Get keys from nested JSONObject fluently
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box;
}
0
I'm trying to extract a value from a nested JSONObject, say "id". I'm using org.json.simple package and my code looks like:
JSONArray entries = (JSONArray) response.get("entries");
JSONObject entry = (JSONObject) entries.get(0);
JSONArray runs = (JSONArray) entry.get("runs");
JSONObject run = (JSONObject) runs.get(0);
String run_id = run.get("id").toString();
where response is a JSONObject.
Is it possible to refactor the code using Fluent Interface Pattern so the code is more readable? For example,
String run_id = response.get("entries")
.get(0)
.get("runs")
.get(0)
.get("id").toString();
Thanks in advance.
java fluent-interface
asked Jan 4 at 10:30
Eric HungEric Hung
16516
add a comment |
0
I'm trying to extract a value from a nested JSONObject, say "id". I'm using org.json.simple package and my code looks like:
JSONArray entries = (JSONArray) response.get("entries");
JSONObject entry = (JSONObject) entries.get(0);
JSONArray runs = (JSONArray) entry.get("runs");
JSONObject run = (JSONObject) runs.get(0);
String run_id = run.get("id").toString();
where response is a JSONObject.
Is it possible to refactor the code using Fluent Interface Pattern so the code is more readable? For example,
String run_id = response.get("entries")
.get(0)
.get("runs")
.get(0)
.get("id").toString();
Thanks in advance.
java fluent-interface
asked Jan 4 at 10:30
Eric HungEric Hung
16516
add a comment |
0
0
0
I'm trying to extract a value from a nested JSONObject, say "id". I'm using org.json.simple package and my code looks like:
JSONArray entries = (JSONArray) response.get("entries");
JSONObject entry = (JSONObject) entries.get(0);
JSONArray runs = (JSONArray) entry.get("runs");
JSONObject run = (JSONObject) runs.get(0);
String run_id = run.get("id").toString();
where response is a JSONObject.
Is it possible to refactor the code using Fluent Interface Pattern so the code is more readable? For example,
String run_id = response.get("entries")
.get(0)
.get("runs")
.get(0)
.get("id").toString();
Thanks in advance.
java fluent-interface
asked Jan 4 at 10:30
Eric HungEric Hung
16516
I'm trying to extract a value from a nested JSONObject, say "id". I'm using org.json.simple package and my code looks like:
JSONArray entries = (JSONArray) response.get("entries");
JSONObject entry = (JSONObject) entries.get(0);
JSONArray runs = (JSONArray) entry.get("runs");
JSONObject run = (JSONObject) runs.get(0);
String run_id = run.get("id").toString();
where response is a JSONObject.
Is it possible to refactor the code using Fluent Interface Pattern so the code is more readable? For example,
String run_id = response.get("entries")
.get(0)
.get("runs")
.get(0)
.get("id").toString();
Thanks in advance.
java fluent-interface
java fluent-interface
asked Jan 4 at 10:30
Eric HungEric Hung
16516
asked Jan 4 at 10:30
Eric HungEric Hung
16516
asked Jan 4 at 10:30
Eric HungEric Hung
16516
asked Jan 4 at 10:30
Eric HungEric Hung
16516
asked Jan 4 at 10:30
Eric HungEric Hung
16516
16516
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
3
Here's a possibility.
class FluentJson {
private Object value;
public FluentJson(Object value) {
this.value = value;
}
public FluentJson get(int index) throws JSONException {
JSONArray a = (JSONArray) value;
return new FluentJson(a.get(index));
}
public FluentJson get(String key) throws JSONException {
JSONObject o = (JSONObject) value;
return new FluentJson(o.get(key));
}
public String toString() {
return value == null ? null : value.toString();
}
public Number toNumber() {
return (Number) value;
}
}
You can use it like this
String run_id = new FluentJson(response)
.get("entries")
.get(0)
.get("runs")
.get(0)
.get("id").toString();
answered Jan 4 at 11:37
Leo AsoLeo Aso
5,57711229
How about returningthisby addingvalue = aandvalue = oafter a and o are assigned respectively?
– Eric Hung
Jan 7 at 4:04
1
That also works.
– Leo Aso
Jan 7 at 9:13
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
});
}
});
draft saved
draft discarded
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%2f54037150%2fget-keys-from-nested-jsonobject-fluently%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
3
Here's a possibility.
class FluentJson {
private Object value;
public FluentJson(Object value) {
this.value = value;
}
public FluentJson get(int index) throws JSONException {
JSONArray a = (JSONArray) value;
return new FluentJson(a.get(index));
}
public FluentJson get(String key) throws JSONException {
JSONObject o = (JSONObject) value;
return new FluentJson(o.get(key));
}
public String toString() {
return value == null ? null : value.toString();
}
public Number toNumber() {
return (Number) value;
}
}
You can use it like this
String run_id = new FluentJson(response)
.get("entries")
.get(0)
.get("runs")
.get(0)
.get("id").toString();
answered Jan 4 at 11:37
Leo AsoLeo Aso
5,57711229
How about returningthisby addingvalue = aandvalue = oafter a and o are assigned respectively?
– Eric Hung
Jan 7 at 4:04
1
That also works.
– Leo Aso
Jan 7 at 9:13
add a comment |
3
Here's a possibility.
class FluentJson {
private Object value;
public FluentJson(Object value) {
this.value = value;
}
public FluentJson get(int index) throws JSONException {
JSONArray a = (JSONArray) value;
return new FluentJson(a.get(index));
}
public FluentJson get(String key) throws JSONException {
JSONObject o = (JSONObject) value;
return new FluentJson(o.get(key));
}
public String toString() {
return value == null ? null : value.toString();
}
public Number toNumber() {
return (Number) value;
}
}
You can use it like this
String run_id = new FluentJson(response)
.get("entries")
.get(0)
.get("runs")
.get(0)
.get("id").toString();
answered Jan 4 at 11:37
Leo AsoLeo Aso
5,57711229
How about returningthisby addingvalue = aandvalue = oafter a and o are assigned respectively?
– Eric Hung
Jan 7 at 4:04
1
That also works.
– Leo Aso
Jan 7 at 9:13
add a comment |
3
3
3
Here's a possibility.
class FluentJson {
private Object value;
public FluentJson(Object value) {
this.value = value;
}
public FluentJson get(int index) throws JSONException {
JSONArray a = (JSONArray) value;
return new FluentJson(a.get(index));
}
public FluentJson get(String key) throws JSONException {
JSONObject o = (JSONObject) value;
return new FluentJson(o.get(key));
}
public String toString() {
return value == null ? null : value.toString();
}
public Number toNumber() {
return (Number) value;
}
}
You can use it like this
String run_id = new FluentJson(response)
.get("entries")
.get(0)
.get("runs")
.get(0)
.get("id").toString();
answered Jan 4 at 11:37
Leo AsoLeo Aso
5,57711229
Here's a possibility.
class FluentJson {
private Object value;
public FluentJson(Object value) {
this.value = value;
}
public FluentJson get(int index) throws JSONException {
JSONArray a = (JSONArray) value;
return new FluentJson(a.get(index));
}
public FluentJson get(String key) throws JSONException {
JSONObject o = (JSONObject) value;
return new FluentJson(o.get(key));
}
public String toString() {
return value == null ? null : value.toString();
}
public Number toNumber() {
return (Number) value;
}
}
You can use it like this
String run_id = new FluentJson(response)
.get("entries")
.get(0)
.get("runs")
.get(0)
.get("id").toString();
answered Jan 4 at 11:37
Leo AsoLeo Aso
5,57711229
answered Jan 4 at 11:37
Leo AsoLeo Aso
5,57711229
answered Jan 4 at 11:37
Leo AsoLeo Aso
5,57711229
answered Jan 4 at 11:37
Leo AsoLeo Aso
5,57711229
5,57711229
How about returningthisby addingvalue = aandvalue = oafter a and o are assigned respectively?
– Eric Hung
Jan 7 at 4:04
1
That also works.
– Leo Aso
Jan 7 at 9:13
add a comment |
How about returningthisby addingvalue = aandvalue = oafter a and o are assigned respectively?
– Eric Hung
Jan 7 at 4:04
1
That also works.
– Leo Aso
Jan 7 at 9:13
How about returning
this by adding value = a and value = o after a and o are assigned respectively?– Eric Hung
Jan 7 at 4:04
How about returning
this by adding value = a and value = o after a and o are assigned respectively?– Eric Hung
Jan 7 at 4:04
1
1
That also works.
– Leo Aso
Jan 7 at 9:13
That also works.
– Leo Aso
Jan 7 at 9:13
add a comment |
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f54037150%2fget-keys-from-nested-jsonobject-fluently%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
