Shell Programming and Scripting

Hi ,

I would like to list or sort by date and time (the files are named in day and time format) where the latest file will be placed at the bottom and the earliest file be placed at the top. Can anybody help me?

My files are named in the following manner.

EG: abc_071128_144121_data
" abc " is random , " 071128 " depicts the date in yymmdd format, " 144121 " depicts the time in the hhmmss format

Input:
abc_071128_144121_data
xxx_071128_123524_data
fff_071128_122837_data
jjj_071129_010544_data
qqq_071129_002603_data
hhh_071128_120728_data
bbb_071128_113329_data

Ouput:
bbb_071128_113329_data
hhh_071128_120728_data
fff_071128_122837_data
xxx_071128_123524_data
abc_071128_144121_data
qqq_071129_002603_data
jjj_071129_010544_data

It's a bit (very) cheap but you could try this:

Code:

#!/bin/sh
for file in *
do
    echo $file | cut -d '_' -f 2,3,4 >> /tmp/sort.$$
done
for file in `sort -n < /tmp/sort.$$`
do
    ls -1 *_${file}
done
rm -f /tmp/sort.$$

Actually, cleaner would be:

Code:

ls -1 * | awk 'BEGIN { FS="_" } { print $2,$3,$4,$1}' | sort -n | awk 'BEGIN { OFS="_" } { print $4,$1,$2,$3}'

The following works for the dataset provided:

Code:

sort -t'_' -k2 -k3  datasetfile

Have not tested it on boundary cases.

Just to throw this out there...if the files are created in a specific time order irrespective of name, then the ls -t command would work nicely. With Solaris, the last modification time is used, but the -u option would use the file's access time.

Check your local man pages.

Hi,

Try sort as follow:

Code:

sort -t "_" +1 filename

Thanks all for the codes.
It's working now !!