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Property '0' is missing in type
Compiler error in Typescript version 3 but not in version 2How to assign any[][] to typescript interfaceDetecting an undefined object propertyHow do I check if an object has a property in JavaScript?How can I merge properties of two JavaScript objects dynamically?How do I remove a property from a JavaScript object?Get the name of an object's typeSorting an array of JavaScript objects by propertySort array of objects by string property valueIterate through object propertiesCan't bind to 'ngModel' since it isn't a known property of 'input'Types of property X are incompatible
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;
I have my interface like this
export interface Details
Name: [
First: string;
Last: string;
];
I have a observable config variable
Configuration: KnockoutObservable<Details> = ko.observable<Details>();
and I would like to assign it a value in the constructor as follows -
config =
Name: [
First: "ABC",
Last: "DEF"
,
First: "LMN",
Last: "XYZ"
]
;
this.Configuration(config);
and I am getting an error:
Types of property 'Name' is incompatible and property '0' is missing
in type.
Type ' First:string; Last:string; []' is not assignable to type '[
First: string; Last:string; ]'
I don't have control on changing the interface as it is being used elsewhere. What is the correct way to initialize this config variable ?
Thanks in advance.
javascript typescript knockout.js observable
add a comment |
I have my interface like this
export interface Details
Name: [
First: string;
Last: string;
];
I have a observable config variable
Configuration: KnockoutObservable<Details> = ko.observable<Details>();
and I would like to assign it a value in the constructor as follows -
config =
Name: [
First: "ABC",
Last: "DEF"
,
First: "LMN",
Last: "XYZ"
]
;
this.Configuration(config);
and I am getting an error:
Types of property 'Name' is incompatible and property '0' is missing
in type.
Type ' First:string; Last:string; []' is not assignable to type '[
First: string; Last:string; ]'
I don't have control on changing the interface as it is being used elsewhere. What is the correct way to initialize this config variable ?
Thanks in advance.
javascript typescript knockout.js observable
add a comment |
I have my interface like this
export interface Details
Name: [
First: string;
Last: string;
];
I have a observable config variable
Configuration: KnockoutObservable<Details> = ko.observable<Details>();
and I would like to assign it a value in the constructor as follows -
config =
Name: [
First: "ABC",
Last: "DEF"
,
First: "LMN",
Last: "XYZ"
]
;
this.Configuration(config);
and I am getting an error:
Types of property 'Name' is incompatible and property '0' is missing
in type.
Type ' First:string; Last:string; []' is not assignable to type '[
First: string; Last:string; ]'
I don't have control on changing the interface as it is being used elsewhere. What is the correct way to initialize this config variable ?
Thanks in advance.
javascript typescript knockout.js observable
I have my interface like this
export interface Details
Name: [
First: string;
Last: string;
];
I have a observable config variable
Configuration: KnockoutObservable<Details> = ko.observable<Details>();
and I would like to assign it a value in the constructor as follows -
config =
Name: [
First: "ABC",
Last: "DEF"
,
First: "LMN",
Last: "XYZ"
]
;
this.Configuration(config);
and I am getting an error:
Types of property 'Name' is incompatible and property '0' is missing
in type.
Type ' First:string; Last:string; []' is not assignable to type '[
First: string; Last:string; ]'
I don't have control on changing the interface as it is being used elsewhere. What is the correct way to initialize this config variable ?
Thanks in advance.
javascript typescript knockout.js observable
javascript typescript knockout.js observable
edited Jun 9 '17 at 2:48
Pablo Cesar Cordova Morales
87031225
87031225
asked Jun 9 '17 at 1:00
rktrkt
3761318
3761318
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
I came across the same issue and got around it by changing the interface to:
interface Details
Name:
First: string;
Last: string;
[];
I know you may not want the interface changed but hope this helps for anyone that is in this situation.
I want to kiss you! I fought this for so long without any answers. Thanks you so much!
– Joseph Hamilton
Mar 29 '18 at 1:39
Ha! I am glad this helped because I definitely found myself in the same boat some time ago.
– RyanA91
Mar 29 '18 at 14:43
Just in case it isn't obvious the solution here is to move the[]
to immediately followinstead of being
[ ]
. This is a change in syntax with recent typescript version.
– Simon_Weaver
May 8 '18 at 18:39
add a comment |
In this type definition:
interface Details
Name: [
First: string;
Last: string;
];
Name
is not an array at compile-time. It's a tuple with one element. Tuples in Typescript are allowed to have extra elements, but they can't have missing elements. As a 1-tuple, Name
is essentially an array which must have at least one element.
However, in this value:
const config =
Name: [
First: "ABC",
Last: "DEF"
,
First: "LMN",
Last: "XYZ"
]
;
Since there is no explicit typing, the Name
property here defaults to array type. Arrays can have any number of elements, including zero - which doesn't fit in a 1-tuple. Hence your error.
Your error can be fixed by giving the compiler a hint that your literal is actually a tuple:
const config: Details = Name: [..., ...] ;
If you do need to be able to take in an array of names, you'll have to do some casting, maybe something like this:
if (names.length > 0)
const config =
Name: names as Details['Name']
;
Configuration(config);
(You could remove the if check if you can determine that the tuple was simply a mistake by whoever wrote the typings.)
Tuples reference:
https://www.typescriptlang.org/docs/handbook/basic-types.html
great answer! just helped me out.
– royse41
Aug 11 '17 at 7:23
add a comment |
Updating the interface to following should fix the issue :
interface Details
Name: Array<
First: string;
Last: string;
>;
2
Can you explain why this fixes the issue please?
– CharliePrynn
Apr 4 '18 at 14:03
I believe, [] in typescript refers to tuple type , Array<> is a closer representation of javascript arrays. That's my understanding .
– Rajaram Nayak
Apr 5 '18 at 13:51
add a comment |
This error can come from incorrectly typing an array (like I just did):
myArray:[]; //Incorrect, results in error message of `Property '0' is missing in type`
myArray: Array<string>; //Correct
myArray: string[]; //Also correct
The reason is that braces denote a tuple in Typescript, not an array.
The Docs
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
I came across the same issue and got around it by changing the interface to:
interface Details
Name:
First: string;
Last: string;
[];
I know you may not want the interface changed but hope this helps for anyone that is in this situation.
I want to kiss you! I fought this for so long without any answers. Thanks you so much!
– Joseph Hamilton
Mar 29 '18 at 1:39
Ha! I am glad this helped because I definitely found myself in the same boat some time ago.
– RyanA91
Mar 29 '18 at 14:43
Just in case it isn't obvious the solution here is to move the[]
to immediately followinstead of being
[ ]
. This is a change in syntax with recent typescript version.
– Simon_Weaver
May 8 '18 at 18:39
add a comment |
I came across the same issue and got around it by changing the interface to:
interface Details
Name:
First: string;
Last: string;
[];
I know you may not want the interface changed but hope this helps for anyone that is in this situation.
I want to kiss you! I fought this for so long without any answers. Thanks you so much!
– Joseph Hamilton
Mar 29 '18 at 1:39
Ha! I am glad this helped because I definitely found myself in the same boat some time ago.
– RyanA91
Mar 29 '18 at 14:43
Just in case it isn't obvious the solution here is to move the[]
to immediately followinstead of being
[ ]
. This is a change in syntax with recent typescript version.
– Simon_Weaver
May 8 '18 at 18:39
add a comment |
I came across the same issue and got around it by changing the interface to:
interface Details
Name:
First: string;
Last: string;
[];
I know you may not want the interface changed but hope this helps for anyone that is in this situation.
I came across the same issue and got around it by changing the interface to:
interface Details
Name:
First: string;
Last: string;
[];
I know you may not want the interface changed but hope this helps for anyone that is in this situation.
answered Sep 26 '17 at 7:51
RyanA91RyanA91
22127
22127
I want to kiss you! I fought this for so long without any answers. Thanks you so much!
– Joseph Hamilton
Mar 29 '18 at 1:39
Ha! I am glad this helped because I definitely found myself in the same boat some time ago.
– RyanA91
Mar 29 '18 at 14:43
Just in case it isn't obvious the solution here is to move the[]
to immediately followinstead of being
[ ]
. This is a change in syntax with recent typescript version.
– Simon_Weaver
May 8 '18 at 18:39
add a comment |
I want to kiss you! I fought this for so long without any answers. Thanks you so much!
– Joseph Hamilton
Mar 29 '18 at 1:39
Ha! I am glad this helped because I definitely found myself in the same boat some time ago.
– RyanA91
Mar 29 '18 at 14:43
Just in case it isn't obvious the solution here is to move the[]
to immediately followinstead of being
[ ]
. This is a change in syntax with recent typescript version.
– Simon_Weaver
May 8 '18 at 18:39
I want to kiss you! I fought this for so long without any answers. Thanks you so much!
– Joseph Hamilton
Mar 29 '18 at 1:39
I want to kiss you! I fought this for so long without any answers. Thanks you so much!
– Joseph Hamilton
Mar 29 '18 at 1:39
Ha! I am glad this helped because I definitely found myself in the same boat some time ago.
– RyanA91
Mar 29 '18 at 14:43
Ha! I am glad this helped because I definitely found myself in the same boat some time ago.
– RyanA91
Mar 29 '18 at 14:43
Just in case it isn't obvious the solution here is to move the
[]
to immediately follow
instead of being [ ]
. This is a change in syntax with recent typescript version.– Simon_Weaver
May 8 '18 at 18:39
Just in case it isn't obvious the solution here is to move the
[]
to immediately follow
instead of being [ ]
. This is a change in syntax with recent typescript version.– Simon_Weaver
May 8 '18 at 18:39
add a comment |
In this type definition:
interface Details
Name: [
First: string;
Last: string;
];
Name
is not an array at compile-time. It's a tuple with one element. Tuples in Typescript are allowed to have extra elements, but they can't have missing elements. As a 1-tuple, Name
is essentially an array which must have at least one element.
However, in this value:
const config =
Name: [
First: "ABC",
Last: "DEF"
,
First: "LMN",
Last: "XYZ"
]
;
Since there is no explicit typing, the Name
property here defaults to array type. Arrays can have any number of elements, including zero - which doesn't fit in a 1-tuple. Hence your error.
Your error can be fixed by giving the compiler a hint that your literal is actually a tuple:
const config: Details = Name: [..., ...] ;
If you do need to be able to take in an array of names, you'll have to do some casting, maybe something like this:
if (names.length > 0)
const config =
Name: names as Details['Name']
;
Configuration(config);
(You could remove the if check if you can determine that the tuple was simply a mistake by whoever wrote the typings.)
Tuples reference:
https://www.typescriptlang.org/docs/handbook/basic-types.html
great answer! just helped me out.
– royse41
Aug 11 '17 at 7:23
add a comment |
In this type definition:
interface Details
Name: [
First: string;
Last: string;
];
Name
is not an array at compile-time. It's a tuple with one element. Tuples in Typescript are allowed to have extra elements, but they can't have missing elements. As a 1-tuple, Name
is essentially an array which must have at least one element.
However, in this value:
const config =
Name: [
First: "ABC",
Last: "DEF"
,
First: "LMN",
Last: "XYZ"
]
;
Since there is no explicit typing, the Name
property here defaults to array type. Arrays can have any number of elements, including zero - which doesn't fit in a 1-tuple. Hence your error.
Your error can be fixed by giving the compiler a hint that your literal is actually a tuple:
const config: Details = Name: [..., ...] ;
If you do need to be able to take in an array of names, you'll have to do some casting, maybe something like this:
if (names.length > 0)
const config =
Name: names as Details['Name']
;
Configuration(config);
(You could remove the if check if you can determine that the tuple was simply a mistake by whoever wrote the typings.)
Tuples reference:
https://www.typescriptlang.org/docs/handbook/basic-types.html
great answer! just helped me out.
– royse41
Aug 11 '17 at 7:23
add a comment |
In this type definition:
interface Details
Name: [
First: string;
Last: string;
];
Name
is not an array at compile-time. It's a tuple with one element. Tuples in Typescript are allowed to have extra elements, but they can't have missing elements. As a 1-tuple, Name
is essentially an array which must have at least one element.
However, in this value:
const config =
Name: [
First: "ABC",
Last: "DEF"
,
First: "LMN",
Last: "XYZ"
]
;
Since there is no explicit typing, the Name
property here defaults to array type. Arrays can have any number of elements, including zero - which doesn't fit in a 1-tuple. Hence your error.
Your error can be fixed by giving the compiler a hint that your literal is actually a tuple:
const config: Details = Name: [..., ...] ;
If you do need to be able to take in an array of names, you'll have to do some casting, maybe something like this:
if (names.length > 0)
const config =
Name: names as Details['Name']
;
Configuration(config);
(You could remove the if check if you can determine that the tuple was simply a mistake by whoever wrote the typings.)
Tuples reference:
https://www.typescriptlang.org/docs/handbook/basic-types.html
In this type definition:
interface Details
Name: [
First: string;
Last: string;
];
Name
is not an array at compile-time. It's a tuple with one element. Tuples in Typescript are allowed to have extra elements, but they can't have missing elements. As a 1-tuple, Name
is essentially an array which must have at least one element.
However, in this value:
const config =
Name: [
First: "ABC",
Last: "DEF"
,
First: "LMN",
Last: "XYZ"
]
;
Since there is no explicit typing, the Name
property here defaults to array type. Arrays can have any number of elements, including zero - which doesn't fit in a 1-tuple. Hence your error.
Your error can be fixed by giving the compiler a hint that your literal is actually a tuple:
const config: Details = Name: [..., ...] ;
If you do need to be able to take in an array of names, you'll have to do some casting, maybe something like this:
if (names.length > 0)
const config =
Name: names as Details['Name']
;
Configuration(config);
(You could remove the if check if you can determine that the tuple was simply a mistake by whoever wrote the typings.)
Tuples reference:
https://www.typescriptlang.org/docs/handbook/basic-types.html
answered Jun 9 '17 at 5:29
dbandstradbandstra
1,05259
1,05259
great answer! just helped me out.
– royse41
Aug 11 '17 at 7:23
add a comment |
great answer! just helped me out.
– royse41
Aug 11 '17 at 7:23
great answer! just helped me out.
– royse41
Aug 11 '17 at 7:23
great answer! just helped me out.
– royse41
Aug 11 '17 at 7:23
add a comment |
Updating the interface to following should fix the issue :
interface Details
Name: Array<
First: string;
Last: string;
>;
2
Can you explain why this fixes the issue please?
– CharliePrynn
Apr 4 '18 at 14:03
I believe, [] in typescript refers to tuple type , Array<> is a closer representation of javascript arrays. That's my understanding .
– Rajaram Nayak
Apr 5 '18 at 13:51
add a comment |
Updating the interface to following should fix the issue :
interface Details
Name: Array<
First: string;
Last: string;
>;
2
Can you explain why this fixes the issue please?
– CharliePrynn
Apr 4 '18 at 14:03
I believe, [] in typescript refers to tuple type , Array<> is a closer representation of javascript arrays. That's my understanding .
– Rajaram Nayak
Apr 5 '18 at 13:51
add a comment |
Updating the interface to following should fix the issue :
interface Details
Name: Array<
First: string;
Last: string;
>;
Updating the interface to following should fix the issue :
interface Details
Name: Array<
First: string;
Last: string;
>;
answered Apr 4 '18 at 14:00
Rajaram NayakRajaram Nayak
7413
7413
2
Can you explain why this fixes the issue please?
– CharliePrynn
Apr 4 '18 at 14:03
I believe, [] in typescript refers to tuple type , Array<> is a closer representation of javascript arrays. That's my understanding .
– Rajaram Nayak
Apr 5 '18 at 13:51
add a comment |
2
Can you explain why this fixes the issue please?
– CharliePrynn
Apr 4 '18 at 14:03
I believe, [] in typescript refers to tuple type , Array<> is a closer representation of javascript arrays. That's my understanding .
– Rajaram Nayak
Apr 5 '18 at 13:51
2
2
Can you explain why this fixes the issue please?
– CharliePrynn
Apr 4 '18 at 14:03
Can you explain why this fixes the issue please?
– CharliePrynn
Apr 4 '18 at 14:03
I believe, [] in typescript refers to tuple type , Array<> is a closer representation of javascript arrays. That's my understanding .
– Rajaram Nayak
Apr 5 '18 at 13:51
I believe, [] in typescript refers to tuple type , Array<> is a closer representation of javascript arrays. That's my understanding .
– Rajaram Nayak
Apr 5 '18 at 13:51
add a comment |
This error can come from incorrectly typing an array (like I just did):
myArray:[]; //Incorrect, results in error message of `Property '0' is missing in type`
myArray: Array<string>; //Correct
myArray: string[]; //Also correct
The reason is that braces denote a tuple in Typescript, not an array.
The Docs
add a comment |
This error can come from incorrectly typing an array (like I just did):
myArray:[]; //Incorrect, results in error message of `Property '0' is missing in type`
myArray: Array<string>; //Correct
myArray: string[]; //Also correct
The reason is that braces denote a tuple in Typescript, not an array.
The Docs
add a comment |
This error can come from incorrectly typing an array (like I just did):
myArray:[]; //Incorrect, results in error message of `Property '0' is missing in type`
myArray: Array<string>; //Correct
myArray: string[]; //Also correct
The reason is that braces denote a tuple in Typescript, not an array.
The Docs
This error can come from incorrectly typing an array (like I just did):
myArray:[]; //Incorrect, results in error message of `Property '0' is missing in type`
myArray: Array<string>; //Correct
myArray: string[]; //Also correct
The reason is that braces denote a tuple in Typescript, not an array.
The Docs
answered Mar 22 at 1:44
Greg GumGreg Gum
11.7k1985136
11.7k1985136
add a comment |
add a comment |
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