If I want to pass a parameter to an awk script file, how can I do that ?
#!/usr/bin/awk -f
{print $1}
Here I want to print the first argument passed to the script from the shell, like:
bash-prompt> echo "test" | ./myawkscript.awk hello
bash-prompt> hello
In awk $1 references the first field in a record not the first argument like it does in bash. You need to use ARGV for this, check out here for the offical word.
Script:
#!/bin/awk -f
BEGIN{
print "AWK Script"
print ARGV[1]
}
Demo:
$ ./script.awk "Passed in using ARGV"
AWK Script
Passed in using ARGV
You can use -v as a command-line option to provide a variable to the script:
Say we have a file script.awk like this:
BEGIN {print "I got the var:", my_var}
Then we run it like this:
$ awk -v my_var="hello this is me" -f script.awk
I got the var: hello this is me
your hash bang defines the script is not shell script, it is an awk script. you cannot do it in bash way within your script.
also, what you did : echo blah|awk ... is not passing paramenter, it pipes the output of echo command to another command.
you could try these way below:
echo "hello"|./foo.awk file -
or
var="hello"
awk -v a="$var" -f foo.awk file
with this, you have var a in your foo.awk, you could use it.
if you want to do something like shell script accept $1 $2 vars, you can write a small shellscript to wrap your awk stuff.
EDIT
No I didn't misunderstand you.
let's take the example:
let's say, your x.awk has:
{print $1}
if you do :
echo "foo" | x.awk file
it is same as:
echo "foo"| awk '{print $1}' file
here the input for awk is only file, your echo foo doesn't make sense. if you do:
echo "foo"|awk '{print $1}' file -
or
echo "foo"|awk '{print $1}' - file
awk takes two input (arguments for awk) one is stdin one is the file, in your awk script you could:
echo "foo"|awk 'NR==FNR{print $1;next}{print $1}' - file
this will print first foo from your echo, then the column1 from file of course this example does nothing actual work, just print them all.
you can of course have more than two inputs, and don't check the NR and FNR, you could use the
ARGC The number of elements in the ARGV array.
ARGV An array of command line arguments, excluding options and the program argument, numbered from zero to ARGC-1
for example :
echo "foo"|./x.awk file1 - file2
then your "foo" is the 2nd arg, you can get it in your x.awk by ARGV[2]
echo "foo" |x.awk file1 file2 file2 -
now it is ARGV[4] case.
I mean, your echo "foo"|.. would be stdin for awk, it could by 1st or nth "argument"/input for awk. depends on where you put the -(stdin). You have to handle it in your awk script.
Related
I need to grep some pattern and further i need to print some output within that. Currently I am using the below command which is working fine. But I like to eliminate using multiple pipe and want to use single awk command to achieve the same output. Is there a way to do it using awk?
root#Server1 # cat file
Jenny:Mon,Tue,Wed:Morning
David:Thu,Fri,Sat:Evening
root#Server1 # awk '/Jenny/ {print $0}' file | awk -F ":" '{ print $2 }' | awk -F "," '{ print $1 }'
Mon
I want to get this output using single awk command. Any help?
You can try something like:
awk -F: '/Jenny/ {split($2,a,","); print a[1]}' file
Try this
awk -F'[:,]+' '/Jenny/{print $2}' file.txt
It is using muliple -F value inside the [ ]
The + means one or more since it is treated as a regex.
For this particular job, I find grep to be slightly more robust.
Unless your company has a policy not to hire people named Eve.
(Try it out if you don't understand.)
grep -oP '^[^:]*Jenny[^:]*:\K[^,:]+' file
Or to do a whole-word match:
grep -oP '^[^:]*\bJenny\b[^:]*:\K[^,:]+' file
Or when you are confident that "Jenny" is the full name:
grep -oP '^Jenny:\K[^,:]+' file
Output:
Mon
Explanation:
The stuff up until \K speaks for itself: it selects the line(s) with the desired name.
[^,:]+ captures the day of week (in this case Mon).
\K cuts off everything preceding Mon.
-o cuts off anything following Mon.
I just want to pass a shell variable that stores name of a file to awk command. When I searched this problem on the net I see many different options but none of them worked for me. I tried the followings:
#!/bin/bash
for i in "$#"
do
case $i in
-p=*|--producedfile=*)
PFILE="${i#*=}"
shift # past argument=value
*)
# unknown option
;;
esac
done
echo "PRODUCEDFILE = ${PFILE}"
awk -v FILE=${PFILE} '{print FILE $0}' #DIDNT WORK
awk '{print FILE $0}' ${PFILE} # DIDNT WORK
awk -v FILE=${PFILE} '{print $0}' FILE #DIDNT WORK
To pass a shell variable to awk, you correctly used -v option.
However, the shift was unnecessary (you're iterating options with for), ;; was missing (you have to terminate each case branch), as well as was the name of the file for awk to process. Fixed, your script looks like:
#!/bin/bash
for i in "$#"; do
case $i in
-p=*|--producedfile=*)
PFILE="${i#*=}"
;;
*)
# unknown option
;;
esac
done
echo "PRODUCEDFILE = ${PFILE}"
awk -v FILE="${PFILE}" '{print FILE, $0}' "${PFILE}"
Note however, awk already makes the name of the currently processed file available in the FILENAME variable. So, you could also write the last line as:
awk '{print FILENAME, $0}' "${PFILE}"
I have this bash statement for printing a specific cell from a .csv file.
set `cat $filename | awk -v FS=',' '{print $2}' | head -5 | tail -n 1`
The '{print $2}' part determines the column and the head -5 part determines the row.
Can I substitute a $counter variable in place of $2 (e.g., '{print $counter}')?
The answer is "yes" -- and there are a couple ways to do what you want. The proper way is to declare an awk variable using -v:
awk -F',' -v c=$counter 'NR==6 { print $c; exit }' "$filename"
(You will forgive me for moving some things around to do everything in awk, for passing "$filename" to awk safely, and for getting rid of set and back ticks -- that were doing nothing for the cause.)
Another way to do this is a bit of a "hackish" way -- leveraging shell quoting rules. This method requires some escaping to ensure that the first $ character (that references the intended field in awk) is not interpreted by the shell... The following works in bash (and POSIX sh):
awk -F',' "NR==6 { print \$$counter; exit }" "$filename"
Yes and all pipes could be removed. Variables are passed to awk with -v var=value.
Give a try to this tested version. Provide a value to the ̀€col and row variables:
set $(awk -F "," -v col=2 -v row=5 'NR==row {print $col; exit}' "${filename}")
$(command) is prefered to `command`, this later is deprecated.
NR is the current line number.
"${filename}" is expanded by the shell to its value: the double quotes will help if the filename contains some special chars.
In an awk file, e.g example.awk, should the header be #!/bin/bash or #!/bin/awk -f?
The reason for my question is that if I try this command in the console I receive the correct file.txt with "line of text":
awk 'BEGIN {print "line of text"}' >> file.txt
but if i try execute the following file with ./example.awk:
#! /bin/awk -f
awk 'BEGIN {print "line of text"}' >> file.txt
it returns an error:
$ ./awk-usage.awk
awk: ./awk-usage.awk:3: awk 'BEGIN {print "line of text"}' >> file.txt
awk: ./awk-usage.awk:3: ^ invalid char ''' in expression
If I change the header to #!/bin/bash or #!/bin/sh it works.
What is my error? What is the reason of that?
Since you explicitly run the awk command, you should use #!/bin/bash. You can use #!/bin/awk if you remove the awk command and include only the awk program (e.g. BEGIN {print "line of text"}), but then you need to append to file using awk syntax (print ... >> file).
awk -f takes a file containing the awk script, so that is completely wrong here.
Your script is a shell script that happens to contains an awk command.
#! /bin/sh tells your shell to execute the file as a shell command with /bin/sh - and it is a shell command. If you replace that with #! /bin/awk -f then the file is executed with awk, basically the same as executing
/bin/awk -f awk 'BEGIN {print "line of text"}' >> file.txt
I am trying to get file extension for a file in shell script. But without any luck.
The command I am using is
file_ext=${filename##*.}
and
file_ext = $filename |awk -F . '{if (NF>1) {print $NF}}'
But both of the commands failed to put value in variable file_ext. But when i try
echo $filename |awk -F . '{if (NF>1) {print $NF}}'
It gives me the desired result. I am new to shell script. Please describe the situation what is happening. And also how should I do it?
Thanks.
to get file extension, just use the shell
$ filename="myfile.ext"
$ echo ${filename##*.}
ext
$ file_ext=${filename##*.} #put to variable
$ echo ${file_ext}
ext
Spaces hurt.
Anyway you should do:
file_ext=$(echo $filename | awk -F . '{if (NF>1) {print $NF}}')
[Edit] Better suggestion by Martin:
file_ext=$(printf '%s' "$filename" | awk -F . '{if (NF>1) {print $NF}}')
That will store in $file_ext the output of the command.
You have to be careful when declaring variables.
variable1="string" # assign a string value
variable3=`command` # assign output from command
variable2=$(command) # assign output from command
Notice that you cannot put a space after the variable, because then it gets interpreted as a normal command.