How to find disk usage more than 70% and find largest in there?

Jokoenom · 10-29-2019, 03:43 AM

Dear friends

I have a problem, how to find disk usage more than 70%, can be found forwarded by finding the largest file in them, then displaying it?

Please help me with the shell script, thanks for the attention.

Turbocapitalist · 10-29-2019, 03:47 AM

Welcome.

That would be done using some combination of du and df plus either AWK or various bashisms.

See

Code:

man du
man df

What have you tried so far and where are you stuck?

pan64 · 10-29-2019, 05:25 AM

there was a similar thread here a few weeks ago, ncdu was suggested.

rtmistler · 10-29-2019, 06:34 AM

I'm sure there have been many similar threads over the years.

Jokoenom,

Welcome to LQ. Please understand that all LQ members, yourself included are free volunteers, we're not paid support.

We're happy to help, but the way this forum works best is if you work along side and solve this problem, with our assistance. The few reasons for this, among many others, are that you'll learn and be capable of doing more and more over time, plus you'll also stand to gain a resolution of your problem or project along exactly the lines you need or would like. Because if someone emits out a copy of an old script that they used, it may have flaws, special cases/directories, or considerations that has nothing to do with your setup. Therefore something like that may cause more problems/confusion than it's worth.

So for starters, do you know anything about shell scripting? And what language would you consider to use, if not just bash?

Here are some BASH references, and in my signature, these references are in a blog, but also that blog covers some techniques for writing and working through shell scripts, which you may find useful, one of the most important statements says something like, "anything you can type into a command line, you can put into a bash script", please take the time to review and update us what you have in mind moving forwards:

Bash Beginners Guide
Advanced Bash Guide
My Bash Blog

Jokoenom · 10-30-2019, 09:54 AM

I have tried this script, but I have not found exactly what I meant, what is needed is only to find the largest file from the filesystem which has 70% utilization.

Turbocapitalist · 10-30-2019, 10:05 AM

AWK is a scripting language. So the script has to be within quotes. Here it is within single quotes ' so that the field numbers do not need to be escaped.

Code:

df -Ph | \
awk '{sub("%","",$5);} $5+1>90 {printf("%02d%\t%s\n",$5,$6);f++} END{if(f){exit(1)}}' \
&& echo OK || echo OMGWTFBBQ

The sub() removes the % from the fifth field.

The $5+1 ensures that the field comparison will be as a number, not a string.

The f is a flag and causes the AWK script to later exit with a non-zero status.

Finding the largest file would be a second script.

Jokoenom · 10-30-2019, 10:32 AM

Finding the largest file will be the second script.

Can it be combined into the first script sir? So if there is a fileusage of more than xx% then look for and display the largest file with the head -10? thank you very much, iam sorry for bothering you.

Jokoenom · 10-30-2019, 10:43 AM

Quote:

Originally Posted by Jokoenom

Finding the largest file will be the second script.

Can it be combined into the first script sir? So if there is a fileusage of more than xx% then look for and display the largest file with the head -10? thank you very much, iam sorry for bothering you.

[root@localhost ~]# df -Ph | awk '{sub("%","",$5);} $5+1>10 {printf("%02d%\t%s\n",$5,$6);f++} END{if(f){exit(1)}}' && echo OK || echo OMGWTFBBQ
17% /
14% /boot
OMGWTFBBQ
[root@localhost ~]# df -Ph | awk '{sub("%","",$5);} $5+1>70 {printf("%02d%\t%s\n",$5,$6);f++} END{if(f){exit(1)}}' && echo OK || echo OMGWTFBBQ
OK