UNIX for Dummies Questions & Answers

hello i'm trying to figure out how to tack the directory name onto the file name when i do the ls -ltrR command please?

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I do the following command ls -ltrR > ls.out

and get the following output:
-----------------------------------------ls.out----------------------------------
./PERL:
total 180
-rw-r--r-- 1 kennellr datawhse 24595 May 21 2004 bob.pl
-rw-r--r-- 1 kennellr datawhse 23965 May 21 2004 rdmtools_DBUtil.pm
-rw-r--r-- 1 kennellr datawhse 38094 May 21 2004 rdmtools_functions.pm
-rw-r--r-- 1 kennellr datawhse 27 May 21 2004 date.ksh
-rw-r--r-- 1 kennellr datawhse 1189 May 24 2004 date.pl

./CONFIG:
total 0

./RUNCNTL:
total 8
drwxr-xr-x 2 kennellr datawhse 512 May 26 2004 RUN3
drwxr-xr-x 2 kennellr datawhse 512 May 26 2004 RUN1
drwxr-xr-x 2 kennellr datawhse 512 May 26 2004 CONFIG
drwxr-xr-x 3 kennellr datawhse 512 May 26 2004 RUN2

./RUNCNTL/RUN3:
total 0
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i then run a script that has the following read command in it:

while read m1 m2 m3 m4 m5 m6 m7 m8 m9
do
echo $m5 $m9
done < ls.out

and it produces the following output:
-----------------------------------------------------------------------------------
24595 bob.pl
23965 rdmtools_DBUtil.pm
38094 rdmtools_functions.pm
27 date.ksh
1189 date.pl
512 RUN3
512 RUN1
512 CONFIG
512 RUN2
-----------------------------------------------------------------------------------
But in order to build commands from this list i need the directory name so does any one know if i can modify the

ls -ltr command so it creates a list of files with the directory name tacked onto the file name like so?

-rw-r--r-- 1 kennellr datawhse 24595 May 21 2004 /PERL/bob.pl

versus

-rw-r--r-- 1 kennellr datawhse 24595 May 21 2004 bob.pl

Not certain you can do that directly with ls, or why that would be necessary, however you can get the output you want other ways.

assuming ksh as your shell.

Code:

DIR=`dirname $PWD`
BASE=`basename $PWD`

 for i in *
 do
    echo ${DIR}/${BASE}/$i | tee -a /tmp/outt
 done

ok thanks i'll try that out!

bk

ok that gave me an idea and i found this in an old script and it seems to to the job sortof now i just need to figure out how to trim the . & : out of the concatenated variable name:

export dir=" "
while read m1 m2 m3 m4 m5 m6 m7 m8 m9 m10
do
echo $m1 | grep ./ > /dev/null
if [ $? = 0 ]; then
dir=$m1
fi
echo $m5" "$dir/$m9
done < ls.out

907 ./pldir:/rdmtp_show_constants01.cgi
2734 ./pldir:/rdmtp_show_chart01.cgi
3693 ./pldir:/rdmtp_search_chart1.cgi
31271 ./pldir:/rdmtp_run_consistency02.cgi
10456 ./pldir:/rdmtp_rbc_showNMI01.cgi
8962 ./pldir:/rdmtp_rates_show01.cgi
57973 ./pldir:/allpl
13050 ./pldir:/rdmtp_financials_show02.cgi
512 ./perl:/rdmtools
438 ./perl:/substitute_char.pl
575 ./perl:/substitute_and_remove_delinator.pl
1013 ./perl:/scan_data.pl
464 ./perl:/remove_deliminator.pl
1444 ./perl:/bcp_out_tables.pl
4774 ./perl:/verbose_add.pl
8708 ./perl:/all
23932 ./perl:/dnr1
23932 ./perl:/dnr1.pl
63694 ./perl:/rdmtools_lib.pm
36180 ./perl:/rdmtools_gap_calculate.cgi
38198 ./perl:/rdmtools_functions.pm
37954 ./perl:/rdmtools_financials_diff.cgi

not sure which '.' you need 'trimmed out', but....

Code:

echo '907 /pldir/rdmtp_show_constants01.cgi' | sed 's#[.:]/#/#g'

thanks vgersh99 works great, have a good weekend all

bk