通过自定义类和添加类型适配器,您可以操纵字符串(json.toString()返回带有“”引号的字符串,因此您可以看到它是否为字符串)。
输出:(类似乎是正确的)
class test.Main$StringPojo pojo{object=foo}
class test.Main$IntPojo pojo{object=1}
class test.Main$StringPojo pojo{object=bar}
class test.Main$StringPojo pojo{object=2}
class test.Main$IntPojo pojo{object=3}
public static void main(final String[] args){
String str = "{\n" +
" \"text\": [\"foo\",1,\"bar\",\"2\",3],\n" +
" \"text1\": \"value1\" }";
GsonBuilder builder = new GsonBuilder();
builder.registerTypeAdapter(pojo.class, new JsonDeserializer<pojo>() {
@Override
public pojo deserialize(JsonElement json, Type typeOfT, JsonDeserializationContext context) throws JsonParseException {
try {
return new IntPojo(Integer.parseInt(json.toString()));
} catch (Exception e) {
return new StringPojo(json.getAsString());
}
}
});
MyPojo myPojo = builder.create().fromJson(str, MyPojo.class);
for (pojo pojo : myPojo.text) {
System.out.println(pojo.getClass() + " " + pojo.object);
}
}
public static abstract class pojo{
protected Object object;
public pojo() {
}
@Override
public String toString() {
return "pojo{" +
"object=" + object +
'}';
}
}
public static class StringPojo extends pojo{
public StringPojo(String str) {
object = str;
}
}
public static class IntPojo extends pojo{
public IntPojo(int intt) {
this.object = intt;
}
}
public static class MyPojo {
List<pojo> text;
String text1;
}