Shell Programming and Scripting

Let awk split on _ and print if the number of fields is less than 6:

Code:

 ls *.zip | awk -F_ 'NF<6'
MYTEST001_RKP_DORALDO_20150402120000.zip

Thanks to all for quick reply. But I am getting some problem because of difference in Unix and Linux option available for command.

I have 2 set of file..

First Set these should be treated seperatly

Code:

EFC10668_LAB_COVANCE_20150415120000.dat
EFC10668_LAB_COVANCE_20150415120000_1.dat
EFC10668_LAB_COVANCE_20150415120000_C.dat

Second Set these should be treated differently

Code:

AVE0005_EFC10668_COVANCE_LABL_20150409.dat
AVE0005_EFC10668_COVANCE_LABL_20150409_C.dat
AVE0005_EFC10668_COVANCE_LABL_20150409_1.dat

In linux i am counting the _ in filename but in HP UX I am able find any option.

Code:

grep -o

currently I am using below If condition to separate but I thing it is not going to work properly.

Code:

if [ "$(ls $n_filezip_name | awk '{FS="_"}  $5~/[0-9]{8}/')" = "" ];

Can someone help on this.

I believe, counting of "_" is quite posixish:

Code:

echo "a_b_c_d" | awk -F_ '{print NF-1}'