Bash：处理目录中的所有文件

Question

How would I apply a logic to all files in a directory and pipe the final output of each input file to a separate file? Something like below

cut -d, -f2 input1.csv | awk 'END{print NR}' > input1_count.csv

(actual command is long, but I used above for simplicity to better understand the logic to all the files in a directory)

Answer 1

You can use a for -loop over the filenames:

shopt -s nullglob # enable nullglob
for f in *.csv; do
   cut -d, -f2 "$f" | awk 'END{print NR}' > "${f%.csv}_count.csv"
done
shopt -u nullglob # disable nullglob

The glob pattern *.csv expands to a null string with nullglob enabled (instead of a literal and probably non-existing filename *.csv ). In the loop, apply the pipeline of commands to each filename and redirect the output to a new filename.

The parameter expansion ${f%.csv} removes the shortest suffix pattern .csv from the filename.

Answer 2

A simple for loop should do the job:

for f in *.csv; do
   cut -d, -f2 "$f" | awk 'END{print NR}' > "${f/%.csv/_count.csv}"
done

This assumes you have .csv files are present in current directory. If files may not exist then use any of these shopt above for loop:

shopt -s failglob    # return error when glob matching fails

or

shopt -s nullglob    # be silent when glob matching fails