RxJava: How to modify only first item before sending it to observer
I have a Rx stream which sends multiple items to the observer.
But I would like to modify the very first item that is sent back. All the other items can be sent, as is.
I know map() would intercept all the items, but I would then have to keep a counter of which item is currently being emitted.
Is there a way to do this only for the very first item?
Thanks.
rx-java rx-java2
asked Nov 28 '18 at 6:29
SunnySunny
2,730144579
add a comment |
I have a Rx stream which sends multiple items to the observer.
But I would like to modify the very first item that is sent back. All the other items can be sent, as is.
I know map() would intercept all the items, but I would then have to keep a counter of which item is currently being emitted.
Is there a way to do this only for the very first item?
Thanks.
rx-java rx-java2
asked Nov 28 '18 at 6:29
SunnySunny
2,730144579
add a comment |
I have a Rx stream which sends multiple items to the observer.
But I would like to modify the very first item that is sent back. All the other items can be sent, as is.
I know map() would intercept all the items, but I would then have to keep a counter of which item is currently being emitted.
Is there a way to do this only for the very first item?
Thanks.
rx-java rx-java2
asked Nov 28 '18 at 6:29
SunnySunny
2,730144579
I have a Rx stream which sends multiple items to the observer.
But I would like to modify the very first item that is sent back. All the other items can be sent, as is.
I know map() would intercept all the items, but I would then have to keep a counter of which item is currently being emitted.
Is there a way to do this only for the very first item?
Thanks.
rx-java rx-java2
rx-java rx-java2
asked Nov 28 '18 at 6:29
SunnySunny
2,730144579
asked Nov 28 '18 at 6:29
SunnySunny
2,730144579
asked Nov 28 '18 at 6:29
SunnySunny
2,730144579
asked Nov 28 '18 at 6:29
SunnySunny
2,730144579
asked Nov 28 '18 at 6:29
SunnySunny
2,730144579
2,730144579
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Try this
Say you have a stream of strings ["First", "Second", "Third", "Fourth"] and you need to modify the very first item only
Observable<String> stringObservable = Observable.just("First", "Second", "Third", "Fourth").publish().refCount();
stringObservable.skip(1)
.startWith(stringObservable.take(1).map(s -> "Modified"))
.subscribe(new Consumer<String>() {
@Override
public void accept(String s) throws Exception {
System.out.println(s);
}
});
the result will be
Modified
Second
Third
Fourth
answered Nov 28 '18 at 7:51
Sarath KnSarath Kn
1,663919
1
It's important to note that introducing two subscriptions may introduce side effects. You can usepublish()to subscribe only once, effectively.stringObservable.publish( shared -> shared.skip(1).startsWith(shared.take(1).map(...))...
– Bob Dalgleish
Nov 28 '18 at 14:44
Yes. noted it. Since publish() will multicast and turn the source into aConnectableobservable, addedpublish().refCount()combination to turn it into a multicasted Observable, which is shared for every subscriber added. Updated the answer
– Sarath Kn
Nov 28 '18 at 15:01
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%2f53513410%2frxjava-how-to-modify-only-first-item-before-sending-it-to-observer%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
Try this
Say you have a stream of strings ["First", "Second", "Third", "Fourth"] and you need to modify the very first item only
Observable<String> stringObservable = Observable.just("First", "Second", "Third", "Fourth").publish().refCount();
stringObservable.skip(1)
.startWith(stringObservable.take(1).map(s -> "Modified"))
.subscribe(new Consumer<String>() {
@Override
public void accept(String s) throws Exception {
System.out.println(s);
}
});
the result will be
Modified
Second
Third
Fourth
answered Nov 28 '18 at 7:51
Sarath KnSarath Kn
1,663919
1
It's important to note that introducing two subscriptions may introduce side effects. You can usepublish()to subscribe only once, effectively.stringObservable.publish( shared -> shared.skip(1).startsWith(shared.take(1).map(...))...
– Bob Dalgleish
Nov 28 '18 at 14:44
Yes. noted it. Since publish() will multicast and turn the source into aConnectableobservable, addedpublish().refCount()combination to turn it into a multicasted Observable, which is shared for every subscriber added. Updated the answer
– Sarath Kn
Nov 28 '18 at 15:01
add a comment |
Try this
Say you have a stream of strings ["First", "Second", "Third", "Fourth"] and you need to modify the very first item only
Observable<String> stringObservable = Observable.just("First", "Second", "Third", "Fourth").publish().refCount();
stringObservable.skip(1)
.startWith(stringObservable.take(1).map(s -> "Modified"))
.subscribe(new Consumer<String>() {
@Override
public void accept(String s) throws Exception {
System.out.println(s);
}
});
the result will be
Modified
Second
Third
Fourth
answered Nov 28 '18 at 7:51
Sarath KnSarath Kn
1,663919
1
It's important to note that introducing two subscriptions may introduce side effects. You can usepublish()to subscribe only once, effectively.stringObservable.publish( shared -> shared.skip(1).startsWith(shared.take(1).map(...))...
– Bob Dalgleish
Nov 28 '18 at 14:44
Yes. noted it. Since publish() will multicast and turn the source into aConnectableobservable, addedpublish().refCount()combination to turn it into a multicasted Observable, which is shared for every subscriber added. Updated the answer
– Sarath Kn
Nov 28 '18 at 15:01
add a comment |
Try this
Say you have a stream of strings ["First", "Second", "Third", "Fourth"] and you need to modify the very first item only
Observable<String> stringObservable = Observable.just("First", "Second", "Third", "Fourth").publish().refCount();
stringObservable.skip(1)
.startWith(stringObservable.take(1).map(s -> "Modified"))
.subscribe(new Consumer<String>() {
@Override
public void accept(String s) throws Exception {
System.out.println(s);
}
});
the result will be
Modified
Second
Third
Fourth
answered Nov 28 '18 at 7:51
Sarath KnSarath Kn
1,663919
Try this
Say you have a stream of strings ["First", "Second", "Third", "Fourth"] and you need to modify the very first item only
Observable<String> stringObservable = Observable.just("First", "Second", "Third", "Fourth").publish().refCount();
stringObservable.skip(1)
.startWith(stringObservable.take(1).map(s -> "Modified"))
.subscribe(new Consumer<String>() {
@Override
public void accept(String s) throws Exception {
System.out.println(s);
}
});
the result will be
Modified
Second
Third
Fourth
answered Nov 28 '18 at 7:51
Sarath KnSarath Kn
1,663919
edited Nov 28 '18 at 15:01
answered Nov 28 '18 at 7:51
Sarath KnSarath Kn
1,663919
answered Nov 28 '18 at 7:51
Sarath KnSarath Kn
1,663919
answered Nov 28 '18 at 7:51
Sarath KnSarath Kn
1,663919
1,663919
1
It's important to note that introducing two subscriptions may introduce side effects. You can usepublish()to subscribe only once, effectively.stringObservable.publish( shared -> shared.skip(1).startsWith(shared.take(1).map(...))...
– Bob Dalgleish
Nov 28 '18 at 14:44
Yes. noted it. Since publish() will multicast and turn the source into aConnectableobservable, addedpublish().refCount()combination to turn it into a multicasted Observable, which is shared for every subscriber added. Updated the answer
– Sarath Kn
Nov 28 '18 at 15:01
add a comment |
1
It's important to note that introducing two subscriptions may introduce side effects. You can usepublish()to subscribe only once, effectively.stringObservable.publish( shared -> shared.skip(1).startsWith(shared.take(1).map(...))...
– Bob Dalgleish
Nov 28 '18 at 14:44
Yes. noted it. Since publish() will multicast and turn the source into aConnectableobservable, addedpublish().refCount()combination to turn it into a multicasted Observable, which is shared for every subscriber added. Updated the answer
– Sarath Kn
Nov 28 '18 at 15:01
1
1
It's important to note that introducing two subscriptions may introduce side effects. You can use
publish() to subscribe only once, effectively. stringObservable.publish( shared -> shared.skip(1).startsWith(shared.take(1).map(...))...– Bob Dalgleish
Nov 28 '18 at 14:44
It's important to note that introducing two subscriptions may introduce side effects. You can use
publish() to subscribe only once, effectively. stringObservable.publish( shared -> shared.skip(1).startsWith(shared.take(1).map(...))...– Bob Dalgleish
Nov 28 '18 at 14:44
Yes. noted it. Since publish() will multicast and turn the source into a
Connectableobservable, added publish().refCount() combination to turn it into a multicasted Observable, which is shared for every subscriber added. Updated the answer– Sarath Kn
Nov 28 '18 at 15:01
Yes. noted it. Since publish() will multicast and turn the source into a
Connectableobservable, added publish().refCount() combination to turn it into a multicasted Observable, which is shared for every subscriber added. Updated the answer– Sarath Kn
Nov 28 '18 at 15:01
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%2f53513410%2frxjava-how-to-modify-only-first-item-before-sending-it-to-observer%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
