How to get Makefile to only produce a file name without extension? - bash

Question

I'm trying to make a Makefile that exports a markdown file to a pdf file that uses the same filename as the original markdown file. I used "basename" command but it produces "inputfile.md.pdf" instead of "inputfile.pdf".
Please see my code below (I adapted a code I found on the Internet. Thank you!):
.PHONY: pdf docx apa format
FILES := $(wildcard ./*.md)
pdf:
for file in $(FILES); do \
pandoc $$file \
--bibliography mypath \
--csl mypath \
--filter pandoc-citeproc \
--template eisvogel \
-o $(basename $$file).pdf; \
open $(basename $$file).pdf; \
done
Anyone who can help me? I'm a novice in Makefile (and programming in general) so any detailed help would be very much appreciated.
I also tried these codes below, but they generated an error message:
-o $(basename -s ".md" $$file).pdf; \
-o $(basename -s .md $$file).pdf; \

Answer 1

The way you write $(basename …) you get the basename make function. This would normally the right thing, but you try to reference a shell variable file in its argument, which is unavailable at the make layer.
In this case, it is probably easiest to call the basename shell utility, at the shell level. Therefore, you need to escape the $ to get shell substitution, like this:
-o "$$(basename -s .md $$file)".pdf; \
open "$$(basename -s .md $$file)".pdf; \
Alternatively, you could try to move the loop to the make layer, perhaps using foreach.