Deserialize a varying number of objects into a list in Java using Jackson
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box;
}
My Spring Boot app makes a call to a REST API and receives a JSON with a varying number of entities. E.g.
{
"content": {
"guest_1": {
"name": {
"firstName": "a",
"lastName": "b"
},
"vip": false
},
"guest_2": {
"name": {
"firstName": "c",
"lastName": "d"
},
"vip": false
},
...more guests omitted...
}
}
There can be 1 to many guests and I don't know their number upfront. As you can see, they aren't in an array, they are objects instead.
I'd like to avoid deserializing into a class like
public class Content {
@JsonProperty("guest_1")
private Guest guest1;
@JsonProperty("guest_2")
private Guest guest2;
// More Guests here each having their own field
}
What I'd like to use is
public class Content {
private List<Guest> guests;
}
The @JsonAnySetter annotation I read about at https://www.baeldung.com/jackson-annotations looks promising but I couldn't get it to work.
3.2. Convert to an object at https://www.baeldung.com/jackson-json-node-tree-model looks also good but it didn't work out either.
I'm not sure if I can make Jackson do this in a declarative way or I should write a custom JsonDeserializer. Could you please help me?
java jackson deserialization json-deserialization
asked Nov 26 '18 at 14:34
Andras SzokeAndras Szoke
375
add a comment |
My Spring Boot app makes a call to a REST API and receives a JSON with a varying number of entities. E.g.
{
"content": {
"guest_1": {
"name": {
"firstName": "a",
"lastName": "b"
},
"vip": false
},
"guest_2": {
"name": {
"firstName": "c",
"lastName": "d"
},
"vip": false
},
...more guests omitted...
}
}
There can be 1 to many guests and I don't know their number upfront. As you can see, they aren't in an array, they are objects instead.
I'd like to avoid deserializing into a class like
public class Content {
@JsonProperty("guest_1")
private Guest guest1;
@JsonProperty("guest_2")
private Guest guest2;
// More Guests here each having their own field
}
What I'd like to use is
public class Content {
private List<Guest> guests;
}
The @JsonAnySetter annotation I read about at https://www.baeldung.com/jackson-annotations looks promising but I couldn't get it to work.
3.2. Convert to an object at https://www.baeldung.com/jackson-json-node-tree-model looks also good but it didn't work out either.
I'm not sure if I can make Jackson do this in a declarative way or I should write a custom JsonDeserializer. Could you please help me?
java jackson deserialization json-deserialization
asked Nov 26 '18 at 14:34
Andras SzokeAndras Szoke
375
add a comment |
My Spring Boot app makes a call to a REST API and receives a JSON with a varying number of entities. E.g.
{
"content": {
"guest_1": {
"name": {
"firstName": "a",
"lastName": "b"
},
"vip": false
},
"guest_2": {
"name": {
"firstName": "c",
"lastName": "d"
},
"vip": false
},
...more guests omitted...
}
}
There can be 1 to many guests and I don't know their number upfront. As you can see, they aren't in an array, they are objects instead.
I'd like to avoid deserializing into a class like
public class Content {
@JsonProperty("guest_1")
private Guest guest1;
@JsonProperty("guest_2")
private Guest guest2;
// More Guests here each having their own field
}
What I'd like to use is
public class Content {
private List<Guest> guests;
}
The @JsonAnySetter annotation I read about at https://www.baeldung.com/jackson-annotations looks promising but I couldn't get it to work.
3.2. Convert to an object at https://www.baeldung.com/jackson-json-node-tree-model looks also good but it didn't work out either.
I'm not sure if I can make Jackson do this in a declarative way or I should write a custom JsonDeserializer. Could you please help me?
java jackson deserialization json-deserialization
asked Nov 26 '18 at 14:34
Andras SzokeAndras Szoke
375
My Spring Boot app makes a call to a REST API and receives a JSON with a varying number of entities. E.g.
{
"content": {
"guest_1": {
"name": {
"firstName": "a",
"lastName": "b"
},
"vip": false
},
"guest_2": {
"name": {
"firstName": "c",
"lastName": "d"
},
"vip": false
},
...more guests omitted...
}
}
There can be 1 to many guests and I don't know their number upfront. As you can see, they aren't in an array, they are objects instead.
I'd like to avoid deserializing into a class like
public class Content {
@JsonProperty("guest_1")
private Guest guest1;
@JsonProperty("guest_2")
private Guest guest2;
// More Guests here each having their own field
}
What I'd like to use is
public class Content {
private List<Guest> guests;
}
The @JsonAnySetter annotation I read about at https://www.baeldung.com/jackson-annotations looks promising but I couldn't get it to work.
3.2. Convert to an object at https://www.baeldung.com/jackson-json-node-tree-model looks also good but it didn't work out either.
I'm not sure if I can make Jackson do this in a declarative way or I should write a custom JsonDeserializer. Could you please help me?
java jackson deserialization json-deserialization
java jackson deserialization json-deserialization
asked Nov 26 '18 at 14:34
Andras SzokeAndras Szoke
375
asked Nov 26 '18 at 14:34
Andras SzokeAndras Szoke
375
edited Nov 26 '18 at 16:03
Andras Szoke
asked Nov 26 '18 at 14:34
Andras SzokeAndras Szoke
375
asked Nov 26 '18 at 14:34
Andras SzokeAndras Szoke
375
asked Nov 26 '18 at 14:34
Andras SzokeAndras Szoke
375
375
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
@JsonAnySetter will work as it allows to specify a POJO type as second parameter. You could recreate the example JSON as, omitting setXXX() and getXXX() methods on POJOs for clarity:
private static class Content {
private Guests content;
}
private static class Guests {
private List<Guest> guests = new ArrayList<>();
@JsonAnySetter
private void addGuest(String name, Guest value) {
guests.add(value);
}
}
private static class Guest {
private Name name;
private boolean vip;
}
private static class Name {
private String firstName;
private String lastName;
}
With your JSON example will produce:
Content root = new ObjectMapper().readValue(json, Content.class);
root.getContent().getGuests().stream()
.map(Guest::getName)
.map(Name::getFirstName)
.forEach(System.out::println); // a, c
answered Nov 26 '18 at 14:58
Karol DowbeckiKarol Dowbecki
26.4k93759
Wow, this works like a charm! Looks like I was close to the solution but missing something. Instead of a List, I ended up putting the values into a Map where the name parameter is the key. Thanks for your quick response!
– Andras Szoke
Nov 26 '18 at 16:05
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53483360%2fdeserialize-a-varying-number-of-objects-into-a-list-in-java-using-jackson%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
@JsonAnySetter will work as it allows to specify a POJO type as second parameter. You could recreate the example JSON as, omitting setXXX() and getXXX() methods on POJOs for clarity:
private static class Content {
private Guests content;
}
private static class Guests {
private List<Guest> guests = new ArrayList<>();
@JsonAnySetter
private void addGuest(String name, Guest value) {
guests.add(value);
}
}
private static class Guest {
private Name name;
private boolean vip;
}
private static class Name {
private String firstName;
private String lastName;
}
With your JSON example will produce:
Content root = new ObjectMapper().readValue(json, Content.class);
root.getContent().getGuests().stream()
.map(Guest::getName)
.map(Name::getFirstName)
.forEach(System.out::println); // a, c
answered Nov 26 '18 at 14:58
Karol DowbeckiKarol Dowbecki
26.4k93759
Wow, this works like a charm! Looks like I was close to the solution but missing something. Instead of a List, I ended up putting the values into a Map where the name parameter is the key. Thanks for your quick response!
– Andras Szoke
Nov 26 '18 at 16:05
add a comment |
@JsonAnySetter will work as it allows to specify a POJO type as second parameter. You could recreate the example JSON as, omitting setXXX() and getXXX() methods on POJOs for clarity:
private static class Content {
private Guests content;
}
private static class Guests {
private List<Guest> guests = new ArrayList<>();
@JsonAnySetter
private void addGuest(String name, Guest value) {
guests.add(value);
}
}
private static class Guest {
private Name name;
private boolean vip;
}
private static class Name {
private String firstName;
private String lastName;
}
With your JSON example will produce:
Content root = new ObjectMapper().readValue(json, Content.class);
root.getContent().getGuests().stream()
.map(Guest::getName)
.map(Name::getFirstName)
.forEach(System.out::println); // a, c
answered Nov 26 '18 at 14:58
Karol DowbeckiKarol Dowbecki
26.4k93759
Wow, this works like a charm! Looks like I was close to the solution but missing something. Instead of a List, I ended up putting the values into a Map where the name parameter is the key. Thanks for your quick response!
– Andras Szoke
Nov 26 '18 at 16:05
add a comment |
@JsonAnySetter will work as it allows to specify a POJO type as second parameter. You could recreate the example JSON as, omitting setXXX() and getXXX() methods on POJOs for clarity:
private static class Content {
private Guests content;
}
private static class Guests {
private List<Guest> guests = new ArrayList<>();
@JsonAnySetter
private void addGuest(String name, Guest value) {
guests.add(value);
}
}
private static class Guest {
private Name name;
private boolean vip;
}
private static class Name {
private String firstName;
private String lastName;
}
With your JSON example will produce:
Content root = new ObjectMapper().readValue(json, Content.class);
root.getContent().getGuests().stream()
.map(Guest::getName)
.map(Name::getFirstName)
.forEach(System.out::println); // a, c
answered Nov 26 '18 at 14:58
Karol DowbeckiKarol Dowbecki
26.4k93759
@JsonAnySetter will work as it allows to specify a POJO type as second parameter. You could recreate the example JSON as, omitting setXXX() and getXXX() methods on POJOs for clarity:
private static class Content {
private Guests content;
}
private static class Guests {
private List<Guest> guests = new ArrayList<>();
@JsonAnySetter
private void addGuest(String name, Guest value) {
guests.add(value);
}
}
private static class Guest {
private Name name;
private boolean vip;
}
private static class Name {
private String firstName;
private String lastName;
}
With your JSON example will produce:
Content root = new ObjectMapper().readValue(json, Content.class);
root.getContent().getGuests().stream()
.map(Guest::getName)
.map(Name::getFirstName)
.forEach(System.out::println); // a, c
answered Nov 26 '18 at 14:58
Karol DowbeckiKarol Dowbecki
26.4k93759
edited Nov 26 '18 at 15:03
answered Nov 26 '18 at 14:58
Karol DowbeckiKarol Dowbecki
26.4k93759
answered Nov 26 '18 at 14:58
Karol DowbeckiKarol Dowbecki
26.4k93759
answered Nov 26 '18 at 14:58
Karol DowbeckiKarol Dowbecki
26.4k93759
26.4k93759
Wow, this works like a charm! Looks like I was close to the solution but missing something. Instead of a List, I ended up putting the values into a Map where the name parameter is the key. Thanks for your quick response!
– Andras Szoke
Nov 26 '18 at 16:05
add a comment |
Wow, this works like a charm! Looks like I was close to the solution but missing something. Instead of a List, I ended up putting the values into a Map where the name parameter is the key. Thanks for your quick response!
– Andras Szoke
Nov 26 '18 at 16:05
Wow, this works like a charm! Looks like I was close to the solution but missing something. Instead of a List, I ended up putting the values into a Map where the name parameter is the key. Thanks for your quick response!
– Andras Szoke
Nov 26 '18 at 16:05
Wow, this works like a charm! Looks like I was close to the solution but missing something. Instead of a List, I ended up putting the values into a Map where the name parameter is the key. Thanks for your quick response!
– Andras Szoke
Nov 26 '18 at 16:05
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53483360%2fdeserialize-a-varying-number-of-objects-into-a-list-in-java-using-jackson%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
