How to conditionally add a property in TypeScript?

Question

typescript

You need to add to the interface:

export interface ModalProps extends CuiModalProps {
    children: ReactElement | ReactElement[]
    setterClosePopup: Dispatch<SetStateAction<boolean>>
    namePopup?: string
    isOpen?: boolean
    withBorder?: boolean
    view?: 'labeled'
    className?: string
}

a label property with a value of string provided that the view property is set to 'labeled', otherwise the label should be set to never

Answer 1

Need more clarity. What else happens viewbesides labled?
This cannot be done through interfaces - it is possible through union types, but there must be a discriminator field that is different for each case.
If there is no such field, a specific type cannot be made, however, judging by the prefix Props, we can assume that these are some props of some React component, then this can be achieved using overloading :

export interface ModalProps extends CuiModalProps {
    children: ReactElement | ReactElement[]
    setterClosePopup: Dispatch<SetStateAction<boolean>>
    namePopup?: string
    isOpen?: boolean
    withBorder?: boolean
    className?: string
}
export interface ModalPropsLabled extends ModalProps {
    view: 'labeled'
    label: string
}

export const Component: ((pros: ModalProps) => ReactNode) & ((pros: ModalPropsLabled) => ReactNode) = (pros: ModalProps | ModalPropsLabled) => { ... }