I have an anonymous function in Express like so:
app.use((err, req, res, next) => {
// ...
});
and I want to set the type of the function to ErrorRequestHandler (Not the return type!).
I can do it like this:
const X: ErrorRequestHandler = (err, req, res, next) => {
// ...
}
app.use(X);
But is there a syntax to do it inline? Like this:
app.use((err, req, res, next) => {
// ...
} : ErrorRequestHandler); // Note: this doesn't work.
You just need to use some extra brackets to help Typescript out.
eg..
app.use(<ErrorRequestHandler>((err, req, res, next) => {
}));
You can of course use as too, to avoid any JSX confusion.
app.use(((err, req, res, next) => {
}) as ErrorRequestHandler);
Notice how above the whole function is enclosed inside brackets, this allows Typescript to apply the typecast to the whole function expression.
Code completion on req, res, & next will then work as expected.
typescript anonymous function type
lass Foo {
save(callback: (n: number) => any) : void {
callback(42);
}
}
var foo = new Foo();
var strCallback = (result: string) : void => {
alert(result);
}
var numCallback = (result: number) : void => {
alert(result.toString());
}
foo.save(strCallback); // not OK
foo.save(numCallback); // OK
app.use, that's built into Express
app.usehas overlaps with parameter count here, so Typescript can't determine which one you want from parameter count only.app.get('/', (req, res) => {})work fine though.