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Typescript & operator


Typescript & operator

By : Jennifer Lasso
Date : November 22 2020, 02:42 PM
help you fix your problem This looks like it's from the Intersection Types portion of the Language Specification. Specifically, the & appears to be an intersection type literal. As for what it does:
code :
interface A { a: number }  
interface B { b: number }

var ab: A & B = { a: 1, b: 1 };  
var a: A = ab;  // A & B assignable to A  
var b: B = ab;  // A & B assignable to B


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TypeScript error cannot use 'in' operator to search for '_id' for typescript and pouchdb

TypeScript error cannot use 'in' operator to search for '_id' for typescript and pouchdb


By : Aye Khine
Date : March 29 2020, 07:55 AM
it should still fix some issue The put method in PouchDb expects a JSON object not a string.
You should just send the object without stringifying it.
code :
var myObj = {"_id" : "1", "description":"hello world"} ;
mydb.put(myObj)
Typescript does not infer about delete operator and spread operator?

Typescript does not infer about delete operator and spread operator?


By : Kavitha D
Date : March 29 2020, 07:55 AM
hope this fix your issue The type of the variable is infered on assignment, and I don't think Typescript does any flow control magic with the delete operator. You could the spread operator to get rid of d:
code :
interface Test {
    a: string;
    b: number;
    c: number;
}

const test = {
    a: 'a',
    b: 1,
    d: []
};

let { d, ...test3} = test

const test2:Test = { ...test3, c:1 };
How to use '! operator' with 'in operator' in TypeScript?

How to use '! operator' with 'in operator' in TypeScript?


By : Arbol
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , The in type-guard (and generally type-guards) will not work on expressions. So the expression ab! will not result in the narrowing of ab. For the type-guard to work, the guarded expression must be a simple parameter/variable/field. The simplest solution is to guard against null explicitly. This is probably a good idea since the field is optional anyway:
code :
function test(ab: Interface['AB']) {
    const ab2 = !ab ? null : ('a' in ab) ? [ab.a] : ab.b;
} 
TypeScript - why is the spread operator used in map?

TypeScript - why is the spread operator used in map?


By : user3337745
Date : March 29 2020, 07:55 AM
I hope this helps . Similar to what previous answers suggest, notice that the spread operator is spreading the contents of item into another curly brace.
It's shallow copying the contents of item into that new object but replacing with the new value for score.
What is the | operator used for in Typescript?

What is the | operator used for in Typescript?


By : Max Bartlett
Date : March 29 2020, 07:55 AM
To fix the issue you can do Credit to @CRice in the 2nd comment of the question for this. I'm just posting it for convenience since s/he didn't.
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