Compare two filenames of first n characters in shell script - bash

Question

My requirement is
localfolder=green.txt,yellow.txt,blue.txt
remotefolder=green_202105050333.txt,yellow_202105050333.txt,blue_202105050333.txt
I want to compare both folders as
IF[[localfolder==remotefolder]] ex: green.txt= green_202105050333.txt(here condition will look always first characters of each file or eliminating the data&time)
then
display the results as "matched"
kindly help me to get the logic here please.
Thanks in advance!!

Answer 1

You can do the next, create a file for each directory(folder), for example if you use ls :
# in local directory
ls localfolder > localFolder.txt
# in some remote directory
ls remoteFolder > remoteFolder.txt
We can suppose that files are the next:
localFolder.txt:
green.txt
yellow.txt
blue.txt
remoteFolder.txt:
green_202105050333.txt
yellow_202105050333.txt
blue_202105050333.txt
gray_202105050333.txt
Now put the files in the same directory and execute the next:
sed -r 's/\.[a-z]+//g' localFolder.txt | xargs -I{} grep {} remoteFolder.txt | sed -r 's/_[0-9]+//g'
The output will be:
green.txt
yellow.txt
blue.txt
Now if you want do it in script, then create a file for example myScript.sh and the content must be:
#!/bin/bash
sed -r 's/\.[a-z]+//g' $1 | xargs -I{} grep {} $2 | sed -r 's/_[0-9]+//g'
Give execution permissions to the script:
chmod +x myScript.sh
And executed like the next:
./diff.sh localFolder.txt remoteFolder.txt
You will have the same result.
This script only works for your case and this is an explain what do each part
# Remove .txt or any extensions for each line in localFolder
sed -r 's/\.[a-z]+//g' localFolder.txt
# The ouput of each line is received by grep and search in remoteFolder
| xargs -I{} grep {} remoteFolder.txt
# Remove the datetime
| sed -r 's/_[0-9]+//g'