Find in array with different types
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
2146
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
2146
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
2146
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
2146
asked Nov 21 '18 at 21:52
domreadydomready
2146
asked Nov 21 '18 at 21:52
domreadydomready
2146
asked Nov 21 '18 at 21:52
domreadydomready
2146
asked Nov 21 '18 at 21:52
domreadydomready
2146
2146
add a comment |
add a comment |
1 Answer
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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
1,4331424
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 |
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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
1,4331424
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
1,4331424
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
1,4331424
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
1,4331424
answered Nov 22 '18 at 10:01
scipperscipper
1,4331424
answered Nov 22 '18 at 10:01
scipperscipper
1,4331424
answered Nov 22 '18 at 10:01
scipperscipper
1,4331424
1,4331424
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 |
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