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.
Related
I have the following single line in my bash script:
echo "foo" | awk -F"=" '{char=system("echo $1 | cut -c1");}{print "this is the result: "$char;}' >> output.txt
I want to print the first letter of "foo" using awk, such that I would get:
this is the result: f
in my output file, but instead, I get:
this is the result: foo
What am i doing wrong?
Thanks
No, this is not the way system command works inside awk.
What's happening in OP's code:
You are giving a shell command in system which is good(for some cases) but there is a problem in this one that you should give it like system("echo " $0" | cut -c1") to get its first character AND you need NOT to have a variable etc to save its value and print it in awk.
You are trying to save its result to a variable but it will not have its value(system command's value) but its status. It doesn't work like shell style in awk in here.
So your variable named char will have 0 value(which is a success status from system command) and when you are printing $char it is printing whole line(because in awk: print $0 means print whole line).
You could do this in a single awk by doing:
echo "foo" | awk '{print substr($0,1,1)}'
OR with GNU awk specifically:
echo "foo" | awk 'BEGIN{FS=""} {print $1}'
you're not using much of awk, same can be done with printf
$ echo "foo" | xargs printf "this is the result: %.1s\n"
this is the result: f
or, directly
$ printf "this is the result: %.1s\n" foo
this is the result: f
I have my script (called test.sh) as follow:
#!/bin/bash
for i in "cat myfile | awk -F',' '{print $1}'"; do
.....
My problem is that my script receives arguments (./tesh.sh arg1 arg2) and '{print $1}' take the script argument (arg1) instead awk result, how can I solve it?
Your original problem is that you wrote the $1 between double-quotes.
"cat myfile | awk -F',' '{print $1}'"
bash variables are still substituted by their value if they are in a double-quoted string, disregarding the fact that they are between single-quotes inside the double-quotes. This is the reason why $1 is being replaced by arg1.
The second problem is that you want to execute the command:
cat myfile | awk -F',' '{print $1}'
but for this you need to use the notation $( command ) or `command`, the latter is however not advised as nesting is difficult.
So, your for-loop should read something like:
#!/usr/bin/env bash
for i in $(awk -F ',' '{print $1}' myfile); do
...
done
I don’t know why I can not give only 0 to awk in a direct statement, e.g. if I want to output the square of a number:
$ echo 4 | awk '$0=$1*$1'
16
$ echo 3 | awk '$0=$1*$1'
9
$ echo 0 | awk '$0=$1*$1'
Why do I get nothing on the last try?
PS. it works if I write $1 in a bracketed statement:
$ echo 0 | awk '{print $1*$1}'
0
No, awk does not ignore a line with 0.
However, your awk command: $0=$1*$1 does not do what you think.
By default awk prints $0 if there is an statement that evaluates to true (not zero).
So, this will always print $0:
awk '1'
And this will never print $0:
awk '0'
To do what you want: to always print $0 after it has been re-calculated, you need to do:
awk '{$0=$1*$1; print}'
And so:
$ echo "0" | awk '{$0=$1*$1; print}'
0
$ echo "2" | awk '{$0=$1*$1; print}'
4
Or, without changing the value of $0, do:
$ echo "2" | awk '{print $0*$0}'
Or (shorter but less readable):
$ echo "2" | awk '{$0=$0*$0}1'
4
And, even shorter:
$ echo "4" | awk '{$0*=$0}1'
16
This last awk script is actually composed of two command lines:
awk '
<default pattern> { $0*=$0 }
1 { <default action> }
'
Which become, replacing the action by print and the condition by all:
awk ' /.*/{$0*=$0}
1 {print $0}'
Both lines are applied to all input lines. For all lines $0 is changed, and for all input lines a print $0 is executed.
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 run my safe shell script to make sure a binary is running
to check a binary is running I do following command
pidof prog.bin | awk '{print NF}'
is some system it gives me 0 when binary not running
and
in some systems it gives me NULL(nothing)
I can check the NULL using -z option but why awk command acting this way ??
Instead of pidof you can use:
pgrep -qf prog.bin
And check its exit status.
As per man pgrep:
-f Match against full argument lists. The default is to match against process names.
-q Do not write anything to standard output.
You can use this,
if [ `pidof 'NetworkManager'` ]; then
echo "Running"
else
echo "Not Running"
fi
One way to handle this sort of thing (undefined variables) in awk is like this:
echo hi | awk '{print a}'
compared with:
echo hi | awk '{print a || 0}'
0
One Liner for If else
[[ $(pidof 'NetworkManager') ]] && echo "Running" || echo "Not Running"
Try this:
pidof prog.bin | awk '{ if (NF!=0) print NF }'
Here's some tests with awk and NF:
$ # regular line of input
$ echo foo | awk '{print NF}'
1
$ # empty line
$ echo | awk '{print NF}'
0
$ # a word on input with no newline
$ printf "%s" nonewline | awk '{print NF}'
1
$ # no input, not even a newline
$ printf %s | awk '{print NF}'
# no output from awk
I suspect the pidof case is the last: not even a newline. To force a newline:
echo $(pidof prog) | ...
printf "%s\n" "$(pidof prog)" | ...