cut 或 awk 命令打印第一行的第一个字段

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

SUSE
VERSION
PATCHLEVEL

Answer 1

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}'

Answer 2

使用NR内置变量指定行号。

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

Answer 3

试试这个：

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

Answer 4

您可以使用head代替cat ：

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

Answer 5

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

如果制表符分隔：

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

Answer 6

Try

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

where NUM is line number

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

Answer 7

awk、sed、管道，这很重

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

Answer 8

You can kill the process which is running the container.

With this command you can list the processes related with the docker container:

ps -aux | grep $(docker ps -a | grep container-name | awk '{print $1}')

Now you have the process ids to kill with kill or kill -9 .

Answer 9

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.

Answer 10

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

1. first pass through exit block, "_" is undefined, so it fails and skips over for row 1.

2. 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.

3. 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=_'

Answer 11

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

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