I'm running this series of commands
passwd=`wc -l /etc/passwd`
echo $passwd
Returns:
34 /etc/passwd
What do I need to do to this so that it will only show the output of wc -l?
Just read from standard input instead of giving wc a file name:
$ passwd=`wc -l < /etc/passwd`
$ echo "$passwd"
86
wc still outputs quite a bit of padding, but the file name is omitted (because wc has no idea what file the data comes from).
Using awk perhaps ?
$ passwd=$(wc -l /etc/passwd | awk '{print $1}')
$ echo $passwd
32
Using cut, from cut (GNU coreutils)
$ passwd=$(wc -l /etc/passwd | cut -d" " -f1)
$ echo $passwd
32
That's the default behaviour of wc:
ยป wc -l /etc/passwd
28 /etc/passwd
There is no way to tell wc not to output the filename.
wc returns also the filename, but there are other ways to do it. Some examples:
passwd=`wc -l /etc/passwd | grep -o [1-9]\*`
or
passwd=`wc -l /etc/passwd | cut -f1 -d' '`
(answer from this question: get just the integer from wc in bash)
Related
New to shell scripting and just writing a little script to check if a process is running, if I use
PROCESS_NUM="ps -ef | grep '$1' | grep -v 'grep' | wc -l"
I get a "ps: illegal argument: |" error although if I echo out PROCESS_NUM and ctrl paste in the line it works just fine manually. Not sure why the | pipe is being troublesome here. Any help is much appreciated!
If you are trying to assig the output of ps -ef | grep '$1' | grep -v 'grep' | wc -l command to PROCESS_NUM variable, you need to use below:
PROCESS_NUM=`ps -ef | grep '$1' | grep -v 'grep' | wc -l`
or
PROCESS_NUM=$(ps -ef | grep '$1' | grep -v 'grep' | wc -l)
Using "", will treat the value inside of it as a string, except for some special character
Following is my script, every time I run this it goes into else part. when I run the TEST2EVAL command it gives me 1
#!/bin/sh
TEST2EVAL='ps auxf | grep some.jar | grep -v grep | wc -l'
if [ "$TEST2EVAL" = 1 ]
then
java -jar /path/to/jar &
else
echo "Running"
fi
Assuming you are trying to find out if any processes are running with some.jar on their command lines you probably want:
if pgrep -f some.jar; then
echo running;
else
echo not running;
fi
In in order save the output of a command in a variable, you have to enclose the command in backticks (`), not single quotes ('). Thus, change the second line of your script to:
TEST2EVAL=`ps auxf | grep some.jar | grep -v grep | wc -l`
You are using the wrong quotes for command substitution: not single quotes:
TEST2EVAL='ps auxf | grep some.jar | grep -v grep | wc -l'
but backquotes:
TEST2EVAL=`ps auxf | grep some.jar | grep -v grep | wc -l`
Better yet, use TEST2EVAL=$(ps auxf | grep some.jar | grep -v grep | wc -l) instead. It's much clearer, supported by all POSIX-compatible shells, and can be nested more easily when necessary.
I have a very basic script that keeps spitting back that the commands are not found. Ive looked all over this site and can not find an answer that works for me. The path to bash is correct. Ive checked the script with od. Ive run dos2unix. None of this helps me.
SCRIPT:
#!/bin/bash
HYBRISPROC=`ps -eo pid,command | grep [h]ybris | grep -v grep | awk '{print $1}'`
echo "Looking for Hybris..."
echo $HYBRISPROC
RESULTS:
./HybrisStopStart.sh: line 5: ps: command not found
./HybrisStopStart.sh: line 5: grep: command not found
./HybrisStopStart.sh: line 5: awk: command not found
./HybrisStopStart.sh: line 5: grep: command not found
Looking for Hybris...
Any ideas? If I run the command just on its own it works fine. Ive tried it as sudo as well and et the same results.
TIA
How about it?
#!/bin/bash
HYBRISPROC=`ps -eo pid,command | grep [h]ybris | grep -v grep | awk '{print $1}'`
echo "Looking for Hybris..."
echo "$HYBRISPROC"
(OR)
#!/bin/bash
HYBRISPROC="ps -eo pid,command | grep [h]ybris | grep -v grep | awk '{print $1}'"
echo "Looking for Hybris..."
bash -c "$HYBRISPROC"
(OR)
#!/bin/bash
HYBRISPROC="ps -eo pid,command | grep [h]ybris | grep -v grep | awk '{print $1}'"
echo "Looking for Hybris..."
eval "$HYBRISPROC"
TOTALY:
you can see the difference:
#!/bin/bash
LS=`ls -l`
echo $LS #nasty way
echo
echo "$LS" #good way
Try to add
PATH="$PATH:/usr/bin:/bin"
before code. Looks like bin directory is not on your path. So the commands are not found.
Hi i wrote the following bash script:
cat /home/xyz/wlandiscovery.sh
#!/bin/bash
DATE=`date +%d-%m-%Y__%H:%M:%S`
#Get the current standard interface e.g. eth0
INTERFACE=`route | grep '*' | awk '{print $8}'`
#Check if mac is available
if /usr/bin/arp-scan --interface $INTERFACE -l -r 5 | grep "xx:xx:xx:xx:xx:xx"
then
echo -e "$DATE AVAILABLE!" >> /home/xyz/wlandiscovery.log
else
echo -e "$DATE NOT AVAILABLE" >> /home/xyz/wlandiscovery.log
fi
exit 0
If i run this and the mac is available i get "AVAILABLE", if i disconnect the device it give "NOT AVAILABLE"...so run as expected.
But if i run it as Cronjob every 5 Minutes I get always "NOT AVAILABLE": (on a Debian system)
crontab -e
#......
*/5 * * * * /bin/bash /home/xyz/wlandiscovery.sh
Whats the problem here?
INTERFACE=`route | grep '*' | awk '{print $8}'`
On my system, route is /usr/sbin/route. /usr/sbin is most likely not in cron's PATH. Specify the full path:
INTERFACE=`/usr/sbin/route | awk '$2 == "*" {print $8}'`
Compare the command line output of the following on your Mac and Debian boxes:
INTERFACE=route | grep '*' | awk '{print $8}'
Is it the same? It should be in order to work.
Then, compare the command line output of:
/usr/bin/arp-scan --interface $INTERFACE -l -r 5 | grep "xx:xx:xx:xx:xx:xx"
Alright, now its working. seems that $PATH with crontab is not equal to $PATH in my Terminal prompt... if i do /sbin/route and /usr/bin/awk and /bin/grep it works.
I have very simple bash script:
#!/bin/bash
echo -n "A: ";
grep -v ">" | grep -o "A" $1 | wc -l;
I type
./script.sh 1.fasta
I got
A: 131
But the curcor is still blicking and my script is not finishing. What's wrong here?
Thank you.
This is the problem command:
grep -v ">" | grep -o "A" $1 | wc -l;
Since first command grep -v ">" is waiting for the input from STDIN as you haven't supplied any file to be searched by grep.
PS: Even grep -o "A" $1 is also problem since piped command will take input from output of the previous command in chain.
Probably you meant:
grep -v ">" "$1" | grep -o "A" | wc -l
Your first grep does not have a file argument so it will read from standard input:
grep -v ">" | grep -o "A" $1 | wc -l;
(read stdin) (read $1)
The reason why you get the 131 is because your second grep does have a file argument so it's getting all lines in $1 that have an A. However it's still waiting around for the end of the first grep to finish (which you can do with CTRL-D).
What you probably wanted to do is this:
grep -v ">" "$1" | grep -o "A" | wc -l
This will find all lines in $1 without a >, then all occurrences of A in that, counting them.