Requirement is that I have to pick the files one by one from xyz.txt file i.e. file1 then file2 then file3 ... for file1 I have to go inside directories listed below file1 i.e. dir1, dir2,dir3, dir4. for file2 I have to go inside directories listed below file2 i.e. dir5,dir6, dir7, dir8. means like a loop. Actually, inside these directories there are some log files. i have to search for a pattern (some records)from these log files and grep these records from the files mentioned in the file xyz.txt ( this part looks easy and i can try for it).
Only thing is I am not able to loop through file xyz.txt
Thanks for the reply. Sorry if i am not clear here.
suppose file1 is as follow ( and other files file2, file3 etc also similar to this, only data is different)
dir1, dir2, dir3,dir4 contain the logs, log1.txt,log2.txt,log3.txt
log1.txt is inside dir1
log2.txt is inside dir2
log3.txt is inside dir3
log4.txt is inside dir4
dir1,dir2,dir3 etc names are always like this, i.e always word "dir" comes at the starting in directory name.
1) Traverse file xyz.txt
2) First, read file1, store the filename(file1) in some variable
3) Traverse through dir1,dir2,dir3,dir4 and cut the first field from 1-14 ( only the numbers) from the logs log1.txt, log2.txt, log3.txt, log4.txt i.e.
Well, not knowing your system nor shell (which you did not mention), based on your examples and requirements, and hoping I understood and interpreted everything correctly, I've come up with this, which works on linux and bash:
Code:
#!/bin/bash
# set -vx
function grepit () {
eval $DIRS" | cut -d\" \" -f1 > pattfile"
grep -f pattfile $fn >$fn.grep
}
FIRST=1
DIRS="cat "
while read line
do if [ "${line:0:3}" != "dir" ]
then [ "$FIRST" -eq "0" ] && grepit
fn=$line
FIRST=0
DIRS="cat "
else DIRS=$DIRS" "$line"/log*.txt"
fi
done <xyz.txt
grepit
rm pattfile
Results will finish up in respective "filename.grep". Pls. test and come back with results.
Thank You RudiC for all your efforts.
Sorry I forgot to mention system and shell.
system is AIX 6.1 and shell is ksh
I will try your code.
in your code
Code:
do if [ "${line:0:3}" != "dir" ]
i am not sure if this line of code will work in ksh.
i guess you are looking for 4 directories always. In my case it could be any number of directories and not 4 always ... You mentioned already " Always four dirs?" i missed it. very sorry for that
---------- Post updated 10-17-12 at 01:35 PM ---------- Previous update was 10-16-12 at 09:26 PM ----------
I tried to run your script by changing the code from
Code:
#!/bin/bash
to
Code:
#!/bin/ksh
and removing that "function" keyword, but it failed in this line
Code:
"${line:0:3}": bad substitution
I guess this doesn't work in ksh.
Last edited by Piyush Jakra; 10-17-2012 at 05:14 AM..
My little meek script will work for more than four directories. In that if clause you'll need to tell files from directories, that's all. Sorry I don't have access to a ksh shell. In bash you could do a check for directory like [ -d "$line" ] or file like [ -f "$line" ]; according to the man page, this should exist in ksh, or, try to find a ksh counterpart.
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10-17-2012
