string replacement using xargs

Question

I am trying to collect file path and timestamp for each file under a certain directory(which is passed as an argument) in makefile

So, it goes like this.

TIMESTAMP_LOG := timestamp.log
TARGET_ROOT := ../../out/root

define collect-timestamp
  $(shell find $(1) | xargs -lfn sh -c 'echo -n fn"," >> $(TIMESTAMP_LOG); stat -c %Y fn >> $(TIMESTAMP_LOG)')
endef

all:
    $(call collect-timestamp, $(TARGET_ROOT))

If I run this, i would the get whole file path and timestamp as below ex) ../../out/root/bin/ls,133030303

but I want to get rid of "../../out/root" in file path.(passing as an argument if possible)

I thought I could do this using sed or shell script(see below) but apparently I am stuck. I tried:

$(shell find $(1) | xargs -Ifn sh -c 'echo -n ${fn##$(1)}"," >> $(TIMESTAMP_LOG); stat -c %Y fn >> $(TIMESTAMP_LOG)')  

$(shell find $(1) | xargs -Ifn sh -c 'sed 's/fn//g' >> $(TIMESTAMP_LOG); stat -c %Y fn >> $(TIMESTAMP_LOG)')

Answer 1

If you use find, you don't need xargs in most cases.

The following should work:

find $(1) -exec stat -c "%n,%Y" {} \; | sed 's#$(1)\/##'

Note that I use $(1) as both parameter to find and in sed substitution command.

Answer 2

You may try

find $(1) | xargs stat -c %Y >> $(TIMESTAMP_LOG)