cut or awk command to print first field of first row

Question

I am trying print the first field of the first row of an output. Here is the case. I just need to print only SUSE from this output.

# cat /etc/*release

SUSE Linux Enterprise Server 11 (x86_64)
VERSION = 11
PATCHLEVEL = 2

Tried with cat /etc/*release | awk {'print $1}' but that print the first string of every row

SUSE
VERSION
PATCHLEVEL

score 244 · Accepted Answer · edited Apr 12 '19 at 16:19

244

Specify NR if you want to capture output from selected rows:

awk 'NR==1{print $1}' /etc/*release

An alternative (ugly) way of achieving the same would be:

awk '{print $1; exit}'

An efficient way of getting the first string from a specific line, say line 42, in the output would be:

awk 'NR==42{print $1; exit}'

edited Apr 12 '19 at 16:19

JohannesM

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answered Mar 05 '14 at 07:07

devnull

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@jaypal I also thought of adding a `tac file | awk 'END{print $1}'` but then realized it might be a bit too much. – devnull Mar 05 '14 at 07:15
Though that would make your *ugly* solution look **remarkably** prettier against the proposed *hideous* solution! – jaypal singh Mar 05 '14 at 07:19
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It not `ugly`. Its even better, since on a large file, it will save lots of time by exit on found and stops processing. – Jotne Mar 05 '14 at 08:02
@Jotne I termed it ugly simply because `NR==1` was _implicit_. Added another example in the answer that probably clarifies a bit more. – devnull Mar 05 '14 at 08:07
@devnull I do know, but OPs requirement is clear `print the first string of the first row of an output` so here the `exit` would be just fine. – Jotne Mar 05 '14 at 08:12
@Jotne You know it, so it's perhaps obvious. To someone new, it might not be. – devnull Mar 05 '14 at 08:20
is `exit` on the third example (after specifying `NR`) really necessary? – Oct 02 '18 at 11:54

score 25 · Answer 2 · answered Mar 05 '14 at 07:07

25

Specify the Line Number using NR built-in variable.

awk 'NR==1{print $1}' /etc/*release

answered Mar 05 '14 at 07:07

jaypal singh

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score 11 · Answer 3 · answered Mar 05 '14 at 08:32

11

try this:

head -1 /etc/*release | awk '{print $1}'

answered Mar 05 '14 at 08:32

developer

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retr0man · Answer 4 · 2021-07-17T22:03:13.233

5

df -h | head -4 | tail -1 | awk '{ print $2 }'

Change the numbers to tweak it to your liking.

Or use a while loop but thats probably a bad way to do it.

edited Jul 17 '21 at 22:03

answered Jul 17 '21 at 21:57

retr0man

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Now that's a nifty hack! – Abhinav Thakur Aug 03 '21 at 08:06

score 4 · Answer 5 · edited Mar 05 '14 at 07:16

4

You could use the head instead of cat:

head -n1 /etc/*release | awk '{print $1}'

edited Mar 05 '14 at 07:16

MLSC

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answered Mar 05 '14 at 07:09

Junaid18183

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score 3 · Answer 6 · edited May 06 '16 at 12:54

3

sed -n 1p /etc/*release |cut -d " " -f1

if tab delimited:

sed -n 1p /etc/*release |cut -f1

edited May 06 '16 at 12:54

Artem Mostyaev

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answered May 06 '16 at 11:50

Raha Pazoki

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score 3 · Answer 7 · answered Sep 24 '16 at 02:02

3

awk, sed, pipe, that's heavy

set `cat /etc/*release`; echo $1

answered Sep 24 '16 at 02:02

Emmanuel

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Jayesh Bhoi · Answer 8 · 2014-03-05T07:30:13.470

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Try

sed 'NUMq;d'  /etc/*release | awk {'print $1}'

where NUM is line number

ex. sed '1q;d'  /etc/*release | awk {'print $1}'

edited Mar 05 '14 at 07:30

answered Mar 05 '14 at 07:15

Jayesh Bhoi

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This would print entire 1st line and not 1st field as OP stated. – jaypal singh Mar 05 '14 at 07:24
Answer saras che pan thodu laambu che. Khali `sed` thi pan kari sakiye - `sed -r '1s/([^ ]+) .*/\1/;q' /etc/*release` `:)` – jaypal singh Mar 05 '14 at 07:37

RARE Kpop Manifesto · Answer 9 · 2022-07-21T10:56:15.690

the most code-golfy way i could think of to print first line only in awk :

awk '_{exit}--_'    # skip the quotations and make it just
                    #   awk _{exit}--_
                    #
                    # if u're feeling adventurous

first pass through exit block, "_" is undefined, so it fails and skips over for row 1.
then the decrementing of the same counter will make it "TRUE" in awk's eyes (anything not empty string or numeric zero is considered "true" in their agile boolean sense). that same counter also triggers default action of print for row 1.
```
—- incrementing… decrementing… it's same thing, 
   merely direction and sign inverted.
```
then finally, at start of row 2, it hits criteria to enter the action block, which instructs it to instantly exit, thus performing essentially the same functionality as

awk '{ print; exit }'

… in a slightly less verbose manner. For a single line print, it's not even worth it to set FS to skip the field splitting part.

using that concept to print just 1st row 1st field :

awk '_{exit} NF=++_'
awk '_++{exit} NF=_'

ufopilot · Answer 10 · 2022-07-21T19:39:26.127

0

awk 'NR==1&&NF=1' file
grep -om1 '^[^ ]\+' file

# multiple files
awk 'FNR==1&&NF=1' file1 file2

edited Jul 21 '22 at 19:39

answered Jul 21 '22 at 19:28

ufopilot

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cut or awk command to print first field of first row

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