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How to display only Owner and directory/sub directory names under particular root

 
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Top Forums UNIX for Dummies Questions & Answers How to display only Owner and directory/sub directory names under particular root
# 1  
Old 09-02-2010
MySQL How to display only Owner and directory/sub directory names under particular root
hai,
I am new to Unix, I have a requirement to display owner name , directory or sub directory name, who's owner name is not equal to "oasitqtc".
(here "oasitqtc" is the owner of the directory or sub directory.)

i have a command (below) which will display all folders and sub folders, but i don't want to display all the information of directory like permission, group...etc.

just i want to display only owner name, directory or sub directory name (who's owner is not equal to "oasitqtc")

Code:
find . -type d -exec ls -ld {} \; > ~/folder_owners.txt

Out put of the above command :

Code:
drwxr-xr-x  2 oasitqtc dba 4096 Dec 28  2004 ./xtr/11.5.0/patch/115/xml/US
drwxr-xr-x  3 oasitbvp dba 4096 Oct  3  2002 ./xtr/11.5.0/reports
drwxr-xr-x  2 oasitbvp dba 4096 Dec 28  2004 ./xtr/11.5.0/reports/US
drwxr-xr-x  2 oasitqtc dba 4096 Oct  3  2002 ./xtr/11.5.0/sql
drwxrwxrwx  3 oasitqtc dba 4096 Nov  4  2004 ./xxau
drwxrwxrwx  19 oasitbvp dba 4096 Aug 17 23:46 ./xxau/11.5.0
drwxrwxrwx  4 abcdxx dba 4096 Aug  4  2005 ./xxau/11.5.0/admin
drwxrwxrwx  3 oacdxx dba 4096 Aug  4  2005 ./xxau/11.5.0/admin/import
drwxrwxrwx  2 oasitqtc dba 4096 Aug  4  2005 ./xxau/11.5.0/admin/import/US

from the above output file, i need to display below data only.
ex
Code:
: oasitbvp  ./xtr/11.5.0/reports
      oasitbvp  ./xtr/11.5.0/reports/US
     oasitbvp   ./xxau/11.5.0
     abcdxx     ./xxau/11.5.0/admin
     oacdxx    ./xxau/11.5.0/admin/import

really appreciate any help.
Thank you.
Last edited by pludi; 09-02-2010 at 03:56 AM.. Reason: code tags, please.
# 2  
Old 09-02-2010
Bug
Try using the below code :


Code:
find . -type d -exec ls -ld {} \; | awk '{print $3,$9}' | grep -v oasitqtc

It should give you desired output

However better code will be with egrep as states below :

Code:
find . -type d -exec ls -ld {} \; | awk '{print $3,$9}' | egrep  -v "^oasitqtc"

Try both Smilie
Last edited by Paarth; 09-02-2010 at 04:26 AM..
This User Gave Thanks to Paarth For This Post:
gagan4599
# 3  
Old 09-02-2010
Quote:
Originally Posted by Paarth
Try using the below code :


Code:
find . -type d -exec ls -ld {} \; | awk '{print $3,$9}' | grep -v oasitqtc

It should give you desired output

However better code will be with egrep as states below :

Code:
find . -type d -exec ls -ld {} \; | awk '{print $3,$9}' | egrep  -v "^oasitqtc"

Try both Smilie
Unnecessary to use grep with awk:
Code:
find . -type d -exec ls -ld {} \; | awk '$3 != "oasitqtc"{print $3, $NF} '

This User Gave Thanks to Franklin52 For This Post:
gagan4599
# 4  
Old 09-02-2010
Thank you very much "Franklin52" your given command is working fine.

Hi Paarth,
"find . -type d -exec ls -ld {} \; | awk '{print $3,$9}' | grep -v oasitqtc" command is not working,
but if i replaced $9 with $NF then the command is working fine.
Thank you very much for your quick response.
# 5  
Old 09-02-2010
It easy:
Code:
find . -type d ! -user oasitqtc

or if you wants find directores and subdirectores whole system
Code:
find / -type d ! -user oasitqtc

or if you wants find directores and subdirectores whole filesystem "/"
Code:
find / -type d ! -user oasitqtc -xdev

# 6  
Old 09-02-2010
Better to use "find" to eliminate what you don't want.
With care you can deal with directory names containing space characters and produce the output in alphabetical order of directory name.


Code:
find . ! -user oasitqtc -type d -print | sort | while read DIR
do
        OWNER=`ls -ld "${DIR}" |awk '{print $3}'`
        echo "${OWNER} ${DIR}"
done

# 7  
Old 09-02-2010
When there is little depth of directories. It is easier to do:
Code:
ls -ldR|awk '$3 !== "oasitqtc" {print $3,$NF}'

now no problem if a directory name include word "oasitqtc"
Methyl, I think it's very good:
Code:
find . ! -user oasitqtc -type d -print | sort | while read DIR
do
        OWNER=`ls -ld "${DIR}" |awk '{print $3}'`
        echo "${OWNER} ${DIR}"
done

 
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