Adding custom column in the output with awk

Question

I am reading file utilization on the server with below command. How can I add the hostname in my output as a first column?

Thanks in advance

df -h | grep % | awk '{OFS="\t";print $6,$5}'

Output:

/apps/inf9b2b  43%
/apps/dbclients        13%
/apps/inf9     77%

As an aside, on some platforms (and maybe some locale settings on platforms on others) the column names contain a %, so grep % or its awk equivalent won't necessarily serve their presumed purpose of filtering out the header. Perhaps you should filter on line number, not by looking for a % sign? With my answer, that would mean changing /%/ to NR>1 — Charles Duffy
– Charles Duffy, Commented Jun 5, 2018 at 20:01

Charles Duffy · Accepted Answer · 2018-06-05 19:58:10Z

1

This is a simple application of How do I use shell variables in an awk script?

df -h | awk -v hostname="$(hostname)" '/%/ {OFS="\t"; print hostname, $6, $5}'

Note that there's no need for an external grep -- just make your pattern match a condition of the awk statement.

answered Jun 5, 2018 at 19:58

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Charles Duffy

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2 Comments

Charles Duffy Over a year ago

CNR -- please be certain you copied it exactly as given. See this code running in the virtual machine at ideone.com/hC7m1I -- no syntax error there.

Charles Duffy Over a year ago

...you'll see that it's including the header line there, but that's because their header uses a % sign; changing it to NR>1 instead of /%/ would avoid the issue.

spectrum · Accepted Answer · 2018-06-05 20:22:13Z

0

You can do df -h | grep % | awk '{OFS="\t";print "hostname\t" $6,$5}'

answered Jun 5, 2018 at 20:22

spectrum

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1 Comment

JonS Over a year ago

hostname /dev/odm 0% <br/> hostname /apps 79% <br/> This it how it looks like now

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