在bash脚本中执行cut命令

Question

Every line printed with the echo includes the forward slashes for the directories that the given files are in. I am trying to cut the forward slashes using the cut command but it is not working. The files are gzipped so they have the .gz extension.

#!/bin/bash

for filename in /data/logs/2017/month_01/201701*
do
echo $filename
cut $filename -d '/' -f1
done

Thanks in advance.

Answer 1

The order of commands is wrong. You need to stream the string input to the cut command via pipe( | ) or here-strings( <<< ).

echo "$filename" | cut  -d '/' -f1

(or)

cut -d '/' -f1 <<<"$filename"

(or) using here-docs

cut -d '/' -f1 <<EOF
$filename
EOF

data

And don't forget to double-quote variables to avoid Word-Splitting done by the shell.

Answer 2

Assuming filename is /a/b/c.gz you just want c.gz ?

Well there's two very easy answers:

basename $filename

The other is:

echo ${filename##*/}

The latter make use bash 's built-in string delete parameter expansion.

Another way of solving your problem, is you could change directory first, ie

#!/bin/bash

pushd /data/logs/2017/month_01
for filename in 201701* 
do
    echo $filename
done
popd

Reference:

• http://wiki.bash-hackers.org/syntax/pe

(EDIT: Fixed typo identified by @123)

Answer 3

As suggested b @lnian cut command used with echo command via pipe sign For getting only file name with your script you need to use.

cut with -f1 option will get first value before / which would give blank so you need to get last value from the filename.

#!/bin/bash

for filename in /data/logs/2017/month_01/201701*
do
echo "$filename" | rev| cut  -d '/' -f1
done

rev command reverse the filename so you will get last value which is your filename