UNIX for Dummies Questions & Answers

Hi,

My requirement goes like this.

I search for a file using its pattern in a directory. This will be a unique file in that directory. Now i want to get the next two files after the one i find based on time stamp.

e.g.

i search for the file SRSCLAR.IN12345 in a directory.

Now the directory has a list of files. For every SRSCLAR file there will be two corresponding msg and log files. all the 3 have the same time stamp. and these files will appear in this format

Feb 12 15:49 SRSCLAR.IN12345
Feb 12 15:49 invoice_msg_20080212154210.txt
Feb 12 15:49 invoice_log_20080212154210.txt

So i need the next files after SRSCLAR.IN12345 file.

Kindly help.

Thanks,
Sunny.

Well, if you can guarantee that there are no other files being written to this directory at about that time, this shouldn't be a problem.

Code:

ls -tr1 | awk '/^filename/ && x=NR,NR==(x+2)'

where filename is your SRSC file. If you need to chop off the first entry (SRSC), add "|tail +2" to the end.

If you need more control, I recommend perl's readdir() and stat() routines, which give you exquisite control over reading the contents of the directory.

Code:

#!/bin/sh                                                                                                                           

FILE="SRSCLAR.IN12345"

TS=`find . -name $FILE -printf "%A@\n"`

find . -name "invoice_msg_*.txt" -printf "%p:%A@\n" | grep ":$TS" | cut -f1 -d:
find . -name "invoice_log_*.txt" -printf "%p:%A@\n" | grep ":$TS" | cut -f1 -d:

alternatively:

Code:

ls -tr1| grep -A2 SRSCLAR.IN12345

if you have a grep command which supports the -A flag (gnu grep does).