CODEWITHSUNDEEP
bashshell
SaltedPork
SaltedPork

Reputation: 377

Assign part of a file name to bash variable?

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?

Upvotes: 4

Views: 2154

Answers (3)

codeforester
codeforester

Reputation: 42989

Using Bash Parameter Expansion to extract the basename and then extract the portion of the filename you want:

fastq="/path/to/12U12345._L001_R1_001.fastq.gz"
file="${fastq##*/}"  # gives 12U12345._L001_R1_001.fastq.gz
string="${file%%.*}" # gives 12U12345

Note that Bash doesn't allow us to nest the parameter expansion. Otherwise, we could have combined statements 2 and 3 above.

Upvotes: 2

Inian
Inian

Reputation: 85530

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

Upvotes: 5

Ronan Boiteau
Ronan Boiteau

Reputation: 10128

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.

Upvotes: 2

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