After defining Partial<T>, encountering an error trying to access an undefined property is unexpected

Question

After defining Partial<T>, encountering an error trying to access an undefined property is unexpected

In my function, I am attempting to standardize certain values by specifying the whole function type as Partial. However, despite declaring the interaction variable as Partial Type, I keep encountering the error message saying "Cannot read property endTime of undefined."

Below is the code snippet for my function:

export function mapOptionsToEditActivity<T extends InteractionType>(
    interaction: Partial<InteractionTypeInteractionMap[T]>,
    interactionType: T,
    matchedRecord: InteractionMatchedRecord,
    user: User,
    activityId: string,
): Partial<CrmInteraction> {
    const mappedInteraction = {
        crmCallId: activityId,
        dateEnd: interaction.endTime,
        dateStart: interaction.startTime,
        userCrmId: user.crmId,
    };

    return mappedInteraction; 
}

The above code clearly indicates that I have defined interaction as Partial Type and the entire function type as Partial as well.

Nevertheless, the issue persists with the following error message:

Cannot read property endTime of undefined

I would greatly appreciate any assistance in pinpointing where the mistake might be occurring.

typescript

Answer 1

Answer №1

It appears that there may be a misunderstanding about the purpose of the 'partial' concept.

interface foo
{
    bar:string
}

In this scenario:

const a:foo = {} // An error will occur because 'foo' requires a 'bar' field

const b:foo = {bar:"123"} // This will work as expected

const c:foo = undefined // This will also result in an error since 'c' is of type 'foo', not 'undefined'

const x:Partial<foo> = {} // This will work because a 'Partial<foo>' may or may not have a 'bar' field

const y:Partial<foo>= {bar:"123"} // This will also work due to the partial nature of 'foo'

const z:Partial<foo>= undefined // Despite being 'undefined', this will still throw an error because it's not recognized as 'foo'


const u:foo|undefined = undefined  // This will work as 'u' can either be 'foo' or 'undefined'

Similarly, TypeScript understands that:

b.bar // Will always work as 'b' cannot be a 'foo' without a valid 'bar'

x.bar // May or may not work depending on the existence of 'bar' within the partial 'foo'

u.bar // Assuming 'u' itself may be 'undefined', precautions should be taken before accessing the property

Various methods exist in TypeScript for handling such situations:

if(u) u.bar;

TypeScript acknowledges that access inside the 'if' block indicates 'u' must be a valid 'foo'

if("bar" in u) u.bar;

Accessing 'bar' only if it exists

u?.bar // Acts as a shorthand for conditional access

Checking for the 'bar' property and returning its value if present

u!.bar

By allowing an assertion operator, you're essentially bypassing type checking under controlled circumstances

However, it's important to note that TypeScript compiles down to plain JavaScript, which means external factors could potentially disrupt the carefully set rules during runtime.

Answer 2

It appears that there may be a misunderstanding about the purpose of the 'partial' concept.

interface foo
{
    bar:string
}

In this scenario:

const a:foo = {} // An error will occur because 'foo' requires a 'bar' field

const b:foo = {bar:"123"} // This will work as expected

const c:foo = undefined // This will also result in an error since 'c' is of type 'foo', not 'undefined'

const x:Partial<foo> = {} // This will work because a 'Partial<foo>' may or may not have a 'bar' field

const y:Partial<foo>= {bar:"123"} // This will also work due to the partial nature of 'foo'

const z:Partial<foo>= undefined // Despite being 'undefined', this will still throw an error because it's not recognized as 'foo'


const u:foo|undefined = undefined  // This will work as 'u' can either be 'foo' or 'undefined'

Similarly, TypeScript understands that:

b.bar // Will always work as 'b' cannot be a 'foo' without a valid 'bar'

x.bar // May or may not work depending on the existence of 'bar' within the partial 'foo'

u.bar // Assuming 'u' itself may be 'undefined', precautions should be taken before accessing the property

Various methods exist in TypeScript for handling such situations:

if(u) u.bar;

TypeScript acknowledges that access inside the 'if' block indicates 'u' must be a valid 'foo'

if("bar" in u) u.bar;

Accessing 'bar' only if it exists

u?.bar // Acts as a shorthand for conditional access

Checking for the 'bar' property and returning its value if present

u!.bar

By allowing an assertion operator, you're essentially bypassing type checking under controlled circumstances

However, it's important to note that TypeScript compiles down to plain JavaScript, which means external factors could potentially disrupt the carefully set rules during runtime.

After defining Partial<T>, encountering an error trying to access an undefined property is unexpected

Answer №1

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