mjhjmtu

I am using the following commandd to get the % of disk used df -h| grep workfld, than getting the top folders that are using the most disk space in %, the last piece would be to use the output/result from du -h /saswork | sort -rh | head -20 in the final grep command to get the USERID of the user who's folder this is. To make code readable here it is broke into steps

df -h| grep saswork
du -h /saswork | sort -rh | head -20
ls -la| grep %OUTPUT_FROM_COMMAND_2%

Whats the proper/easiest way to do this so I can do this repeatedly for say the top 5 folders and get the userIDs of those folders?

so results from second command would look something like this:
569G/saswork/SAS_work438800007078_prdsasgridn03/SAS_work9A0700007078_prdsasgridn03
569G/saswork/SAS_work438800007078_prdsasgridn03

There would be the top 20 results though. The piece I want to use as input in the 3rd command would be the folder name in bold above is what I would need to go into the third command, so the third command would look like this:

ls -la| grep SAS_work438800007078_prdsasgridn03

This commands output will give me the userID. I want to run this for each foldername that is produced from the output of the second command that is run.

Adding for more exact example

Input: df -h| grep saswork
Output: /dev/mapper/vg_saswork--prd-lv_sas--saswork 3.1T 2.2T 861G 73% 
/saswork

Input: du -h /saswork | sort -rh | head -20
Output: 
569G/saswork/SAS_work438800007078_prdsasgridn03/SAS_work9A0700007078_prdsasgridn03

569G /saswork/SAS_work438800007078_prdsasgridn03
526G/saswork/SAS_work445F00002079_prdsasgridn01/SAS_work189900002079_prdsasgridn01
526G /saswork/SAS_work445F00002079_prdsasgridn01
165G/saswork/SAS_workBD3300007B7E_prdsasgridn04/SAS_work154B00007B7E_prdsasgridn04
165G/saswork/SAS_workBD3300007B7E_prdsasgridn04
134G/saswork/SAS_work36E800005097_prdsasgridn04/SAS_workA86000005097_prdsasgridn04
134G /saswork/SAS_work36E800005097_prdsasgridn04
110G /saswork/SAS_workB87B00002026_prdsasgridn01/SAS_workD37900002026_prdsasgridn01
110G /saswork/SAS_workB87B00002026_prdsasgridn01
105G/saswork/SAS_work55C800001BDA_prdsasgridn01/SAS_work849500001BDA_prdsasgridn01
105G /saswork/SAS_work55C800001BDA_prdsasgridn01
57G/saswork/SAS_work3FB700003AAF_prdsasgridn03/SAS_work826800003AAF_prdsasgridn03
57G /saswork/SAS_work3FB700003AAF_prdsasgridn03
55G/saswork/SAS_work8744000068D9_prdsasgridn01/SAS_work8CA9000068D9_prdsasgridn01
55G /saswork/SAS_work8744000068D9_prdsasgridn01
46G/saswork/SAS_work400B00002BFF_prdsasgridn02/SAS_work668100002BFF_prdsasgridn02
46G /saswork/SAS_work400B00002BFF_prdsasgridn02
40G/saswork/SAS_work67780000280E_prdsasgridn02/SAS_work91E90000280E_prdsasgridn02

Input: ls -la| grep **SAS_work438800007078_prdsasgridn03** NOTE: The 
foldername SAS_work438800007078_prdsasgridn03 came from one of the results 
from the second command output. That's where I need to pull it from for each 
one.
Output: drwx------. 3 **g6753** ereapp 3864 Jul 12 12:25 
AS_work438800007078_prdsasgridn03
Note - bold in this line is the ID of the developer that I need.

To keep the result of a command in a variable you just need to wrap the command between `` or $():

yourvar=`date +%Y`

or

yourvar=$(date +%Y)

yourvar will have the value of the current year (2018). You can also execute the result directly in your command:

ls -la| grep `date +%Y`

or

ls -la| grep $(date +%Y)

To achive what you want I would use a for loop:

for i in `du -h |sort -rh|awk 'print $2'|sed "s/.///g"`;do ls -la |grep $i|awk 'print $3 " " $9';done

Bear in mind that the command is a rough example and it may need to be tweeked to avoid doubles (since if your directory has subdirectories you will have various outputs for the same parent).

If I understand your question correctly, you want to see a list of the owner of the largest directories (in descending order). In other words:

1. You want to sort directories based on their size


2. Find the user ID of each of the top 20 largest directories.

If you want see the output in the ls -l format, then try this:

(du -sh ./*; ls -lh --color=no) | awk ' if($1 == "total") X = 1 else if (!X) SIZES[$2] = $1 else sub($5 "[ ]*", sprintf("%-7s ", SIZES["./" $9]), $0); print $0 ' | sort --key=5,5hr

Sample output:

drwxrwxrwx 1 root root 1017G Jun 20 15:44 Raw_data_files
drwxrwxrwx 1 root root 188G May 12 11:34 Old_data
drwxrwxrwx 1 root root 8.8G Jul 12 15:28 backups
drwxrwxrwx 1 root root 1.4G Jun 29 15:32 randomized_sets
-rwxrwxrwx 1 root root 124M May 22 10:41 archive_web.html.7z
-rwxrwxrwx 1 root root 113M Jan 15 14:51 example1.tif
-rwxrwxrwx 1 root root 113M Apr 15 13:27 example3.pdf
-rwxrwxrwx 1 root root 1.0M Apr 15 09:17 sample_info.xlsx
-rwxrwxrwx 1 root root 1.0M Apr 27 09:20 sample_info2.xlsx
-rwxrwxrwx 1 root root 1.0M Jun 12 09:18 sample_run.R

As you can see, this gives you a sorted list (based on size) of all directories and files. But you are only interested in directories/folders and their size and user id. So if you add a simple pipe to all that (the command above) and use the grep command (i.e. add | grep "^d" --color=never" to the command above), you will then only get directories listed, and files will be no longer listed.

drwxrwxrwx 1 root root 1017G Jun 20 15:44 Raw_data_files
drwxrwxrwx 1 root root 188G May 12 11:34 Old_data
drwxrwxrwx 1 root root 8.8G Jul 12 15:28 backups
drwxrwxrwx 1 root root 1.4G Jun 29 15:32 randomized_sets

So in this case, the complete command will be as follows:

(du -sh ./*; ls -lh --color=no) | awk ' if($1 == "total") X = 1 else if (!X) SIZES[$2] = $1 else sub($5 "[ ]*", sprintf("%-7s ", SIZES["./" $9]), $0); print $0 ' | sort --key=5,5hr | grep "^d" --color=never"

All you have to do next is to only add head -20 to command above to see only the top 20 largest directories.

All credit goes to Stack Overflow user Sebi. See this thread on SO for more information: Using ls to list directories and their total sizes