Need to cut filename in LINUX ksh

Shell Programming and Scripting
1

Hi,
I need to cut filename in Linux ksh.
for example file name is c_xxxx_cp_200908175035.zip. I need to get variable with only c_xxxx_cp value.

2

what have you tried?

3

dir_name=${zip_file%.*}
dir_name_1=${dir_name#*JULIY/}

This will give me file name without ".zip", but still I have to get rid of timestamp.

4

the last "_" underscore is before the extension. therefore you can use % which means "shortest match from the back"

${zip_file%_*}

please read again Advance Bash guide (google) for details on string manipulation

5

Thanks a lot.

6

Even Simpler :

echo "c_xxxx_cp_200908175035.zip" | cut -d "." -f 1 | cut -d "_" -f 1,2,3

7

certainly not simpler. you don't need to call 2 cuts.

8

fname="c_xxxx_cp_200908175035.zip"

file_without_ext=$(basename $fname .zip)

echo $file_without_ext

output:
c_xxxx_cp_200908175035

-----Post Update-----

fname="test_high_prio123.zip"

file_name_wo_ext=$(basename $fname .zip)

echo $file_name_wo_ext

O/P:

test_high_prio123

9 
echo c_xxxx_cp_200908175035.zip | sed 's/_[^_]*$//'
10

Or using ksh93 extended patterns:

fname="test_high_prio123_200908175035.zip"
echo "${fname//_{12,12}([[:digit:]]).*/}
test_high_prio123
11 
echo "c_xxxx_cp_200908175035.zip"|cut -d_ -f1-3

c_xxxx_cp