Isn't that exactly what the nil-coalescing operator ?? does?
func query(name: String?) -> String {
return "{ param: \(name ?? "null") }"
}
If you use the JSONSerialization class from the Foundation library then `nil`
is bridged to `NSNull` and translated to "null" automatically (see SE-0140).
Regards,
Martin
> Am 08.02.2017 um 15:04 schrieb Maxim Veksler via swift-evolution
> <[email protected]>:
>
> Hello,
>
> Let's assume I have an optional "name" parameter, based on the optional
> status of the parameters I would like to compose string with either the
> unwrapped value of name, or the string "null". The use case for is syntactic
> sugar to compose a GraphQL queries.
>
> A (sampled) illustration of the code that currently solves it looks as
> following:
>
> func query(name: String?) {
> let variables_name = name != nil ? "\"\(name!)\"" : "null"
> return "{ param: \(variables_name) }"
> }
>
> Based on optional status the following output is expected
>
> let name = "Max"
> query(name: name)
> // { param: "Max" }
>
> let name: String? = nil
> query(name: name)
> // { param: null }
>
> I think it might be useful to have an conditional unwrap operator !? that
> would enable syntax sugar uch as the following built into the language:
>
> func query(name: String?) {
> return "{ param: \(name !? "\"\(name)\"": "null") }"
> }
>
> This expression is expected to produce same output as the examples above. It
> means check the Optional state of name: String?, in case it has a value,
> unwrap it and make the unwrapped value accessible under the same name to the
> true condition part of the expression, otherwise return the false condition
> part of the expression.
>
> The effectively removes the need to have the "if != nil" and the forced
> unwrap syntactical code, and IMHO improves code readability and
> expressiveness.
>
> I'm wondering what the community thinks of the displayed use case, and
> proposed solution?
>
>
> -m
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