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Here is my piece of code:
CONFIG= ifconfig lo;
MAC_ADDRESS="$("${CONFIG}" | awk '/HWaddr/ {print $NF}' | sed -e 's/://g')";
MAC_ADDRESS2="$("${CONFIG}" | awk '/HWaddr/ {print $NF}' | sed -e 's/://g')";
echo "$MAC_ADDRESS"
echo "$MAC_ADDRESS2"
Here I am trying to set ifconfig lo into a single variable and try passing into another one where it is needed.
So my expectation is, it should pass like below
MAC_ADDRESS="$(ifconfig lo | awk '/HWaddr/ {print $NF}' | sed -e 's/://g')";
MAC_ADDRESS2="$(ifconfig lo | awk '/HWaddr/ {print $NF}' | sed -e 's/://g')";
Here I get permission denied exception. Any other possible way to do?
Expected Output:
JUKUJSDJDJSJDJSJ
KIIPPSKKSKDKDKKS
note: this answer assumes bash. Other shells have different behaviour.
If you attempt something similar like this:
bash-4.2$ command="ls"
bash-4.2$ $command
bash-4.2$ "$command"
Then you will notice that you execute ls in both cases. However, if you attempt the following:
bash-4.2$ command="ls -l"
bash-4.2$ $command
will work as expected (this fails in zsh), but if you quote it like
bash-4.2$ "$command"
bash: ls -l: command not found
it fails.
You attempt a similar thing but you have quoted the command in your command-substitution $(command).
In short: do not
CONFIG= ifconfig lo;
MAC_ADDRESS="$("${CONFIG}" | awk '/HWaddr/ {print $NF}' | sed -e 's/://g')";
but do
CONFIG="ifconfig lo"
MAC_ADDRESS="$(${CONFIG} | awk '/HWaddr/ {print $NF}' | sed -e 's/://g')";
I also fixed your definition of CONFIG as it needs <double-quotes> and spaces are not allowed before and after the <equal>-sign (=)
Related
This question already has answers here:
How do I use shell variables in an awk script?
(7 answers)
Closed 1 year ago.
good morning, I am a newbie in the world of scripts and I have this problem and I don't know why it happens to me and why I have it wrong, thanks in advance.
For example, I have this command that searches for users with X less letters:
cut -d: -f1 /etc/passwd | awk 'length($1) <= 4'
It works correctly but when I substitute a 4 for a variable with the same value it doesn't do it well:
number=4
echo -e $(cut -d: -f1 /etc/passwd | awk 'length($1) <= $number')
The same error happens to me here too, when I search for users who have an old password
awk -F: '{if($3<=18388)print$1}' < /etc/shadow
Works, but when I use the variable it stops working
variable=18388
awk -F: '{if($3<=$variable)print$1}' < /etc/shadow
Consider using awk's ability to import variables via the -v option, eg:
number=4
cut -d: -f1 /etc/passwd | awk -v num="${number}" 'length($1) <= num'
Though if the first field from /etc/passwd contains white space, and you want to consider the length of the entire field, you should replace $1 with $0, eg:
number=4
cut -d: -f1 /etc/passwd | awk -v num="${number}" 'length($0) <= num'
Even better, eliminate cut and the subprocess (due to the pipe) and use awk for the entire operation by designating the input field separator as a colon:
number=4
awk -F':' -v num="${number}" 'length($1) <= num { print $1 }' /etc/passwd
You need to use double quotes:
echo -e $(cut -d: -f1 /etc/passwd | awk "length(\$1) <= $number")
In single quotes, variables are not interpreted.
Updated: Here is an illustrative example:
> X=1; echo '$X'; echo "$X"
$X
1
Update 2: as rightfully pointed out in the comment, when using double quotes one need to make sure that the variables that are meant for awk to interpret are escaped, so they are not interpreted during script evaluation. Command updated above.
Alternatively, this task could be done using only a single GNU sed one-liner:
num=4
sed -E "s/:.*//; /..{$num}/d" /etc/passwd
Another one-liner, using only grep would be
grep -Po "^[^:]{1,$num}(?=:)" /etc/passwd
but this requires a grep supporting perl regular expressions, for the lookahead construct (?=...).
This question already has answers here:
Command not found error in Bash variable assignment
(5 answers)
Closed 4 years ago.
Im having trouble calling a variable that should bring out the output of a command.
#!/bin/bash
ipAddresses = 'ifconfig | awk -v OFS=": " -v RS= '$1!="lo" && split($0, a, /inet addr:/) > 1{sub(/ .*/, "", a[2]); print $1, a[2]}''
echo -e "Sus direcciones IP son: \n " $(ipAddresses)
Appreciating any advice
Variable assignments cannot have space around the = in the shell. Also, you don't want single quotes there, you want either backticks or $(). The single quotes should only be for your awk command. Your awk is needlessly complicated as well, and you are using command substitution ($()) when printing, but ipAdresses is a variable, not a command.
Try something like this:
#!/bin/bash
ipAddresses=$(ifconfig | sed 's/^ *//' | awk -F'[: ]' '/^ *inet addr:/{print $3}')
printf 'Sus direcciones IP son:\n%s\n' "$ipAddresses"
But that is really not portable. You didn't mention your OS, but I am assuming it's a Linux and the output suggests Ubuntu (I don't have addr after inet in the output of ifconfig on my Arch, for example).
If you are running Linux, you could use grep instead:
ipAddresses=$(ifconfig | grep -oP 'inet addr:\K\S+')
ip is generally replacing ifconfig, so try this instead:
ipAddresses=$(ip addr | awk '/inet /{print $2}')
or
ipAddresses=$(ip addr | grep -oP 'inet \K\S+')
Or, to remove the trailing /N:
ipAddresses=$(ip addr | grep -oP 'inet \K[\d.]+')
And you don't need the variable anyway, you can just:
printf 'Sus direcciones IP son:\n%s\n' "$(ip addr | awk '/inet /{print $2}')"
I am not sure about your intention, since they are not stated, so I am trying to guess them from the script.
Option 1: you are trying to get IP address to into the variable ipAddresses and that is not happenning.
Start by changing single quotes around the long command and debug the command.
Option 2: you are storing a command in variable ipAddresses that you want to execute on the second line.
For both of the options you need to use the the value of the variable through $ipAdresses on the second line.
Also fix the assignment to following formart:
varName="value" # Note no spaces around = sign
Replace the final $(ipAddresses) with ${ipAddresses} or just "$ipAddresses", but also save the output of your command using $().
Check Difference between ${} and $() in Bash.
A basic example:
#!/bin/sh
OUTPUT=$(uname -a)
echo "The output: $OUTPUT"
This question already has answers here:
How do I set a variable to the output of a command in Bash?
(15 answers)
Closed 2 years ago.
I want to assign the output of a shell command to a variable.
If I directly echo the command, the code will execute correctly:
for ((i=0; i<${#result[#]}; i++)); do
echo ${result[$i]} | awk '{print $1}'
done
But, if I assign it to a variable,
size=`${result[$i]} | awk '{print $1}'`
echo $size
Or
size=$(${result[$i]} | awk '{print $1}')
echo $size
They are not working.
How can I fix it?
You missed the echo
size=$(echo ${result[$i]} | awk '{print $1}')
Here the output the the echo is passed as input to the awk
The $() or back ticks just run the command and assign it to a variable, so when you just write
${result[$i]} | awk '{print $1}'
it won't give you anything as nothing is passed as input to the awk command.
This question already has an answer here:
Shell script, saving the command value to a variable
(1 answer)
Closed 9 years ago.
Can anyone help me out with this problem?
I'm trying to save the awk output into a variable.
variable = `ps -ef | grep "port 10 -" | grep -v "grep port 10 -"| awk '{printf "%s", $12}'`
printf "$variable"
EDIT: $12 corresponds to a parameter running on that process.
Thanks!
#!/bin/bash
variable=`ps -ef | grep "port 10 -" | grep -v "grep port 10 -" | awk '{printf $12}'`
echo $variable
Notice that there's no space after the equal sign.
You can also use $() which allows nesting and is readable.
I think the $() syntax is easier to read...
variable=$(ps -ef | grep "port 10 -" | grep -v "grep port 10 -"| awk '{printf "%s", $12}')
But the real issue is probably that $12 should not be qouted with ""
Edited since the question was changed, This returns valid data, but it is not clear what the expected output of ps -ef is and what is expected in variable.
as noted earlier, setting bash variables does not allow whitespace between the variable name on the LHS, and the variable value on the RHS, of the '=' sign.
awk can do everything and avoid the "awk"ward extra 'grep'. The use of awk's printf is to not add an unnecessary "\n" in the string which would give perl-ish matcher programs conniptions. The variable/parameter expansion for your case in bash doesn't have that issue, so either of these work:
variable=$(ps -ef | awk '/port 10 \-/ {print $12}')
variable=`ps -ef | awk '/port 10 \-/ {print $12}'`
The '-' int the awk record matching pattern removes the need to remove awk itself from the search results.
variable=$(ps -ef | awk '/[p]ort 10/ {print $12}')
The [p] is a neat trick to remove the search from showing from ps
#Jeremy
If you post the output of ps -ef | grep "port 10", and what you need from the line, it would be more easy to help you getting correct syntax
This question already has answers here:
sed unknown option to `s' in bash script [duplicate]
(4 answers)
Closed 2 years ago.
I am trying
grep searchterm myfile.csv | sed 's/replaceme/withthis/g'
and getting
unknown option to `s'
What am I doing wrong?
Edit:
As per the comments the code is actually correct. My full code resembled something like the following
grep searchterm myfile.csv | sed 's/replaceme/withthis/g'
# my comment
And it appears that for some reason my comment was being fed as input into sed. Very strange.
use the --expression option
grep searchterm myfile.csv | sed --expression='s/replaceme/withthis/g'
use "-e" to specify the sed-expression
cat input.txt | sed -e 's/foo/bar/g'
To make sed catch from stdin , instead of from a file, you should use -e.
Like this:
curl -k -u admin:admin https://$HOSTNAME:9070/api/tm/3.8/status/$HOSTNAME/statistics/traffic_ips/trafc_ip/ | sed -e 's/["{}]//g' |sed -e 's/[]]//g' |sed -e 's/[\[]//g' |awk 'BEGIN{FS=":"} {print $4}'
If you are trying to do an in-place update of text within a file, this is much easier to reason about in my mind.
grep -Rl text_to_find directory_to_search 2>/dev/null | while read line; do sed -i 's/text_to_find/replacement_text/g' $line; done
Open the file using vi myfile.csv
Press Escape
Type :%s/replaceme/withthis/
Type :wq and press Enter
Now you will have the new pattern in your file.