反序列化映射< Object，Object>与GSon [英] Deserializing Map<Object, Object> with GSon

问题描述

我有一个包含混合类型的Map，就像这个简单的例子一样

  final Map< String，Object> map = new LinkedHashMap< String，Object>（）; 
 map.put（a，1）; 
 map.put（b，a）; 
 map.put（c，2）; 
 final Gson gson = new Gson（）; 
 final String string = gson.toJson（map）; 
 final类型type = new TypeToken< LinkedHashMap< String，Object>>（）{}。getType（）; 
 final Map< Object，Object> map2 = gson.fromJson（string，type）; 
 for（final Entry< Object，Object> entry：map2.entrySet（））{
 System.out.println（entry.getKey（）+：+ entry.getValue（））; 
} 
  

我得到的是 Object s，no 整数 s，no String s。输出如下所示：

  a：java.lang.Object@48d19bc8 
b：java.lang.Object@394a8cd1 
c：java.lang.Object@4d630ab9 
  

我可以以某种方式修复它吗？我希望这种简单的情况会在默认情况下正确处理。

^{我知道关于类型的信息不能总是被保留下来， 1 和1在JSON中的含义完全相同。但是，返回简单的无内容对象对我来说毫无意义。}

更新：序列化版本（即<

$ p $ {a：1，code> string above）看起来不错：

b：a，c：2}

解决方案

< Gson并不聪明。而是提供一个清晰和静态的Javabean类风格的数据结构，以便Gson理解单独的属性应该被反序列化到什么类型。

例如

  public class Data {
 private Integer a; 
私人字符串b; 
私人整数c; 
 // ... 
} 
  

结合

  Data data1 = new Data（1，a，2）; 
 String json = gson.toJson（data1）; 
 Data data2 = gson.fromJson（json，Data.class）; 
  

---

更新：as根据评论，键集看起来不是固定的（尽管事先不知道结构，你似乎可以手动将其转换）。您可以创建自定义解串器。这里有一个quick'n'dirty的例子。

  public class ObjectDeserializer实现了JsonDeserializer< Object> {
 $ b $ @Override 
 public Object deserialize（JsonElement element，Type type，JsonDeserializationContext context）throws JsonParseException {
 String value = element.getAsString（）; 
尝试{
 return Long.valueOf（value）; 
} catch（NumberFormatException e）{
返回值; 
 
 
 
 $ b $ / code> 

你使用如下：



  final Gson gson = new GsonBuilder（）。registerTypeAdapter（Object.class，new ObjectDeserializer（））。create （）; 
 // ... 
  




I have a Map containing a mixture of types like in this simple example

final Map<String, Object> map = new LinkedHashMap<String, Object>();
map.put("a", 1);
map.put("b", "a");
map.put("c", 2);
final Gson gson = new Gson();
final String string = gson.toJson(map);
final Type type = new TypeToken<LinkedHashMap<String, Object>>(){}.getType();
final Map<Object, Object> map2 = gson.fromJson(string, type);
for (final Entry<Object, Object> entry : map2.entrySet()) {
    System.out.println(entry.getKey() + " : " + entry.getValue());
}

What I get back are plain Objects, no Integers, no Strings. The output looks like

a : java.lang.Object@48d19bc8
b : java.lang.Object@394a8cd1
c : java.lang.Object@4d630ab9

Can I fix it somehow? I'd expect that such simple cases will be handled correctly by default.

^{I know that the information about the type can't always be preserved, and possibly 1 and "1" means exactly the same in JSON. However, returning plain content-less objects just makes no sense to me.}

Update: The serialized version (i.e. the string above) looks fine:

{"a":1,"b":"a","c":2}

解决方案

Gson isn't that smart. Rather provide a clear and static data structure in flavor of a Javabean class so that Gson understands what type the separate properties are supposed to be deserialized to.

E.g.

public class Data {
    private Integer a;
    private String b;
    private Integer c;
    // ...
}

in combination with

Data data1 = new Data(1, "a", 2);
String json = gson.toJson(data1);
Data data2 = gson.fromJson(json, Data.class);

---

Update: as per the comments, the keyset seems to be not fixed (although you seem to be able to convert it manually afterwards without knowing the structure beforehand). You could create a custom deserializer. Here's a quick'n'dirty example.

public class ObjectDeserializer implements JsonDeserializer<Object> {

    @Override
    public Object deserialize(JsonElement element, Type type, JsonDeserializationContext context) throws JsonParseException {
        String value = element.getAsString();
        try {
            return Long.valueOf(value);
        } catch (NumberFormatException e) {
            return value;
        }
    }

}

which you use as follows:

final Gson gson = new GsonBuilder().registerTypeAdapter(Object.class, new ObjectDeserializer()).create();
// ... 

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