I have a issue with my Script, i am just trying to fingure out if my screen session is running or not (line 19).
The rest of the script is working.
#!/bin/bash
echo $PATH // /usr/local/bin:/usr/bin:/bin:/usr/local/games:/usr/games
echo "0"
content=$(wget http://interwebs.com/index.php?page=count -q -O -)
z=$(($content / 5))
z=$(($z + 1))
echo $z // 4
lockfile=/var/tmp/mylock
if ( set -o noclobber; echo "$$" > "$lockfile") 2> /dev/null; then
trap 'rm -f "$lockfile"; exit $?' INT TERM EXIT
# do stuff here
x=1
count=0
while [ $x -le $z ]
do
$req ="$(ps -ef | grep -i mystatus$count | grep -v grep)"
if [ "$req" = "" ]; then
# run bash script
screen -amds mystatus$count /usr/bin/wget --spider interwebs.com/index.php?page=cronwhatsoever$(( $count +1))-$(( $count +5))
else
echo "Cron running"
fi
x=$(( $x + 1 ))
count=$(( $count +5))
done
# clean up after yourself, and release your trap
rm -f "$lockfile"
trap - INT TERM EXIT
else
echo "Lock Exists: $lockfile owned by $(cat $lockfile)"
fi
sleep 15
It returns line 19: =: command not found. Actually running:
ps -ef | grep -i bukkit | grep -v grep
Works without issues if i run it directly in my Terminal, so any idea how to solve this issue?
I guess it something PATH related but grep is located in /bin/grep.
$req ="$(ps -ef | grep -i mystatus$count | grep -v grep)"
should be
req="$(ps -ef | grep -i mystatus$count | grep -v grep)"
Don't use $ on the left-hand side of an assignment, and you must not have spaces around the =
Related
Need to execute on MacOS.
Most of the solution is giving status running or stopped but for Not Responding state not having any solution.
tried solutions like this
pgrep "$1" 2>&1 > /dev/null
echo $?
if [ $? -eq 0 ]
then
{
echo " "$1" PROCESS RUNNING "
ps -ef | grep $1 | grep -v grep | awk '{print $2}'| xargs kill -9
}
else
{
echo " NO $1 PROCESS RUNNING"
};fi
i have a code:
L12(){
echo -e "/tftpboot/log/archive/L12/*/*$sn*L12*.log /tftpboot/log/diag/*$sn*L12*.log"
command="| grep -v hdd"
}
getlog(){
echo $(ls -ltr $(${1}) 2>/dev/null `${command}` | tail -1)
}
however $command does not seem to be inserting | grep -v hdd correctly
i need $command to be either empty or | grep
is there a simple solution to my issue or should i go for different approach
edit:
there may be another problem in there
i am loading a few "modules"
EVAL.sh
ev(){
case "${1}" in
*FAIL*) paint $red "FAIL";;
*PASS*) paint $green "PASS";;
*)echo;;
esac
result=${1}
}
rackinfo.sh (the "main script")
#! /bin/bash
#set -x
n=0
for src in $(ls modules/)
do
source modules/$src && ((n++))
## debugging
# source src/$src || ((n++)) || echo "there may be an issue in $src"
done
## debugging
# x=($n - $(ls | grep src | wc -l))
# echo -e "$x plugin(s) failed to laod correctly"
# echo -e "loaded $n modules"
########################################################################
command=cat
tests=("L12" "AL" "BI" "L12-3")
while read sn
do
paint $blue "$sn\t"
for test in ${tests[#]}
do
log="$(ev "$(getlog ${test})")"
if [[ -z ${log} ]]
then
paint $cyan "${test} "; paint $red "!LOG "
else
paint $cyan "${test} ";echo -ne "$log "
fi
done
echo
done <$1
the results i get are still containing "hdd" for L12()
Set command to cat as a default.
Also, it's best to use an array for commands with arguments, in case any of the arguments is multiple words.
There's rarely a reason to write echo $(command). That's essentially the same as just writing command.
#default command does nothing
command=(cat)
L12(){
echo -e "/tftpboot/log/archive/L12/*/*$sn*L12*.log /tftpboot/log/diag/*$sn*L12*.log"
command=(grep -v hdd)
}
getlog(){
ls -ltr $(${1}) 2>/dev/null | "${command[#]}" | tail -1)
}
I am checking to see if a process on a remote server has been killed. The code I'm using is:
if [ `ssh -t -t -i id_dsa headless#remoteserver.com "ps -auxwww |grep pipeline| wc -l" | sed -e 's/^[ \t]*//'` -lt 3 ]
then
echo "PIPELINE STOPPED SUCCESSFULLY"
exit 0
else
echo "PIPELINE WAS NOT STOPPED SUCCESSFULLY"
exit 1
fi
However when I execute this I get:
: integer expression expected
PIPELINE WAS NOT STOPPED SUCCESSFULLY
1
The actual value returned is "1" with no whitespace. I checked that by:
vim <(ssh -t -t -i id_dsa headless#remoteserver.com "ps -auxwww |grep pipeline| wc -l" | sed -e 's/^[ \t]*//')
and then ":set list" which showed only the integer and a line feed as the returned value.
I'm at a loss here as to why this is not working.
If the output of the ssh command is truly just an integer preceded by optional tabs, then you shouldn't need the sed command; the shell will strip the leading and/or trailing whitespace as unnecessary before using it as an operand for the -lt operator.
if [ $(ssh -tti id_dsa headless#remoteserver.com "ps -auxwww | grep -c pipeline") -lt 3 ]; then
It is possible that result of the ssh is not the same when you run it manually as when it runs in the shell. You might try saving it in a variable so you can output it before testing it in your script:
result=$( ssh -tti id_dsa headless#remoteserver.com "ps -auxwww | grep -c pipeline" )
if [ $result -lt 3 ];
The return value you get is not entirely a digit. Maybe some shell-metacharacter/linefeed/whatever gets into your way here:
#!/bin/bash
var=$(ssh -t -t -i id_dsa headless#remoteserver.com "ps auxwww |grep -c pipeline")
echo $var
# just to prove my point here
# Remove all digits, and look wether there is a rest -> then its not integer
test -z "$var" -o -n "`echo $var | tr -d '[0-9]'`" && echo not-integer
# get out all the digits to use them for the arithmetic comparison
var2=$(grep -o "[0-9]" <<<"$var")
echo $var2
if [[ $var2 -lt 3 ]]
then
echo "PIPELINE STOPPED SUCCESSFULLY"
exit 0
else
echo "PIPELINE WAS NOT STOPPED SUCCESSFULLY"
exit 1
fi
As user mbratch noticed I was getting a "\r" in the returned value in addition to the expected "\n". So I changed my sed script so that it stripped out the "\r" instead of the whitespace (which chepner pointed out was unnecessary).
sed -e 's/\r*$//'
I am trying to check if a process (assume it is called some_process) is running on a server. If it is, then echo 1, otherwise echo 0.
This is the command that I am using but it only works partially (more info below). Note that I need to write the script in one line.
ps aux | grep some_proces[s] > /tmp/test.txt && if [ $? -eq 0 ]; then echo 1; else echo 0; fi
Note: The [s] in some_proces[s] is to prevent grep from returning itself.
If some_process is running, then "1" gets echoed, which is fine. However, if some_process is not running, nothing gets echoed.
There is no need to explicitly check $?. Just do:
ps aux | grep some_proces[s] > /tmp/test.txt && echo 1 || echo 0
Note that this relies on echo not failing, which is certainly not guaranteed. A more reliable way to write this is:
if ps aux | grep some_proces[s] > /tmp/test.txt; then echo 1; else echo 0; fi
&& means "and if successful"; by placing your if statement on the right-hand side of it, you ensure that it will only run if grep returns 0. To fix it, use ; instead:
ps aux | grep some_proces[s] > /tmp/test.txt ; if [ $? -eq 0 ]; then echo 1; else echo 0; fi
(or just use a line-break).
Use grep -vc to ignore grep in the ps output and count the lines simultaneously.
if [[ $(ps aux | grep process | grep -vc grep) > 0 ]] ; then echo 1; else echo 0 ; fi
You can make full use of the && and || operators like this:
ps aux | grep some_proces[s] > /tmp/test.txt && echo 1 || echo 0
For excluding grep itself, you could also do something like:
ps aux | grep some_proces | grep -vw grep > /tmp/test.txt && echo 1 || echo 0
pgrep -q some_process && echo 1 || echo 0
more oneliners here
Being relatively new to anything other than bash scripting, I have created a script to
check if a process is running
output PID's to the shell
if not prompt user input and start etc/etc.
I've moved onto positional parameters and can't see where I'm going wrong:
if [ "$1" == "" ]; then
proc_finder
elif [ $1 != "" ];then
case $1 in
-p | --process )
shift
z=$(ps aux |grep $1 |grep -v grep > /dev/null)
if [ ! -z "$z" ]; then
echo "YES"
else
echo "NO"
fi
;;
* )
echo "Usage -p (process)"
esac
fi
This always seems to return yes even when putting in -p test for example. I know im doing something fundamentally wrong, looking at the verbose output the grep -v grep is being done last hence I believe it always returnes an exit state of 0.
Shouldn't that be if [ $? -eq 0 ]?
EDIT 1
You can try this:
z=`ps aux | grep $1 | grep -v grep > /dev/null`
if [ ! -z "$z" ]; then
echo "YES"
else
echo "NO"
fi
If $z is not empty (-z: test for zero-length string) this implies the process was found with the ps command.
EDIT 2
The ps ... grep ... grep is being redirect to /dev/null. That means z will contain nothing. remove the redirection and z should have some output.
z=`ps aux | grep $1 | grep -v grep`
EDIT 3
Alternatively, you can just do this:
ps aux | grep $1 | grep -v grep > /dev/null 2>&1
if [ $? -eq 0 ]; then
echo "YES"
else
echo "NO"
fi
In this case, you are not saving the grep output. That's good if you don't really need it.