I have a directory structure in my system (linux) as shown below


/test/
---/test/stest1/
------/test/stest1/stest1_t2
---------/test/stest1/stest1_t2/stest1_t2_t1
---/test/stest2/
------/test/stest2/stest2_t2
---------/test/stest2/stest2_t2/stest2_t2_t1

out put should be like

directorynamewner:groupermissions

E.g.
I have two owners steve and scott
I have one group msngrp
I want to write output as below


/test/:steve:msngrp:rwx:rwx:r:
/test/stest1/:steve:msngrp:rwx:rs:r:
/test/stest1/stest1_t2:steve:msngrp:rwx:rs:r:
/test/stest1/stest1_t2/stest1_t2_t1:steve:msngrp:rwx:rs:r:
/test/stest2/:scott:msngrp:rwx:rs:r:
/test/stest2/stest2_t2:scott:msngrp:rwx:rs:r:
/test/stest2/stest2_t2/stest2_t2_t1:scott:msngrp:rwx:rs:r:




Can any one please help me for this

Thank you
linux_kv

Help you with WHAT? You haven't asked a question.

If you mean you want a SCRIPT to do this, yes, we can HELP you with it...but we're not going to write it for you. Post what you've written, and where you're stuck, and we can help you. Otherwise, check Google...there are thousands of scripting tutorials that you can easily find.

test/ ls -l -R > text.file

Will provide the output in long list format and lists subdirectories recursively- including permissions, group and ownership - not exactly in the same order you listed - output to a "text.file" -

Is this info. helpful?

Please find my script. recursive function is not working correctly(ending). If I don't use recursively it is working fine.
Please correct my mistake.



#!/bin/ksh

fdirinfo()
{
if [ "$1" != "" ] #if parameter exists, use as base folder
then cd "$1"
fi
cpath=$(pwd)
a=$(ls -R)
for files1 in $a; do

#echo "filename111= $filename1"

filename1=$(echo $files1|sed -e 's/[^-][^\/]*\///g')

filetype1=$(ls -l | grep "$filename1$" | awk '{print substr($1,1,1)}')

#echo "filename222= $filename1 filetype= $filetype1 ls -l | grep $filename1 | awk '{print substr($1,1,1)}'"

if [ "$filetype1" = "d" ]
then

owner1=$(ls -l | grep "$filename1$" | awk '{print $3}')
ownerpermission1=$(ls -l | grep "$filename1$" | awk '{print substr($1,2,3)}')
ownerpermission1=$(echo $ownerpermission1|sed -e 's/-//g')

group1=$(ls -l | grep "$filename1$" | awk '{print $4}')
grouppermission1=$(ls -l | grep "$filename1$" | awk '{print substr($1,5,3)}')
grouppermission1=$(echo $grouppermission1|sed -e 's/-//g')

otherpermission1=$(ls -l | grep "$filename1$" | awk '{print substr($1,8,1)}')

echo "$cpath/$filename1:$owner1:$group1:$ownerpermission1:$grouppermission1:$otherpermission1:"
#fdirinfo filename1
cd $cpath
fi
done
}
fdirinfo $1

Ok, that seems a little complicated. Try this one instead:
The above assumes you only want to have files listed in your output and not directories (remove "-type f" from find if you want to include directories). Also, it's probably better to use octal modes (permissions) when scripting. Keep in mind that you really don't need use separate fields for $ownerpermission1:$grouppermission1 because an octal value like the one in the above script already contains both (a file with mode 644 equals user=r+w,group=r,world=r).

Take a look at the -printf predicate of find, e.g.

Very useful, thanks! Didn't realize find had this feature. So the solution to his problem is reduced to a single command, awesome .. and the final script would be (with line-feed and octal modes):

Thanks for the replies I will try your suggestions however I fixed my script
I made a mistake fdirinfo "filename1" instead of fdirinfo "$filename1"
here is my script


dirName="."

function fdirinfo()
{
if !(test -d "$1") #check the parameter is directory
then echo $1; return;
fi

cd "$1" #change to directory

for filename1 in * #loop trhough all the files
do
if test -d "$filename1" #if directory
then

owner1=$(ls -l | grep "$filename1$" | awk '{print $3}') #get the owner
ownerpermission1=$(ls -l | grep "$filename1$" | awk '{print substr($1,2,3)}') #get the permissions
ownerpermission1=$(echo $ownerpermission1|sed -e 's/-//g') #replace any -

group1=$(ls -l | grep "$filename1$" | awk '{print $4}') #get the group
grouppermission1=$(ls -l | grep "$filename1$" | awk '{print substr($1,5,3)}') #get the permissions
grouppermission1=$(echo $grouppermission1|sed -e 's/-//g') #replace any -

otherpermission1=$(ls -l | grep "$filename1$" | awk '{print substr($1,8,1)}')

echo "$(pwd)/$filename1:$owner1:$group1:$ownerpermission1:$grouppermission1:$otherpermission1:" #make string


fdirinfo "$filename1" #call recursively
cd .. #change directory to previous one
fi
done
}




if [ "$1" != "" ] #if parameter exists, use as base folder
then dirName=$1
fi


fdirinfo "$dirName"

I used your suggestions and wrote



find $dirName \( ! -regex '.*/\..*' \) -type d -printf "%p:%u:%g:%M:\n" | grep -v "lost+found" | sed -e 's/:/|/3'| awk '{print substr($0,0,index($0,"|")-1) ":" substr($0,index($0,"|")+2,3) ":" substr($0,index($0,"|")+5,3) ":" substr($0,index($0,"|")+8,3) }' | sed -e 's/-//g' -e 's/:$//' | awk '{print $0 ":" } ' >$2



#this command give output in a specific format (dir with fullpath : owner:groupwnerpermissions:grouppermissions and redirects to a file
#find command gets all the directories (non hidden) and printf prints path user group and permssions
#grep command ignores lost+found directory
#sed replaces third : as |
#awk prints anything before | and substring of remaining portion
#sed replace any -
#and replaces if last char is : this is required because sometimes replace all the - will leave :
#awk prints all the output and also : at the end