Mixing
Java Generic Map not converted to proper type in runtime for Json converstion
I have one method which works fine when i use actual class, but don't give expected output when using generics.
Below is the method which works fine when using ABC class
public static List<ABC> getMemberViewRepresentation(Response response) throws JSONException, IOException {
JSONObject jsonObj = new JSONObject(response.readEntity(String.class));
ObjectMapper mapper = new ObjectMapper();
JSONObject memberViewObj = (JSONObject)jsonObj.get("members");
TypeReference<HashMap<String, ABC>> typeRef = new TypeReference<HashMap<String, ABC>>() {};
Map<String, ABC> map = mapper.readValue(memberViewObj.toString(), typeRef);
return new ArrayList<>(map.values());
}
This methods gives proper output which has list of type ABC.
But i want to write code such that i pass class dynamically so that this method can be used by anyone. So i have written below code, also i tried other way but none of them seems to work.
public static<T> List<T> getMemberViewRepresentation(Response response) throws JSONException, IOException {
JSONObject jsonObj = new JSONObject(response.readEntity(String.class));
ObjectMapper mapper = new ObjectMapper();
JSONObject memberViewObj = (JSONObject)jsonObj.get("members");
TypeReference<HashMap<String, T>> typeRef = new TypeReference<HashMap<String, T>>() {};
Map<String, T> map = mapper.readValue(memberViewObj.toString(), typeRef);
return new ArrayList<>(map.values());
}
I am calling in this way
List<ABC> nodes = ResponseUtil.getMemberViewRepresentation(response);
But the output of above method is not same. List is not of type ABC
@JsonIgnoreProperties(ignoreUnknown = true)
public class ABC {
@JsonProperty("id")
private int id;
@JsonProperty("uid")
private String uid;
public int getId() {
return id;
}
public String getUid() {
return uid;
}
}
Any idea how to keep the method dynamic such that i can pass Class details separately
java
asked Oct 19 '18 at 8:07
abhishek chaubalabhishek chaubal
211
add a comment |
I have one method which works fine when i use actual class, but don't give expected output when using generics.
Below is the method which works fine when using ABC class
public static List<ABC> getMemberViewRepresentation(Response response) throws JSONException, IOException {
JSONObject jsonObj = new JSONObject(response.readEntity(String.class));
ObjectMapper mapper = new ObjectMapper();
JSONObject memberViewObj = (JSONObject)jsonObj.get("members");
TypeReference<HashMap<String, ABC>> typeRef = new TypeReference<HashMap<String, ABC>>() {};
Map<String, ABC> map = mapper.readValue(memberViewObj.toString(), typeRef);
return new ArrayList<>(map.values());
}
This methods gives proper output which has list of type ABC.
But i want to write code such that i pass class dynamically so that this method can be used by anyone. So i have written below code, also i tried other way but none of them seems to work.
public static<T> List<T> getMemberViewRepresentation(Response response) throws JSONException, IOException {
JSONObject jsonObj = new JSONObject(response.readEntity(String.class));
ObjectMapper mapper = new ObjectMapper();
JSONObject memberViewObj = (JSONObject)jsonObj.get("members");
TypeReference<HashMap<String, T>> typeRef = new TypeReference<HashMap<String, T>>() {};
Map<String, T> map = mapper.readValue(memberViewObj.toString(), typeRef);
return new ArrayList<>(map.values());
}
I am calling in this way
List<ABC> nodes = ResponseUtil.getMemberViewRepresentation(response);
But the output of above method is not same. List is not of type ABC
@JsonIgnoreProperties(ignoreUnknown = true)
public class ABC {
@JsonProperty("id")
private int id;
@JsonProperty("uid")
private String uid;
public int getId() {
return id;
}
public String getUid() {
return uid;
}
}
Any idea how to keep the method dynamic such that i can pass Class details separately
java
asked Oct 19 '18 at 8:07
abhishek chaubalabhishek chaubal
211
add a comment |
I have one method which works fine when i use actual class, but don't give expected output when using generics.
Below is the method which works fine when using ABC class
public static List<ABC> getMemberViewRepresentation(Response response) throws JSONException, IOException {
JSONObject jsonObj = new JSONObject(response.readEntity(String.class));
ObjectMapper mapper = new ObjectMapper();
JSONObject memberViewObj = (JSONObject)jsonObj.get("members");
TypeReference<HashMap<String, ABC>> typeRef = new TypeReference<HashMap<String, ABC>>() {};
Map<String, ABC> map = mapper.readValue(memberViewObj.toString(), typeRef);
return new ArrayList<>(map.values());
}
This methods gives proper output which has list of type ABC.
But i want to write code such that i pass class dynamically so that this method can be used by anyone. So i have written below code, also i tried other way but none of them seems to work.
public static<T> List<T> getMemberViewRepresentation(Response response) throws JSONException, IOException {
JSONObject jsonObj = new JSONObject(response.readEntity(String.class));
ObjectMapper mapper = new ObjectMapper();
JSONObject memberViewObj = (JSONObject)jsonObj.get("members");
TypeReference<HashMap<String, T>> typeRef = new TypeReference<HashMap<String, T>>() {};
Map<String, T> map = mapper.readValue(memberViewObj.toString(), typeRef);
return new ArrayList<>(map.values());
}
I am calling in this way
List<ABC> nodes = ResponseUtil.getMemberViewRepresentation(response);
But the output of above method is not same. List is not of type ABC
@JsonIgnoreProperties(ignoreUnknown = true)
public class ABC {
@JsonProperty("id")
private int id;
@JsonProperty("uid")
private String uid;
public int getId() {
return id;
}
public String getUid() {
return uid;
}
}
Any idea how to keep the method dynamic such that i can pass Class details separately
java
asked Oct 19 '18 at 8:07
abhishek chaubalabhishek chaubal
211
I have one method which works fine when i use actual class, but don't give expected output when using generics.
Below is the method which works fine when using ABC class
public static List<ABC> getMemberViewRepresentation(Response response) throws JSONException, IOException {
JSONObject jsonObj = new JSONObject(response.readEntity(String.class));
ObjectMapper mapper = new ObjectMapper();
JSONObject memberViewObj = (JSONObject)jsonObj.get("members");
TypeReference<HashMap<String, ABC>> typeRef = new TypeReference<HashMap<String, ABC>>() {};
Map<String, ABC> map = mapper.readValue(memberViewObj.toString(), typeRef);
return new ArrayList<>(map.values());
}
This methods gives proper output which has list of type ABC.
But i want to write code such that i pass class dynamically so that this method can be used by anyone. So i have written below code, also i tried other way but none of them seems to work.
public static<T> List<T> getMemberViewRepresentation(Response response) throws JSONException, IOException {
JSONObject jsonObj = new JSONObject(response.readEntity(String.class));
ObjectMapper mapper = new ObjectMapper();
JSONObject memberViewObj = (JSONObject)jsonObj.get("members");
TypeReference<HashMap<String, T>> typeRef = new TypeReference<HashMap<String, T>>() {};
Map<String, T> map = mapper.readValue(memberViewObj.toString(), typeRef);
return new ArrayList<>(map.values());
}
I am calling in this way
List<ABC> nodes = ResponseUtil.getMemberViewRepresentation(response);
But the output of above method is not same. List is not of type ABC
@JsonIgnoreProperties(ignoreUnknown = true)
public class ABC {
@JsonProperty("id")
private int id;
@JsonProperty("uid")
private String uid;
public int getId() {
return id;
}
public String getUid() {
return uid;
}
}
Any idea how to keep the method dynamic such that i can pass Class details separately
java
java
asked Oct 19 '18 at 8:07
abhishek chaubalabhishek chaubal
211
asked Oct 19 '18 at 8:07
abhishek chaubalabhishek chaubal
211
asked Oct 19 '18 at 8:07
abhishek chaubalabhishek chaubal
211
asked Oct 19 '18 at 8:07
abhishek chaubalabhishek chaubal
211
asked Oct 19 '18 at 8:07
abhishek chaubalabhishek chaubal
211
211
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
I think, what you need is:
List<ABC> nodes = ResponseUtil.<ABC>getMemberViewRepresentation(response);
...to infer the generic parameter(s) into a static method (WITHOUT typed parameter(s)).
Note the extra <ABC> before method invocation!
Normally you don't need it, because a regular generic method goes like:
public static<T> void foo(T someInput) {...} ...and providing someInput, T is known/easy to infer at runtime. In your case Response has nothing to do with ABC...so can't be inferred (with anything else than ?).
The first method works as expected, because there is no "generic method", while the "class generic type" seems to be inferred correctly (somewhere else in your code/configuration).
The second doesn't, because the "generic method type" is missing (and i assume inferred as <?> respectively <java.lang.Object> ...so you get a List<Object> returned and assigned to your List<ABC>, which is not as expected, but at least "fail free").
You can argument, that we can "infer" ABC from the return type (you expect a List<ABC>, where a List<T> is delivered), but unfortunately this is not how it works/this information is not available at that point/needs deeper study.
answered Nov 20 '18 at 11:32
xerx593xerx593
2,4791429
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I think, what you need is:
List<ABC> nodes = ResponseUtil.<ABC>getMemberViewRepresentation(response);
...to infer the generic parameter(s) into a static method (WITHOUT typed parameter(s)).
Note the extra <ABC> before method invocation!
Normally you don't need it, because a regular generic method goes like:
public static<T> void foo(T someInput) {...} ...and providing someInput, T is known/easy to infer at runtime. In your case Response has nothing to do with ABC...so can't be inferred (with anything else than ?).
The first method works as expected, because there is no "generic method", while the "class generic type" seems to be inferred correctly (somewhere else in your code/configuration).
The second doesn't, because the "generic method type" is missing (and i assume inferred as <?> respectively <java.lang.Object> ...so you get a List<Object> returned and assigned to your List<ABC>, which is not as expected, but at least "fail free").
You can argument, that we can "infer" ABC from the return type (you expect a List<ABC>, where a List<T> is delivered), but unfortunately this is not how it works/this information is not available at that point/needs deeper study.
answered Nov 20 '18 at 11:32
xerx593xerx593
2,4791429
add a comment |
I think, what you need is:
List<ABC> nodes = ResponseUtil.<ABC>getMemberViewRepresentation(response);
...to infer the generic parameter(s) into a static method (WITHOUT typed parameter(s)).
Note the extra <ABC> before method invocation!
Normally you don't need it, because a regular generic method goes like:
public static<T> void foo(T someInput) {...} ...and providing someInput, T is known/easy to infer at runtime. In your case Response has nothing to do with ABC...so can't be inferred (with anything else than ?).
The first method works as expected, because there is no "generic method", while the "class generic type" seems to be inferred correctly (somewhere else in your code/configuration).
The second doesn't, because the "generic method type" is missing (and i assume inferred as <?> respectively <java.lang.Object> ...so you get a List<Object> returned and assigned to your List<ABC>, which is not as expected, but at least "fail free").
You can argument, that we can "infer" ABC from the return type (you expect a List<ABC>, where a List<T> is delivered), but unfortunately this is not how it works/this information is not available at that point/needs deeper study.
answered Nov 20 '18 at 11:32
xerx593xerx593
2,4791429
add a comment |
I think, what you need is:
List<ABC> nodes = ResponseUtil.<ABC>getMemberViewRepresentation(response);
...to infer the generic parameter(s) into a static method (WITHOUT typed parameter(s)).
Note the extra <ABC> before method invocation!
Normally you don't need it, because a regular generic method goes like:
public static<T> void foo(T someInput) {...} ...and providing someInput, T is known/easy to infer at runtime. In your case Response has nothing to do with ABC...so can't be inferred (with anything else than ?).
The first method works as expected, because there is no "generic method", while the "class generic type" seems to be inferred correctly (somewhere else in your code/configuration).
The second doesn't, because the "generic method type" is missing (and i assume inferred as <?> respectively <java.lang.Object> ...so you get a List<Object> returned and assigned to your List<ABC>, which is not as expected, but at least "fail free").
You can argument, that we can "infer" ABC from the return type (you expect a List<ABC>, where a List<T> is delivered), but unfortunately this is not how it works/this information is not available at that point/needs deeper study.
answered Nov 20 '18 at 11:32
xerx593xerx593
2,4791429
I think, what you need is:
List<ABC> nodes = ResponseUtil.<ABC>getMemberViewRepresentation(response);
...to infer the generic parameter(s) into a static method (WITHOUT typed parameter(s)).
Note the extra <ABC> before method invocation!
Normally you don't need it, because a regular generic method goes like:
public static<T> void foo(T someInput) {...} ...and providing someInput, T is known/easy to infer at runtime. In your case Response has nothing to do with ABC...so can't be inferred (with anything else than ?).
The first method works as expected, because there is no "generic method", while the "class generic type" seems to be inferred correctly (somewhere else in your code/configuration).
The second doesn't, because the "generic method type" is missing (and i assume inferred as <?> respectively <java.lang.Object> ...so you get a List<Object> returned and assigned to your List<ABC>, which is not as expected, but at least "fail free").
You can argument, that we can "infer" ABC from the return type (you expect a List<ABC>, where a List<T> is delivered), but unfortunately this is not how it works/this information is not available at that point/needs deeper study.
answered Nov 20 '18 at 11:32
xerx593xerx593
2,4791429
edited Nov 20 '18 at 11:47
answered Nov 20 '18 at 11:32
xerx593xerx593
2,4791429
answered Nov 20 '18 at 11:32
xerx593xerx593
2,4791429
answered Nov 20 '18 at 11:32
xerx593xerx593
2,4791429
2,4791429
add a comment |
add a comment |
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