Originally Posted by princetd001

i executed your script, i got result , but not placed as comma seperated

can you help me to resolve it .

I tried your line of code : echo ${id/~(+E)^[0]+/} $name
it does not work for me , it gives sh[10]: ${id/~(+E)^[0]+/}: bad substitution


script
#!/usr/bin/env ksh
OUTFILE=test.txt
find 20[0-1][0-9] -type f | while read path
do
name=${path##*/}
name=${name%.trns*}
id=${name%_*}
id=${id##*_}
id=${id##*000}
echo "id: $id"
echo "file name: $name"
done > ${OUTFILE}
exit




MY SCRIPT RESULT
id: 20532
file name: sasmm_fsbc_durds_id00020532_t20120112192606.dat
id: 20533
file name: sasmm_fsbc_durds_id00020533_t20120212192606.dat
id: 20534
file name: sasmm_fsbc_durds_id00020534_t20120312192606.dat


This i show i got the result

but I need it in comma seprated csv file.

I need to address one more issue ; that is each folder may contain another file with same name but extention is different, as given below

sasmm_fsbc_durds_id00020227_t20120901005046.aud.trnsfr.gz
sasmm_fsbc_durds_id00020227_t20120901005046.dat.trnsfr.gz
sasmm_fsbc_durds_id00020228_t20120901015112.aud.trnsfr.gz
sasmm_fsbc_durds_id00020228_t20120901015112.dat.trnsfr.gz
sasmm_fsbc_durds_id00020229_t20120901025124.aud.trnsfr.gz

i want to omit file with file names ends with "aud.trnsfr.gz" ,only consider file name with "dat.trnsfr.gz"


how do I omit file ends with "aud.trnsfr.gz" while loop through folders and files?