rxjs - Zipping value with corresponging flatMap result -

let's have stream of files filesstream , function uploadfile returns stream of 1 value (rx.observable.frompromise(...)). files on stream can uploaded via simple flatmap:

filesstream.flatmap(uploadfile) 

what need zip files filesstream corresponding responses uploadfile:

zippedstream.subscribe(     (file, response) => console.log("file " + file.name + " uploaded: " + response.message) ) 

i've come quite messy approach work

var zipppedstream = filesstream.flatmap(     (file) => uploadfile(file).zip(rx.observable.just(file), (r, f) => [r, f]) ) 

but don't since need unpack two-valued array in subscribe , looks heavy. how or i'm missing something?

you can bypass using zip , use flatmap overload instead accomplish want here

let zippedstream = filestream.flatmap(  (file) => uploadfile(file),  (file, uploaded) => [file, uploaded]);  zippedstream.subscribe(  ([file, response]) => console.log("file " + file.name + " uploaded: " + response.message));