The string format is
Executed: variable_name
What is the simplest way to get the *variable_name* sub-string?
foo="Executed: variable_name"
echo ${foo##* } # Strip everything up to and including the rightmost space.
or
set -- $foo
echo $2
Unlike other solutions using awk, sed, and whatnot, these don't fork other programs and save thousands of CPU cycles since they execute completely in your shell. They are also more portable (unlike ${i/* /} which is a bashism).
With sed:
echo "Executed: variable_name" | sed 's/[^:]*: //'
Using awk:
echo 'Executed: variable_name' | awk -F' *: *' '{print $2}'
variable_name
If you have the string in a variable:
$ i="Executed: variable_name"
$ echo ${i/* /}
variable_name
If you have the string as output of a command
$ cmd
Executed: variable_name
$ cmd | awk '{print $NF}'
variable_name
Note that 'NF' means "number of fields", so $NF is always the last field on the line. Fields are assumed to be separated by spaces (unless -F is specified)
If your variable_name could have spaces in it, the
-F' *: *'
mentioned previously ensures that only the ": " is used as a field separator. However, this will preserve spaces at the end of the line if there are any.
If the line is mixed in with other output, you might need to filter.
Either grep..
$ cmd | grep '^Executed: ' | awk '{print $NF}'
or more clever awk..
$ cmd | awk '/^Executed: /{print $NF}'
Related
I am having trouble saving variables from file using grep/sed/awk.
The text in file.txt is on the form:
NUM_ITER = 1000 # Number of iterations
NUM_STEP = 1000
And I would like to save these to bash variables without the comments.
So far, I have attempted this:
grep -oP "^NUM_ITER[ ]*=\K.*#" file.txt
which yields
1000 #
Any suggestions?
I would use awk, like this:
awk -F'[=[:blank:]#]+' '$1 == "NUM_ITER" {print $2}' file
To store it in a variable:
NUM_ITER=$(awk -F'[=[:blank:]#]+' '$1 == "NUM_ITER" {print $2}' file)
As long as a line can only contain a single match, this is easy with sed.
sed -n '# Remove comments
s/[ ]*#.*//
# If keyword found, remove keyword and print value
s/^NUM_ITER[ ]*=[ ]*//p' file.txt
This can be trimmed down to a one-liner if you remove the comments.
sed -n 's/[ ]*#.*//;s/^NUM_ITER[ ]*=[ ]*//p' file.txt
The -n option turns off printing, and the /p flag after the final substitution says to print that line after all only if the substitution was successful.
I'm having some trouble passing bash script variables into awk command-line.
Here is pseudocode:
for FILE in $INPUT_DIR/*.txt; do
filename=`echo $FILE | sed -n 's/^.*\(chr[0-9A-Z]*\).*.vcf$/\1/p'`
OUTPUT_FILE=$OUTPUT_DIR/$filename.snps.txt
egrep -v "^#" $FILE | awk '{print $2,$4,$5}' > $OUTPUT_FILE
done
The final line where I awk the columns, I would like it to be flexible or user input. For example, the user could want columns 6,7,and 8 as well, or column 133 and 138, or column 245 through 248. So how do I custom this so I can have that 'print $2 .... $5' be a user input thing? For example the user would run this script like : bash script.sh input_dir output_dir [user inputs whatever string of columns], and then I would get those columns in the output. I tried passing it in, but I guess I'm not getting the syntax right.
With awk, you should declare the variable before use it. This is better than the escape method (awk '{print $'$var'}'):
awk -v var1="$col1" -v var2="$col2" 'BEGIN {print var1,var2 }'
Where $col1 and $col2 would be the input variables.
Maybe you can try an input variable as string with "$2,$4,$5" and print this variable to get the values (I am not sure if this works)
The following test works for me:
A="\$3" ; ls -l | awk "{ print $A }"
I'd like to run:
grep nfs /etc/fstab | awk '{print $2}'
[root#nyproxy5 ~]# grep nfs /etc/fstab | awk '{print $2}'
/proxy_logs
/proxy_dump
/sync_logs
[root#nyproxy5 ~]#
And to get the output in one line delimited by space.
How can I do that?
If you don't mind a space (and no newline) at the end, you could do use this awk script:
awk '/nfs/{printf "%s ", $2}' /etc/fstab
For lines that match the pattern /nfs/, the second column is printed followed by a space. As a general rule, piping grep into awk is unnecessary as awk can do the pattern matching itself.
If you would like a newline at the end, you could use the END block:
awk '/nfs/{printf "%s ", $2}END{print ""}' /etc/fstab
This prints an empty string, followed by the output record separator (which is a newline). This will mean that you always have a newline in the output even if no matching records were found. If that's a problem, you could use a flag:
awk '/nfs/{f=1;printf "%s ", $2}END{if(f)print ""}' /etc/fstab
The flag f is set to true if the pattern is ever matched, causing the newline to be printed.
Newlines or any other character could be removed or replaced with tr command:
grep nfs /etc/fstab | awk '{print $2}' | tr -c '\n' ' '
If you want to get rid of tabs also:
| tr -c '\t' ' '
I am working on a script where I am grepping lines that contains -abc_1.
I need to extract string that appear just after this string as follow :
option : -abc_1 <some_path>
I have used following code :
grep "abc_1" | awk -F " " {print $4}
This code is failing if there are more spaces used between string , e.g :
option : -abc_1 <some_path>
It will be helpful if I can extract the path somehow without bothering of spaces.
thanks
This should do:
echo 'option : -abc_1 <some_path>' | awk '/abc_1/ {print $4}'
<some_path>
If you do not specify field separator, it uses one ore more blank as separator.
PS you do not need both grep and awk
With sed you can do the search and the filter in one step:
sed -n 's/^.*abc_1 *: *\([^ ]*\).*$/\1/p'
The -n option suppresses printing, but the p command at the end still prints if a successful substitution was made.
perl -lne ' print $1 if(/-abc_1 (.*)/)' your_file
Tested Here
Or if you want to use awk:
awk '{for(i=1;i<=NF;i++)if($i="-abc_1")print $(i+1)}' your_file
try this grep only way:
grep -Po '^option\s*:\s*-abc_1\s*\K.*' file
or if the white spaces were fixed:
grep -Po '^option : -abc_1 \K.*' file
I am trying to make a single line ssh call from a ruby script. My script takes a hostname, and then sets out to return the hostname's machine info.
return_value = %x{ ssh #{hostname} "#{number_of_users}; #{number_of_processes};
#{number_of_processes_running}; #{number_of_processes_sleeping}; "}
Where the variables are formatted like this.
number_of_users = %Q(users | wc -w | cat | awk '{print "Number of Users: "\$1}')
number_of_processes = %Q(ps -el | awk '{print $2}' | wc -l | awk '{print "Number of Processes: "$1}')
I have tried both %q, %Q, and just plain "" and I cannot get the awk to print anything before the output. I either get this error (if I include the colon)
awk: line 1: syntax error at or near :
or if I don't include the slash in front of $1 I just get empty output for that line. Is there any solution for this? I thought it might be because I was using %q, but it even happens with just double quotes.
Use backticks to capture the output of the command and return the output as a string:
number_of_users = `users | wc -w | cat | awk '{print "Number of Users:", $1}'`
puts number_of_users
Results on my system:
48
But you can improve your pipeline:
users | awk '{ print "Number of Users:", NF }'
ps -e | awk 'END { print "Number of Processes:", NR }'
So the solution to this problem is:
%q(users | wc -w | awk '{print \"Number of Users: \"\$1}')
Where you have to use %q, not %, not %Q, and not ""
You must backslash double quotes and the dollar sign in front of any awk variables
If somebody could improve upon this answer by explaining why, that would be most appreciated
Though as Steve pointed out I could have improved my code using users | awk '{ print \"Number of Users:\", NF }'
In which case there is no need to backslash the NF.