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bashinputoutputfilenames

Output filename from input in bash

I have this script:

#!/bin/bash
FASTQFILES=~/Programs/ncbi-blast-2.2.29+/DB_files/*.fastq
FASTAFILES=~/Programs/ncbi-blast-2.2.29+/DB_files/*.fasta
clear
for file in $FASTQFILES
do cat $FASTQFILES | perl -e '$i=0;while(<>){if(/^\@/&&$i==0){s/^\@/\>/;print;}elsif($i==1){print;$i=-3}$i++;}' > ~/Programs/ncbi-blast-2.2.29+/DB_files/"${FASTQFILES%.*}.fasta"
mv $FASTAFILES ~/Programs/ncbi-blast-2.2.29+/db/

done

I'm trying it to grab the files defined in $FASTQFILES, do the .fastq to .fasta conversion, name the output with the same filename of the input, and move it to a new folder. E.g., ~/./DB_files/HELLO.fastq should give a converted ~/./db/HELLO.fasta

The problem is that the output of the conversion is a properly formatted hidden file called .fasta in the first folder instead of the expected one named HELLO.fasta. So there is nothing to mv. I think I'm messing up in the ${FASTQFILES%.*}.fasta argument but I can't seem to fix it.

Solution

  • I see three problems:

    1. One part of your trouble is that you use cat $FASTQFILES instead of cat $file.

    2. You also need to fix the I/O redirection at the end of that line to > ~/Programs/ncbi-blast-2.2.29+/DB_files/"${file%.fastq}.fasta".

    3. The mv command needs to be executed outside the loop.

    In fact, when processing a single file at a time, you don't need to use cat at all (UUOC — Useless Use Of Cat). Simply provide "$file" as an argument to the Perl script.