TypeScript：将参数传递给函数调用之前，检查是否已定义参数的必需属性

Question

This doesn't compile ( playground ):

function myFunction(params: {
    a: Date,
    b?: Date
}) {
    if (params.b) {
        myFunctionInternal(params); // ERROR!
    }
}

function myFunctionInternal(params: {
    a: Date,
    b: Date
}) {}

Is there a more elegant workaround than params as any ?

Answer 1

The problem is that the type guard impacts just the type of the field ( params.b will have the undefined removed) not the type of whole object ( param will continue to have the type { a: Date, b?: Date } )

Not sure I would call it more elegant, but we can create a type guard that removes the undefined from a type field:

type RemoveOptionalFromField<T, K extends keyof T> = Pick<T, Exclude<keyof T, K>> & { [P in K]-?: T[P] }

function notNull<T, K extends keyof T>(o: T | RemoveOptionalFromField<T, K>, key: K) : o is RemoveOptionalFromField<T, K> {
    return !!o[key];
}

function myFunction(params: {
    a: Date,
    b?: Date
}) {
    if (notNull(params, 'b')) {
        params.b.getDate()
        myFunctionInternal(params);
    }
}

We could even create a version that takes any number of keys:

function allNotNull<T, K extends keyof T>(o: T | RemoveOptionalFromField<T, K>, ...keys: K[]) : o is RemoveOptionalFromField<T, K> {
    return keys.every(k => !!o[k]);
}
function myFunction(params: {
    a?: Date,
    b?: Date
}) {
    if (allNotNull(params, 'b', 'a')) {
        params.b.getDate()
        myFunctionInternal(params);
    }
}

Answer 2

the error message said property 'b' is optional in type '{ a: Date; b?: Date; }' but required in type '{ a: Date; b: Date; }'

It can be solve like this

myFunctionInternal(params as {a,b}); 

or

myFunctionInternal({a:params.a ,b:params.b});