I am having a problem getting my shellscript working using backticks. Here is an example version of the script I am having an issue with:
#!/bin/sh
ECHO_TEXT="Echo this"
ECHO_CMD="echo ${ECHO_TEXT} | awk -F' ' '{print \$1}'"
result=`${ECHO_CMD}`;
echo $result;
result=`echo ${ECHO_TEXT} | awk -F' ' '{print \$1}'`;
echo $result;
The output of this script is:
sh-3.2$ ./test.sh
Echo this | awk -F' ' '{print $1}'
Echo
Why does the first backtick using a variable for the command not actually execute the full command but only returns the output of the first command along with the second command? I am missing something in order to get the first backtick to execute the command?
You need to use eval to get it working
result=`eval ${ECHO_CMD}`;
in place of
result=`${ECHO_CMD}`;
Without eval
${ECHO_TEXT} | awk -F' ' '{print \$1}
which will be expanded to
Echo this | awk -F' ' '{print \$1}
will be treated as argument to echo and will be output verbatim. With eval that line will actually be run.
You Hi,
you need to know eval command.
See :
#!/bin/sh
ECHO_TEXT="Echo this"
ECHO_CMD="echo ${ECHO_TEXT} | awk -F' ' '{print \$1}'"
result="`eval ${ECHO_CMD}`"
echo "$result"
result="`echo ${ECHO_TEXT} | awk -F' ' '{print $1}'`"
echo "$result"
Take a look to the doc :
help eval
In your first example echo is parsing the parameters - the shell never sees them. In the second example it works because the shell is doing the parsing and knows what to do with a pipe. If you change ECHO_CMD to be "bash echo ..." it will work.
Bash is escaping your command for you. Try
ECHO_TEXT="Echo this"
ECHO_CMD='echo ${ECHO_TEXT} | awk -F" " "'"{print \$1}"'"'
result=`${ECHO_CMD}`;
echo $result;
result=`echo ${ECHO_TEXT} | awk -F' ' '{print \$1}'`;
echo $result;
Or even better, try set -x on the first line, so you see what bash is doing
Related
My code is :
#!/bin/bash
strversion=`apache2ctl -v | awk '{print $3}' | sed 's/(Debian)//g;s/Server//g;s/built//g;s/2022-06-09T04:26:43//g'`
echo ${strversion%}
exit 0
i get this:
Apache/2.4.54
but i will have to look
Apache version 2.4.54
That's because the variable expansion is not quoted. An unquoted variable is subject to Word Splitting (and Filename Expansion)
echo "Apache2 - ${strversion%')'}"
# ...^...........................^
See also Security implications of forgetting to quote a variable in bash/POSIX shells
Try the following:
strversion=$(apache2ctl -v | head -1 | awk -F '[ /]' '{printf "%s version %s", $3, $4}')
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'm trying to cut the below string starting on the single quote:
name1=O'Reilly
so it leaves:
name2=Reilly
That's easy from the command line with the following commands:
echo $name | cut -d\' -f
echo $name | awk -F\' '{print $2}'
However when I run these commands from a script the string remains unaltered. I've been looking into problems with using single quotes as a delimiter but couldn't find anything. Any way to solve this issue?
That does not change the string the variable expands to, it just outputs the result of string manipulation.
If you want to create a new reference for variable name, use command substitution to save the result of cut/awk operation as variable name:
% name="O'Reilly"
% echo "$name" | awk -F\' '{print $2}'
Reilly
% name=$(echo "$name" | awk -F\' '{print $2}')
% echo "$name"
Reilly
On the other hand, if you want to declare the input as one (name1), and save the output as a different variable (name2):
% name1="O'Reilly"
% name2=$(echo "$name1" | awk -F\' '{print $2}')
% echo "$name2"
Reilly
This might be easier to get using Parameter expansion though:
$ name="O'Reilly"
$ echo "${name#*\'}"
Reilly
$ name="${name#*\'}"
$ echo "$name"
Reilly
I'm trying to pass a variable to nawk in a bash script, but it's not actually printing the $commentValue variable's contents. Everything works great, except the last part of the printf statement. Thanks!
echo -n "Service Name: "
read serviceName
echo -n "Comment: "
read commentValue
for check in $(grep "CURRENT SERVICE STATE" $nagiosLog |grep -w "$serviceName" | nawk -F": " '{print $2}' |sort -u ) ; do
echo $check | nawk -F";" -v now=$now '{ printf( "[%u]=ACKNOWLEDGE_SVC_PROBLEM;"$1";"$2";2;1;0;admin;$commentValue"\n", now)}' >> $nagiosCommand
done
$commentValue is inside an invocation to nawk, so it is considered as a variable in nawk, not a variable in bash. Since you do not have such a variable in nawk, you won't get anything there. You should first pass the variable "inside" nawk using the -v switch just like you did for the now variable; i.e.:
... | nawk -F";" -v now=$now -v "commentValue=$commentValue"
Note the quotes - they are required in case $commentValue contains whitespace.
Expecting this to print out abc - but I get nothing, every time, nothing.
echo abc=xyz | g="$(awk -F "=" '{print $1}')" | echo $g
A pipeline isn't a set of separate assignments. However, you could rewrite your current code as follows:
result=$(
echo 'abc=xyz' | awk -F '=' '{print $1}'
)
echo "$result"
However, a more Bash-centric solution without intermediate assignments could take advantage of a here-string. For example:
awk -F '=' '{print $1}' <<< 'abc=xyz'
Other solutions are possible, too, but this should be enough to get you started in the right direction.