java <T> 无法解析为类型？
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<T> cannot be resolved to a type?
提问by Siddharth Trikha
I want to convert a json string to a List<Someclass>using Hymanson json library.
我想将 json 字符串转换为List<Someclass>使用 Hymanson json 库的字符串。
public static List<T> toList(String json, Class<T> type, ObjectMapperProperties objectMapperProperties){
ObjectMapper objectMapper = ObjectMapperFactory.getObjectMapper(objectMapperProperties);
try {
return objectMapper.readValue(json, objectMapper.getTypeFactory().constructCollectionType(List.class, type));
} catch (JsonParseException e) {
e.printStackTrace();
} catch (JsonMappingException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
So if I pass Attribute.classas type, then it must return a List<Attribute>.
所以如果我通过Attribute.classas type，那么它必须返回一个List<Attribute>.
However, this gives me a compile time error
但是，这给了我一个编译时错误
T cannot be resolved to a type
I guess generics part is not clear to me here. Any help is appreciated.
我想这里的泛型部分对我来说不清楚。任何帮助表示赞赏。
回答by Sach141
you need to declare Tfirst in your generic method, In your case it would be :
您需要T在您的泛型方法中首先声明，在您的情况下，它将是：
public static <T> List<T> toList(String json, Class<T> type, ObjectMapperProperties objectMapperProperties)
for more info please check oracle documentation for generic methods:
有关更多信息，请查看通用方法的 oracle 文档：
https://docs.oracle.com/javase/tutorial/extra/generics/methods.html
https://docs.oracle.com/javase/tutorial/extra/generics/methods.html
