How to convert Option<i64> to a custom type Option<XXX> using query_as! macro

[tonitrnel]

For example, I implemented an ID which is essentially a wrapper around i64, and also implemented the From<i64> trait for it:

#[derive(..., sqlx::Type)]
pub struct ID(i64);

impl From<i64> for ID { ... }
impl From<ID> for i64 { ... }

When using query_as!, it can be directly converted in non-Option cases. The macro generated code looks like this:

let sqlx_query_as_id = row.try_get_unchecked::<i64, _>(0usize)?.into();

But for Option cases, Option<i64> doesn't automatically convert to Option<ID>:

let sqlx_query_as_id = row.try_get_unchecked::<Option<i64>, _>(0usize)?.into(); // ❌ Fails

Why isn't the macro generating code in this form instead ?

let sqlx_query_as_id = row.try_get_unchecked::<Option<i64>, _>(0usize)?.map(Into::into); // ✅ Expected

Did I overlook something ?

[joeydewaal]

You can override the type inference in the macros. This example shows how to force a custom type that is optional.

let mut conn = PgConnection::connect("...").await?;

struct Tweet {
    id: Option<ID>
}

let user: Tweet = sqlx::query_as!(Tweet, r#"select id as "id?: ID" from tweet"#)
    .fetch_one(&mut conn)
    .await?;

For other ways to override types in the macro's, take a look at this table.