how can I combine the following two awk commands into a single line:
awk -F= '$1=="ID" { print $2 ;}' /etc/*release && awk -F= '$1=="VERSION_ID" { print $2 ;}' /etc/*release | xargs
Im trying to get the linux distribution and os in a single line, in the format distribution+version.
For example: ubuntu20.04, rhel7.5
With bash:
source /etc/os-release; echo "$ID $VERSION_ID"
You can capture each part into a variable and then print them out once you have processed the file:
awk -F= '$1=="ID"{id=$2}$1=="VERSION_ID"{vid=$2}END{print id,vid}' /etc/*release
I assume you don't want the quotes (as in your example), in that case:
awk -F '["=]' '$1=="ID" {printf("%s,",$3)} $1=="VERSION_ID" {printf("%s\n",$3)}' < /etc/*release
Related
I have many files with different names that end with txt.
rtfgtq56.txt
fgutr567.txt
..
So I am running this command
for i in *txt
do
awk -F "\t" '{print $2}' $i | grep "K" | awk '{print}' ORS=';' | awk -F "\t" '{OFS="\t"; print $i, $1}' > ${i%.txt*}.k
done
My problem is that I want to add the name of every file in the first column, so I run this part:
awk -F "\t" '{OFS="\t"; print $i, $1}' > ${i%.txt*}
$i means the file that are in the for loop,
but it did not work because awk can't read the $i in the for loop.
Do you know how I can solve it?
You want to refactor eveything into a single Awk script anyway, and take care to quote your shell variables.
for i in *.txt
do
awk -F "\t" '/K/{a = a ";" $2}
END { print FILENAME, substr(a, 1) }' "$i" > "${i%.txt*}.k"
done
... assuming I untangled your logic correctly. The FILENAME Awk variable contains the current input file name.
More generally, if you genuinely want to pass a variable from a shell script to Awk, you can use
awk -v awkvar="$shellvar" ' .... # your awk script here
# Use awkwar to refer to the Awk variable'
Perhaps see also useless use of grep.
Using the -v option of awk, you can create an awk Variable based on a shell variable.
awk -v i="$i" ....
Another possibility would be to make i an environment variable, which means that awk can access it via the predefined ENVIRON array, i.e. as ENVIRON["i"].
/Home/in/test_file.txt
echo /Home/in/test_file.txt | awk -F'/' '{ print $2,$3 }'
Gives the result as:
Home in
But I need /Home/in/ as the result .I have to get all except test_file.txt
How to achieve this?
$ echo '/Home/in/test_file.txt' | awk '{sub("/[^/]+$","")} 1'
/Home/in
$ echo '/Home/in/test_file.txt' | awk '{sub("[^/]+$","")} 1'
/Home/in/
$ echo '/Home/in/test_file.txt' | sed 's:/[^/]*$::'
/Home/in
$ echo '/Home/in/test_file.txt' | sed 's:[^/]*$::'
/Home/in/
$ dirname '/Home/in/test_file.txt'
/Home/in
Your attempt awk -F'/' '{ print $2,$3 }' didn't do what you wanted as -F'/' is telling awk to split the input into fields at every / and then print $2,$3 is telling awk to print the 2nd and 3rd fields separated by a blank char (the default value for OFS). You could do:
$ echo '/Home/in/test_file.txt' | awk 'BEGIN{FS=OFS="/"} { print "",$2,$3,"" }'
/Home/in/
to get the expected output but it'd be the wrong approach since it's removing the field you don't want AND removing the input separators AND then adding new output separators which happen to the have the same value as the input separators rather than simply removing the field you don't want like the other solutions above do.
echo /Home/in/test_file.txt | awk -F'/[^/]*$' '{ print $1 }'
..will print the everything but the trailing slash
There are several ways to achieve this:
Using dirname:
$ dirname /home/in/test_file.txt
/home/in
Using Shell substitution:
$ var="/home/in/test_file.txt"
$ echo "${var%/*}"
/home/in
Using sed: (See Ed Morton)
Using AWK:
$ echo "/home/in/test_file.txt" | awk -F'/' '{OFS=FS;$NF=""}1'
/home/in/
Remark: all these work since you can't have a filename with a forward slash (Is it possible to use "/" in a filename?)
Note: all but dirname will fail if you just have a single file_name without a path. While dirname foo will return ./ all others will return foo
awk behaves as it should.
When you define slash / as a separator, the fields in your expression become the content between the separators.
If you need the separator to be printed as well, you need to do it explicitly, like:
echo /Home/in/test_file.txt | awk -F'/' '{ printf "%s/%s/",$2,$3 }'
replace your last field with an empty string and
put the slash back in as the (builtin) Output Field Separator (OFS)
echo /Home/in/test_file.txt | awk -F'/' -vOFS='/' '{$NF="";print}
In a bash script, I need to do this:
cat<<EOF> ins.exe
grep 'pattern' file | awk '{print $2}' > results
EOF
The problem is that $2 is interpreted as a variable and the file ins.exe ends up containing
"grep 'pattern' file | awk '{print }' > results", without the $2.
I've tried using
echo "grep 'pattern' file | awk '{print $2}' > results" >> ins.exe
But it's the same problem.
How can I fix this?
Just escape the $:
cat<<EOF> ins.exe
awk '/pattern/ { print \$2 }' file > results
EOF
No need to pipe grep to awk, by the way.
With bash, you have another option as well, which is to use <<'EOF'. This means that no expansions will occur within the string.
I spent on this 2 hours and get nothing. I want to get $1 and $2 as a first command line input of shell script, but I couldn't manage this. And $3 and $0 would be columns in awk. I try different methods but nothing works for me.
awk -F':' -v "limit=1000" '{ if ( $3 >=limit ) gsub("~/$1/",~/$2/); print \$0}' file.txt
the cleanest method is to explicitly pass the values from shell to awk with awk's -v option:
awk -F: -v limit=1000 -v patt="~/$1/" -v repl="~/$2/" '
$3 >=limit {gsub(patt,repl); print}
' file.txt
When your awk line is part of a script file, and you want to use $1 and $2 from the script in your awk command, you should temporary stop the literal string with a single quote and start it again.
awk -F':' -v "limit=1000" '{ if ( $3 >=limit ) gsub("~/'$1'/",~/'$2'/); print $0}' file.txt
You didn't post any sample input or expected output so this is a guess but you probably want something like this:
awk -F':' -v limit=1000 -v arg1="$1" -v arg2="$2" '$3 >= limit{gsub("~/" arg1 "/","~/" arg2 "/"); print}' file.txt
I have this command which executes correctly if run directly on the terminal.
awk '/word/ {print NR}' file.txt | head -n 1
The purpose is to find the line number of the line on which the word 'word' first appears in file.txt.
But when I put it in a script file, it doens't seem to work.
#! /bin/sh
if [ $# -ne 2 ]
then
echo "Usage: $0 <word> <filename>"
exit 1
fi
awk '/$1/ {print NR}' $2 | head -n 1
So what did I do wrong?
Thanks,
Replace the single quotes with double quotes so that the $1 is evaluated by the shell:
awk "/$1/ {print NR}" $2 | head -n 1
In the shell, single-quotes prevent parameter-substitution; so if your script is invoked like this:
script.sh word
then you want to run this AWK program:
/word/ {print NR}
but you're actually running this one:
/$1/ {print NR}
and needless to say, AWK has no idea what $1 is supposed to be.
To fix this, change your single-quotes to double-quotes:
awk "/$1/ {print NR}" $2 | head -n 1
so that the shell will substitute word for $1.
You should use AWK's variable passing feature:
awk -v patt="$1" '$0 ~ patt {print NR; exit}' "$2"
The exit makes the head -1 unnecessary.
you could also pass the value as a variable to awk:
awk -v varA=$1 '{if(match($0,varA)>0){print NR;}}' $2 | head -n 1
Seems more cumbersome than the above, but illustrates passing vars.