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
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 6 years ago.
I am not sure how to save the output of a command via bash into a variable:
PID = 'ps -ef | grep -v color=auto | grep raspivid | awk '{print $2}''
Do I have to use a special character for the apostrophe or for the pipes?
Thanks!
To capture the output of a command in shell, use command substitution: $(...). Thus:
pid=$(ps -ef | grep -v color=auto | grep raspivid | awk '{print $2}')
Notes
When making an assignment in shell, there must be no spaces around the equal sign.
When defining shell variables for local use, it is best practice to use lower case or mixed case. Variables that are important to the system are defined in upper case and you don't want to accidentally overwrite one of them.
Simplification
If the goal is to get the PID of the raspivid process, then the grep and awk can be combined into a single process:
pid=$(ps -ef | awk '/[r]aspivid/{print $2}')
Note the simple trick that excludes the current process from the output: instead of searching for raspivid we search for [r]aspivid. The string [r]aspivid does not match the regular expression [r]aspivid. Hence the current process is removed from the output.
The Flexibility of awk
For the purpose of showing how awk can replace multiple calls to grep, consider this scenario: suppose that we want to find lines that contain raspivid but that do not contain color=auto. With awk, both conditions can be combined logically:
pid=$(ps -ef | awk '/raspivid/ && !/color=auto/{print $2}')
Here, /raspivid/ requires a match with raspivid. The && symbol means logical "and". The ! before the regex /color=auto/ means logical "not". Thus, /raspivid/ && !/color=auto/ matches only on lines that contain raspivid but not color=auto.
A more straightforward approach:
pid=$(pgrep raspivid)
... or a little different
echo pgrep [t]eleport
I am trying print the first field of the first row of an output. Here is the case. I just need to print only SUSE from this output.
# cat /etc/*release
SUSE Linux Enterprise Server 11 (x86_64)
VERSION = 11
PATCHLEVEL = 2
Tried with cat /etc/*release | awk {'print $1}' but that print the first string of every row
SUSE
VERSION
PATCHLEVEL
Specify NR if you want to capture output from selected rows:
awk 'NR==1{print $1}' /etc/*release
An alternative (ugly) way of achieving the same would be:
awk '{print $1; exit}'
An efficient way of getting the first string from a specific line, say line 42, in the output would be:
awk 'NR==42{print $1; exit}'
Specify the Line Number using NR built-in variable.
awk 'NR==1{print $1}' /etc/*release
try this:
head -1 /etc/*release | awk '{print $1}'
df -h | head -4 | tail -1 | awk '{ print $2 }'
Change the numbers to tweak it to your liking.
Or use a while loop but thats probably a bad way to do it.
You could use the head instead of cat:
head -n1 /etc/*release | awk '{print $1}'
sed -n 1p /etc/*release |cut -d " " -f1
if tab delimited:
sed -n 1p /etc/*release |cut -f1
Try
sed 'NUMq;d' /etc/*release | awk {'print $1}'
where NUM is line number
ex. sed '1q;d' /etc/*release | awk {'print $1}'
awk, sed, pipe, that's heavy
set `cat /etc/*release`; echo $1
the most code-golfy way i could think of to print first line only in awk :
awk '_{exit}--_' # skip the quotations and make it just
# awk _{exit}--_
#
# if u're feeling adventurous
first pass through exit block, "_" is undefined,
so it fails and skips over for row 1.
then the decrementing of the same counter will make
it "TRUE" in awk's eyes (anything not empty string
or numeric zero is considered "true" in their agile boolean sense). that same counter also triggers default action of print for row 1.
—- incrementing… decrementing… it's same thing,
merely direction and sign inverted.
then finally, at start of row 2, it hits criteria to
enter the action block, which instructs it to instantly
exit, thus performing essentially the same functionality as
awk '{ print; exit }'
… in a slightly less verbose manner. For a single line print, it's not even worth it to set FS to skip the field splitting part.
using that concept to print just 1st row 1st field :
awk '_{exit} NF=++_'
awk '_++{exit} NF=_'
awk 'NR==1&&NF=1' file
grep -om1 '^[^ ]\+' file
# multiple files
awk 'FNR==1&&NF=1' file1 file2
You can kill the process which is running the container.
With this command you can list the processes related with the docker container:
ps -aux | grep $(docker ps -a | grep container-name | awk '{print $1}')
Now you have the process ids to kill with kill or kill -9.
This question already has answers here:
Using awk with variables
(3 answers)
Closed 8 years ago.
I have following variable set in my unix environment. If i try to use it in awk command its not working but the same command is working when i dont use $b variable
$b="NEW"
when i try following command it is not working
echo "$a" | tr [a-z] [A-Z] |awk -v RS=, '/TABLE/&&/CREATE/&&/`echo ${b}`/{print $NF}'
But, if i replace the $b value to NEW as below its working
echo "$a" | tr [a-z] [A-Z] |awk -v RS=, '/TABLE/&&/CREATE/&&/NEW/{print $NF}'
You cannot use a bash var inside awk like that. Instead, use:
echo "$a" | tr [a-z] [A-Z] | awk -v RS=, -v myvar=$b '/TABLE/&&/CREATE/&& $0~myvar {print $NF}'
See an example:
$ var="hello"
$ awk -v text=$var 'BEGIN{print text}'
hello
Also, to me it works with tr 'a-z' 'A-Z' instead of tr [a-z] [A-Z]. And based on Mark Setchell suggestion, you can skip it by using the IGNORECASE = 1:
echo "$a" | awk -v RS=, -v myvar=$b 'BEGIN{IGNORECASE=1} /TABLE/&&/CREATE/&& $0~myvar {print $NF}'
Regarding your question:
if i replace the $b value to NEW as below its working
It works because the value of your variable is NEW and what you end up doing is using that in the regex, which is exactly how it is supposed to be done.
about your second question:
can not use unix $variable in awk command
You cannot use shell variables in awk like that. You need to create an awk variable by using -v option and assigning your bash variable.
awk -v awkvar="$bashvar" '/ /{ ... }'
This makes your existing syntax as:
echo "$a" | tr [a-z] [A-Z] | awk -v RS=, -v var="$b" '/TABLE/&&/CREATE/&&/var/{print $NF}'
This again won't work because inside /../ variables are not interpolated, meaning they are considered literally. So, you need to do:
echo "$a" | tr [a-z] [A-Z] |awk -v RS=, -v var="$b" '/TABLE/&&/CREATE/&&$0~var{print $NF}'