Generating the output file name from the input file in bash script

Question

[Fluorine]

I'm writing a shell script that would convert all my .sam files to .bam files. I'm using a loop for all files in the folder that end with .sam. Now my question is how do I write in the script that the output files will have the same name as the input files, but end with .bam?

Answer 1

[Jorge Amigo]

for file in *.sam; do samtools view -bS $file > ${file/%.sam/.bam}; done

[5heikki]

Learn something new everyday. I always used basename but this is better in so many ways.

[Jorge Amigo]

strictly speaking, if you want to make sure that there's no other ".sam" string in the filename that could avoid this code to replace the extension, you can always force the string replacement to start from back:

for file in *.sam; do samtools view -bS $file > ${file/%.sam/.bam}; done

rather than simply

for file in *.sam; do samtools view -bS $file > ${file/.sam/.bam}; done

[Fluorine]

Thank you! That saved so much time :)