Find in array with different types
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box;
}
I have two subClasses that will be extended to many classes. Now i will find a class with a specific type (in my case this is a property on the subClass).
After creating many instances i have a helper function that will return the first class with a match of needed type.
So far everything works fine.
My Problem
I do not get the correct ts-type when i return my class on my Helper funtion. I always get a typescript error but i have no idea how to solve that problem.
Is there anyone show me the right way to do it?
Or is there some tutorial? (I found some tutorials but only withoud extends and no dynamic store array with many different types)
My example code on the typescript playground with the ts-error
typescript
asked Nov 21 '18 at 21:52
domreadydomready
2246
add a comment |
I have two subClasses that will be extended to many classes. Now i will find a class with a specific type (in my case this is a property on the subClass).
After creating many instances i have a helper function that will return the first class with a match of needed type.
So far everything works fine.
My Problem
I do not get the correct ts-type when i return my class on my Helper funtion. I always get a typescript error but i have no idea how to solve that problem.
Is there anyone show me the right way to do it?
Or is there some tutorial? (I found some tutorials but only withoud extends and no dynamic store array with many different types)
My example code on the typescript playground with the ts-error
typescript
asked Nov 21 '18 at 21:52
domreadydomready
2246
add a comment |
I have two subClasses that will be extended to many classes. Now i will find a class with a specific type (in my case this is a property on the subClass).
After creating many instances i have a helper function that will return the first class with a match of needed type.
So far everything works fine.
My Problem
I do not get the correct ts-type when i return my class on my Helper funtion. I always get a typescript error but i have no idea how to solve that problem.
Is there anyone show me the right way to do it?
Or is there some tutorial? (I found some tutorials but only withoud extends and no dynamic store array with many different types)
My example code on the typescript playground with the ts-error
typescript
asked Nov 21 '18 at 21:52
domreadydomready
2246
I have two subClasses that will be extended to many classes. Now i will find a class with a specific type (in my case this is a property on the subClass).
After creating many instances i have a helper function that will return the first class with a match of needed type.
So far everything works fine.
My Problem
I do not get the correct ts-type when i return my class on my Helper funtion. I always get a typescript error but i have no idea how to solve that problem.
Is there anyone show me the right way to do it?
Or is there some tutorial? (I found some tutorials but only withoud extends and no dynamic store array with many different types)
My example code on the typescript playground with the ts-error
typescript
typescript
asked Nov 21 '18 at 21:52
domreadydomready
2246
asked Nov 21 '18 at 21:52
domreadydomready
2246
asked Nov 21 '18 at 21:52
domreadydomready
2246
asked Nov 21 '18 at 21:52
domreadydomready
2246
asked Nov 21 '18 at 21:52
domreadydomready
2246
2246
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
I mentioned two things:
You declared the array with type
Building
const buildingsStore: (Building) = [storehouse, barracks];
and then in the filter expecting building to have the property type:
return type === building.type;
This does obviously not work, because Building does not have type.
- I can't sadly explain what exactly goes on in TypeScript, but removing the return type of the function and type cast the result instead works (has somethign to do with Type Inference):
Working code here
EDIT
I updated my code regarding to the type property: here
answered Nov 22 '18 at 10:01
scipperscipper
1,4431424
Thanks, it works fine. I do not think this is the best solution but a big step to the next optimal solution.
– domready
Nov 22 '18 at 10:48
You are welcome. Type Inference isn't that bad, btw.
– scipper
Nov 22 '18 at 10:49
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%2f53420973%2ffind-in-array-with-different-types%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
I mentioned two things:
You declared the array with type
Building
const buildingsStore: (Building) = [storehouse, barracks];
and then in the filter expecting building to have the property type:
return type === building.type;
This does obviously not work, because Building does not have type.
- I can't sadly explain what exactly goes on in TypeScript, but removing the return type of the function and type cast the result instead works (has somethign to do with Type Inference):
Working code here
EDIT
I updated my code regarding to the type property: here
answered Nov 22 '18 at 10:01
scipperscipper
1,4431424
Thanks, it works fine. I do not think this is the best solution but a big step to the next optimal solution.
– domready
Nov 22 '18 at 10:48
You are welcome. Type Inference isn't that bad, btw.
– scipper
Nov 22 '18 at 10:49
add a comment |
I mentioned two things:
You declared the array with type
Building
const buildingsStore: (Building) = [storehouse, barracks];
and then in the filter expecting building to have the property type:
return type === building.type;
This does obviously not work, because Building does not have type.
- I can't sadly explain what exactly goes on in TypeScript, but removing the return type of the function and type cast the result instead works (has somethign to do with Type Inference):
Working code here
EDIT
I updated my code regarding to the type property: here
answered Nov 22 '18 at 10:01
scipperscipper
1,4431424
Thanks, it works fine. I do not think this is the best solution but a big step to the next optimal solution.
– domready
Nov 22 '18 at 10:48
You are welcome. Type Inference isn't that bad, btw.
– scipper
Nov 22 '18 at 10:49
add a comment |
I mentioned two things:
You declared the array with type
Building
const buildingsStore: (Building) = [storehouse, barracks];
and then in the filter expecting building to have the property type:
return type === building.type;
This does obviously not work, because Building does not have type.
- I can't sadly explain what exactly goes on in TypeScript, but removing the return type of the function and type cast the result instead works (has somethign to do with Type Inference):
Working code here
EDIT
I updated my code regarding to the type property: here
answered Nov 22 '18 at 10:01
scipperscipper
1,4431424
I mentioned two things:
You declared the array with type
Building
const buildingsStore: (Building) = [storehouse, barracks];
and then in the filter expecting building to have the property type:
return type === building.type;
This does obviously not work, because Building does not have type.
- I can't sadly explain what exactly goes on in TypeScript, but removing the return type of the function and type cast the result instead works (has somethign to do with Type Inference):
Working code here
EDIT
I updated my code regarding to the type property: here
answered Nov 22 '18 at 10:01
scipperscipper
1,4431424
answered Nov 22 '18 at 10:01
scipperscipper
1,4431424
answered Nov 22 '18 at 10:01
scipperscipper
1,4431424
answered Nov 22 '18 at 10:01
scipperscipper
1,4431424
1,4431424
Thanks, it works fine. I do not think this is the best solution but a big step to the next optimal solution.
– domready
Nov 22 '18 at 10:48
You are welcome. Type Inference isn't that bad, btw.
– scipper
Nov 22 '18 at 10:49
add a comment |
Thanks, it works fine. I do not think this is the best solution but a big step to the next optimal solution.
– domready
Nov 22 '18 at 10:48
You are welcome. Type Inference isn't that bad, btw.
– scipper
Nov 22 '18 at 10:49
Thanks, it works fine. I do not think this is the best solution but a big step to the next optimal solution.
– domready
Nov 22 '18 at 10:48
Thanks, it works fine. I do not think this is the best solution but a big step to the next optimal solution.
– domready
Nov 22 '18 at 10:48
You are welcome. Type Inference isn't that bad, btw.
– scipper
Nov 22 '18 at 10:49
You are welcome. Type Inference isn't that bad, btw.
– scipper
Nov 22 '18 at 10:49
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%2f53420973%2ffind-in-array-with-different-types%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
