Assign part of a file name to bash variable?

Question

I have a file and its name looks like:

12U12345._L001_R1_001.fastq.gz

I want to assign to a variable just the 12U12345 part.

So far I have:

variable=`basename $fastq | sed {s'/_S[0-9]*_L001_R1_001.fastq.gz//'}`

Note: $fastq is a variable with the full path to the file in it.

This solution currently returns the full file name, any ideas how to get this right?

Answer 1

Just use the built-in parameter expansion provided by the shell, instead of spawning a separate process

fastq="12U12345._L001_R1_001.fastq.gz"
printf '%s\n' "${fastq%%.*}"
12U12345

or use printf() itself to store to a new variable in one-shot

printf -v numericPart '%s' "${fastq%%.*}"
printf '%s\n' "${numericPart}"

Also bash has a built-in regular expression comparison operator, represented by =~ using which you could do

fastq="12U12345._L001_R1_001.fastq.gz"
regex='^([[:alnum:]]+)\.(.*)'

if [[ $fastq =~ $regex ]]; then
    numericPart="${BASH_REMATCH[1]}"
    printf '%s\n' "${numericPart}"
fi

Answer 2

You could use cut:

$> fastq="/path/to/12U12345._L001_R1_001.fastq.gz"
$> variable=$(basename "$fastq" | cut -d '.' -f 1)
$> echo "$variable"
12U12345

Also, please note that:

• It's better to wrap your variable inside quotes. Otherwise you command won't work with filenames that contain space(s).

• You should use $() instead of the backticks.