "G":"val7"
   }
}

如本 post 中所述