if I have a string like "sn":"1$$$$12056597.3,2595585.69$$", how can I use awk to split "1$$$$"
I tried
**cat $filename | awk -F "\"1\$\$\$\$" '{ print $2 }'**
**cat $filename | awk -F "\"1$$$$" '{ print $2 }'**
but all failed
any number of $ use
echo '"1$$$$12056597.3,2595585.69$$"' | awk -F '"1[$]+' '{ print $2 }'
exactly 4 use
echo '"1$$$$12056597.3,2595585.69$$"' | awk -F '"1[$]{4}' '{ print $2 }'
to help debug problems with escape characters in the shell you can use the built-in shell command set which will print the arguments that are being passed to awk after the shell has interpreted any escape characters and replaced shell variables
In this case the shell first interprets \$ as an escape for a plain $
set -x
echo '"1$$$$12056597.3,2595585.69$$"'|awk -F "\"1\$\$\$\$" '{ print $2 }'
+ echo '"1$$$$12056597.3,2595585.69$$"'
+ awk -F '"1$$$$' '{ print $2 }'
You can use \$ so the \$ get to awk, but \$ is interpreted in awk regular expressions as a $ anyway. At least awk is nice enough to warn you...
echo '"1$$$$12056597.3,2595585.69$$"'|awk -F "\"1\\$\\$\\$\\$" '{ print $2 }'
+ echo '"1$$$$12056597.3,2595585.69$$"'
+ awk -F '"1\$\$\$\$' '{ print $2 }'
awk: warning: escape sequence `\$' treated as plain `$'
Turn off debugging with
set +x
echo '"1$$$$12056597.3,2595585.69$$"' | awk -F '"1[$]+' '{ print $2 }' |sed 's/.\{3\}$//'
Or if you want to split both float digit:
echo '"1$$$$12056597.3,2595585.69$$"' | awk -F '"1[$]+' '{ print $2 }' |sed 's/.\{3\}$//' |awk 'BEGIN {FS=","};{print $1}'
And
echo '"1$$$$12056597.3,2595585.69$$"' | awk -F '"1[$]+' '{ print $2 }' |sed 's/.\{3\}$//' |awk 'BEGIN {FS=","};{print $2}'
Related
If given the string '1234',56789, how can I use awk to split by the sequence ',? Here  represents a literal newline character.
Right now I have,
echo $LINE | awk -F'\\\\n',' '{ print $1}'
The split doesn't happen with this. Any advice?
Try to print all fields using the value of -F
echo "1234\n',56789," | awk -F "[',]+" -v ORS="" '{$1=$1}1'
line="1234\n',56789,"; echo "$line" | awk -F "[',]+" -v ORS="" '{$1=$1; print $0}'
Output
1234\n 56789
To print a specific field
echo "1234\n',56789," | awk -F "[',]+" -v ORS="" '{$1=$1; print $1}'
line="1234\n',56789,"; echo "$line" | awk -F "[',]+" -v ORS="" '{$1=$1; print $1}'
Output
1234\n
I have a linux script for selecting the node.
For example:
4
40*r13n15:40*r10n61:40*r11n18:40*r09n15
The correct result should be:
r13n15
r10n61
r11n18
r09n15
My linux script content is like:
hostNum=`bjobs -X -o "nexec_host" $1 | grep -v NEXEC`
hostSer=`bjobs -X -o "exec_host" $1 | grep -v EXEC`
echo $hostNum
echo $hostSer
for i in `seq 1 $hostNum`
do
echo $hostSer | awk -F ':' '{print '$i'}' | awk -F '*' '{print $2}'
done
But unlucky, I got nothing about node information.
I have tried:
echo $hostSer | awk -F ':' '{print "'$i'"}' | awk -F '*' '{print $2}'
and
echo $hostSer | awk -F ':' '{print '"$i"'}' | awk -F '*' '{print $2}'
But there are wrong. Who can give me a help?
One more awk:
$ echo "$variable" | awk 'NR%2==0' RS='[*:\n]'
r13n15
r10n61
r11n18
r09n15
By setting the record separtor(RS) to *:\n , the string is broken into individual tokens, after which you can just print every 2nd line(NR%2==0).
You can use multiple separators in awk. Please try below:
h='40*r13n15:40*r10n61:40*r11n18:40*r09n15'
echo "$h"| awk -F '[:*]' '{ for (i=2;i<=NF;i+=2) print $i }'
**edited to make it generic based on the comment from RavinderSingh13.
I'm trying to map a file to a variable and take from it only the part after a '/'
For now I have this:
mapfile VAR < path_to_file
echo $VAR | awk -F '/' '{ print $2 }'
How to combine those two commands into one? I can't find any examples.
you can use
awk -F '/' '{ print $2 }' path_to_file
what you do is actually same as
mapfile VAR < test.txt && echo $VAR | awk -F '/' '{ print $2 }'
Below is a part of my script where I used different lines for different variable.In the first two lines command used is similar though not same , is there a way to reduce these five lines:
oldv=$( sed -n "${c}p" ~/grepoutput | awk '{print $3 }')
line=$(sed -n "${c}p" ~/grepoutput | awk -F":" '{print $2 }')
newv=$( sed -n "${c}p" ~/vpkglist | awk '{print $3 }' )
oldr=$( sed -n "${c}p" ~/vpkglist | awk '{print $4 }' )
newr=$( sed -n "${c}p" ~/vpkglist | awk '{print $5 }' )
In the first two lines it`s 'grepoutput' is the same file and bottom three lines 'vpkglist' remains same.
echo "this is a test:foo,bar,baz']" | grep -o -E "test:.*" | awk -F: '{ print $2 }'
foo,bar,baz']
I get '] printed at the end, how to print only the word characters and common, nothing else, in this case I need to extract only foo,bar,baz
You can use a single awk for this:
echo "this is a test:foo,bar,baz']" | awk -F 'test:' '{sub(/[^,[:alnum:]].*/, "", $2); print $2}'
foo,bar,baz
Or, you can use a single sed:
echo "this is a test:foo,bar,baz']" | sed 's/.*test://; s/[^,[:alnum:]].*//'
foo,bar,baz
echo "this is a test:foo,bar,baz']"| awk -F: '{sub(/baz../,"baz"); print $2}'
outputs
foo,bar,baz
Using gnu grep pearl regex
$ echo "this is a test:foo,bar,baz']" | grep -oP "(?<=test:)(\w,*)+"
foo,bar,baz