Mixing
What's the Jackson deserialization equivalent of @JsonUnwrapped?
Say I have the following class:
public class Parent {
public int age;
@JsonUnwrapped
public Name name;
}
Producing JSON:
{
"age" : 18,
"first" : "Joey",
"last" : "Sixpack"
}
How do I deserialize this back into the Parent class? I could use @JsonCreator
@JsonCreator
public Parent(Map<String,String> jsonMap) {
age = jsonMap.get("age");
name = new Name(jsonMap.get("first"), jsonMap.get("last"));
}
But this also effectively adds @JsonIgnoreProperties(ignoreUnknown=true) to the Parent class, as all properties map to here. So if you wanted unknown JSON fields to throw an exception, you'd have to do that yourself. In addition, if the map values could be something other than Strings, you'd have to do some manual type checking and conversion. Is there a way for Jackson to handle this case automatically?
Edit:
I might be crazy, but this actually appears to work despite never being explicitly mentioned in the documentation: http://fasterxml.github.io/jackson-annotations/javadoc/2.2.0/com/fasterxml/jackson/annotation/JsonUnwrapped.html
I was pretty sure it didn't work for me previously. Still, the proposed @JsonCreator approach might be preferred when custom logic is required to deserialize unwrapped polymorphic types.
java json jackson deserialization
asked May 15 '13 at 16:13
ShaunShaun
74051537
add a comment |
Say I have the following class:
public class Parent {
public int age;
@JsonUnwrapped
public Name name;
}
Producing JSON:
{
"age" : 18,
"first" : "Joey",
"last" : "Sixpack"
}
How do I deserialize this back into the Parent class? I could use @JsonCreator
@JsonCreator
public Parent(Map<String,String> jsonMap) {
age = jsonMap.get("age");
name = new Name(jsonMap.get("first"), jsonMap.get("last"));
}
But this also effectively adds @JsonIgnoreProperties(ignoreUnknown=true) to the Parent class, as all properties map to here. So if you wanted unknown JSON fields to throw an exception, you'd have to do that yourself. In addition, if the map values could be something other than Strings, you'd have to do some manual type checking and conversion. Is there a way for Jackson to handle this case automatically?
Edit:
I might be crazy, but this actually appears to work despite never being explicitly mentioned in the documentation: http://fasterxml.github.io/jackson-annotations/javadoc/2.2.0/com/fasterxml/jackson/annotation/JsonUnwrapped.html
I was pretty sure it didn't work for me previously. Still, the proposed @JsonCreator approach might be preferred when custom logic is required to deserialize unwrapped polymorphic types.
java json jackson deserialization
asked May 15 '13 at 16:13
ShaunShaun
74051537
4
Are you certain@JsonUnwrappedis working for deserialization? I just tried it and am gettingCould not read JSON: Unrecognized field...errors when I try to deserialize the flattened JSON.
– E-Riz
Apr 23 '14 at 18:47
add a comment |
Say I have the following class:
public class Parent {
public int age;
@JsonUnwrapped
public Name name;
}
Producing JSON:
{
"age" : 18,
"first" : "Joey",
"last" : "Sixpack"
}
How do I deserialize this back into the Parent class? I could use @JsonCreator
@JsonCreator
public Parent(Map<String,String> jsonMap) {
age = jsonMap.get("age");
name = new Name(jsonMap.get("first"), jsonMap.get("last"));
}
But this also effectively adds @JsonIgnoreProperties(ignoreUnknown=true) to the Parent class, as all properties map to here. So if you wanted unknown JSON fields to throw an exception, you'd have to do that yourself. In addition, if the map values could be something other than Strings, you'd have to do some manual type checking and conversion. Is there a way for Jackson to handle this case automatically?
Edit:
I might be crazy, but this actually appears to work despite never being explicitly mentioned in the documentation: http://fasterxml.github.io/jackson-annotations/javadoc/2.2.0/com/fasterxml/jackson/annotation/JsonUnwrapped.html
I was pretty sure it didn't work for me previously. Still, the proposed @JsonCreator approach might be preferred when custom logic is required to deserialize unwrapped polymorphic types.
java json jackson deserialization
asked May 15 '13 at 16:13
ShaunShaun
74051537
Say I have the following class:
public class Parent {
public int age;
@JsonUnwrapped
public Name name;
}
Producing JSON:
{
"age" : 18,
"first" : "Joey",
"last" : "Sixpack"
}
How do I deserialize this back into the Parent class? I could use @JsonCreator
@JsonCreator
public Parent(Map<String,String> jsonMap) {
age = jsonMap.get("age");
name = new Name(jsonMap.get("first"), jsonMap.get("last"));
}
But this also effectively adds @JsonIgnoreProperties(ignoreUnknown=true) to the Parent class, as all properties map to here. So if you wanted unknown JSON fields to throw an exception, you'd have to do that yourself. In addition, if the map values could be something other than Strings, you'd have to do some manual type checking and conversion. Is there a way for Jackson to handle this case automatically?
Edit:
I might be crazy, but this actually appears to work despite never being explicitly mentioned in the documentation: http://fasterxml.github.io/jackson-annotations/javadoc/2.2.0/com/fasterxml/jackson/annotation/JsonUnwrapped.html
I was pretty sure it didn't work for me previously. Still, the proposed @JsonCreator approach might be preferred when custom logic is required to deserialize unwrapped polymorphic types.
java json jackson deserialization
java json jackson deserialization
asked May 15 '13 at 16:13
ShaunShaun
74051537
asked May 15 '13 at 16:13
ShaunShaun
74051537
edited May 15 '13 at 16:35
Shaun
asked May 15 '13 at 16:13
ShaunShaun
74051537
asked May 15 '13 at 16:13
ShaunShaun
74051537
asked May 15 '13 at 16:13
ShaunShaun
74051537
74051537
4
Are you certain@JsonUnwrappedis working for deserialization? I just tried it and am gettingCould not read JSON: Unrecognized field...errors when I try to deserialize the flattened JSON.
– E-Riz
Apr 23 '14 at 18:47
add a comment |
4
Are you certain@JsonUnwrappedis working for deserialization? I just tried it and am gettingCould not read JSON: Unrecognized field...errors when I try to deserialize the flattened JSON.
– E-Riz
Apr 23 '14 at 18:47
4
4
Are you certain
@JsonUnwrapped is working for deserialization? I just tried it and am getting Could not read JSON: Unrecognized field... errors when I try to deserialize the flattened JSON.– E-Riz
Apr 23 '14 at 18:47
Are you certain
@JsonUnwrapped is working for deserialization? I just tried it and am getting Could not read JSON: Unrecognized field... errors when I try to deserialize the flattened JSON.– E-Riz
Apr 23 '14 at 18:47
add a comment |
3 Answers
3
active
oldest
votes
You can use @JsonCreator with @JsonProperty for each field:
@JsonCreator
public Parent(@JsonProperty("age") Integer age, @JsonProperty("firstName") String firstName,
@JsonProperty("lastName") String lastName) {
this.age = age;
this.name = new Name(firstName, lastName);
}
Jackson does type checking and unknown field checking for you in this case.
answered May 15 '13 at 16:27
hoazhoaz
7,55543349
2
Good point. This does answer the question of how to preserve jackson type/field checking in JsonCreators in general, which might be the only useful question remaining. =)
– Shaun
May 15 '13 at 16:37
add a comment |
It does work for deserialization as well, although it's not mentioned in the docs explicitly, like you said. See the unit test for deserialization of @JsonUnwrapped here for confirmation - https://github.com/FasterXML/jackson-databind/blob/d2c083a6220f2875c97c29f4823d9818972511dc/src/test/java/com/fasterxml/jackson/databind/struct/TestUnwrapped.java#L138
answered Aug 27 '18 at 18:56
spinlokspinlok
2,493817
add a comment |
For those who googled here like me, trying to resolve issue when deserializing unwrapepd Map, there is a solution with @JsonAnySetter:
public class CountryList
{
Map<String, Country> countries = new HashMap<>();
@JsonAnySetter
public void setCountry(String key, Country value)
{
countries.put(key, value);
}
}
answered Jun 21 '18 at 18:48
Jan MaresJan Mares
494514
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can use @JsonCreator with @JsonProperty for each field:
@JsonCreator
public Parent(@JsonProperty("age") Integer age, @JsonProperty("firstName") String firstName,
@JsonProperty("lastName") String lastName) {
this.age = age;
this.name = new Name(firstName, lastName);
}
Jackson does type checking and unknown field checking for you in this case.
answered May 15 '13 at 16:27
hoazhoaz
7,55543349
2
Good point. This does answer the question of how to preserve jackson type/field checking in JsonCreators in general, which might be the only useful question remaining. =)
– Shaun
May 15 '13 at 16:37
add a comment |
You can use @JsonCreator with @JsonProperty for each field:
@JsonCreator
public Parent(@JsonProperty("age") Integer age, @JsonProperty("firstName") String firstName,
@JsonProperty("lastName") String lastName) {
this.age = age;
this.name = new Name(firstName, lastName);
}
Jackson does type checking and unknown field checking for you in this case.
answered May 15 '13 at 16:27
hoazhoaz
7,55543349
2
Good point. This does answer the question of how to preserve jackson type/field checking in JsonCreators in general, which might be the only useful question remaining. =)
– Shaun
May 15 '13 at 16:37
add a comment |
You can use @JsonCreator with @JsonProperty for each field:
@JsonCreator
public Parent(@JsonProperty("age") Integer age, @JsonProperty("firstName") String firstName,
@JsonProperty("lastName") String lastName) {
this.age = age;
this.name = new Name(firstName, lastName);
}
Jackson does type checking and unknown field checking for you in this case.
answered May 15 '13 at 16:27
hoazhoaz
7,55543349
You can use @JsonCreator with @JsonProperty for each field:
@JsonCreator
public Parent(@JsonProperty("age") Integer age, @JsonProperty("firstName") String firstName,
@JsonProperty("lastName") String lastName) {
this.age = age;
this.name = new Name(firstName, lastName);
}
Jackson does type checking and unknown field checking for you in this case.
answered May 15 '13 at 16:27
hoazhoaz
7,55543349
answered May 15 '13 at 16:27
hoazhoaz
7,55543349
answered May 15 '13 at 16:27
hoazhoaz
7,55543349
answered May 15 '13 at 16:27
hoazhoaz
7,55543349
7,55543349
2
Good point. This does answer the question of how to preserve jackson type/field checking in JsonCreators in general, which might be the only useful question remaining. =)
– Shaun
May 15 '13 at 16:37
add a comment |
2
Good point. This does answer the question of how to preserve jackson type/field checking in JsonCreators in general, which might be the only useful question remaining. =)
– Shaun
May 15 '13 at 16:37
2
2
Good point. This does answer the question of how to preserve jackson type/field checking in JsonCreators in general, which might be the only useful question remaining. =)
– Shaun
May 15 '13 at 16:37
Good point. This does answer the question of how to preserve jackson type/field checking in JsonCreators in general, which might be the only useful question remaining. =)
– Shaun
May 15 '13 at 16:37
add a comment |
It does work for deserialization as well, although it's not mentioned in the docs explicitly, like you said. See the unit test for deserialization of @JsonUnwrapped here for confirmation - https://github.com/FasterXML/jackson-databind/blob/d2c083a6220f2875c97c29f4823d9818972511dc/src/test/java/com/fasterxml/jackson/databind/struct/TestUnwrapped.java#L138
answered Aug 27 '18 at 18:56
spinlokspinlok
2,493817
add a comment |
It does work for deserialization as well, although it's not mentioned in the docs explicitly, like you said. See the unit test for deserialization of @JsonUnwrapped here for confirmation - https://github.com/FasterXML/jackson-databind/blob/d2c083a6220f2875c97c29f4823d9818972511dc/src/test/java/com/fasterxml/jackson/databind/struct/TestUnwrapped.java#L138
answered Aug 27 '18 at 18:56
spinlokspinlok
2,493817
add a comment |
It does work for deserialization as well, although it's not mentioned in the docs explicitly, like you said. See the unit test for deserialization of @JsonUnwrapped here for confirmation - https://github.com/FasterXML/jackson-databind/blob/d2c083a6220f2875c97c29f4823d9818972511dc/src/test/java/com/fasterxml/jackson/databind/struct/TestUnwrapped.java#L138
answered Aug 27 '18 at 18:56
spinlokspinlok
2,493817
It does work for deserialization as well, although it's not mentioned in the docs explicitly, like you said. See the unit test for deserialization of @JsonUnwrapped here for confirmation - https://github.com/FasterXML/jackson-databind/blob/d2c083a6220f2875c97c29f4823d9818972511dc/src/test/java/com/fasterxml/jackson/databind/struct/TestUnwrapped.java#L138
answered Aug 27 '18 at 18:56
spinlokspinlok
2,493817
answered Aug 27 '18 at 18:56
spinlokspinlok
2,493817
answered Aug 27 '18 at 18:56
spinlokspinlok
2,493817
answered Aug 27 '18 at 18:56
spinlokspinlok
2,493817
2,493817
add a comment |
add a comment |
For those who googled here like me, trying to resolve issue when deserializing unwrapepd Map, there is a solution with @JsonAnySetter:
public class CountryList
{
Map<String, Country> countries = new HashMap<>();
@JsonAnySetter
public void setCountry(String key, Country value)
{
countries.put(key, value);
}
}
answered Jun 21 '18 at 18:48
Jan MaresJan Mares
494514
add a comment |
For those who googled here like me, trying to resolve issue when deserializing unwrapepd Map, there is a solution with @JsonAnySetter:
public class CountryList
{
Map<String, Country> countries = new HashMap<>();
@JsonAnySetter
public void setCountry(String key, Country value)
{
countries.put(key, value);
}
}
answered Jun 21 '18 at 18:48
Jan MaresJan Mares
494514
add a comment |
For those who googled here like me, trying to resolve issue when deserializing unwrapepd Map, there is a solution with @JsonAnySetter:
public class CountryList
{
Map<String, Country> countries = new HashMap<>();
@JsonAnySetter
public void setCountry(String key, Country value)
{
countries.put(key, value);
}
}
answered Jun 21 '18 at 18:48
Jan MaresJan Mares
494514
For those who googled here like me, trying to resolve issue when deserializing unwrapepd Map, there is a solution with @JsonAnySetter:
public class CountryList
{
Map<String, Country> countries = new HashMap<>();
@JsonAnySetter
public void setCountry(String key, Country value)
{
countries.put(key, value);
}
}
answered Jun 21 '18 at 18:48
Jan MaresJan Mares
494514
answered Jun 21 '18 at 18:48
Jan MaresJan Mares
494514
answered Jun 21 '18 at 18:48
Jan MaresJan Mares
494514
answered Jun 21 '18 at 18:48
Jan MaresJan Mares
494514
494514
add a comment |
add a comment |
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4
Are you certain
@JsonUnwrappedis working for deserialization? I just tried it and am gettingCould not read JSON: Unrecognized field...errors when I try to deserialize the flattened JSON.– E-Riz
Apr 23 '14 at 18:47