crontab unable to find the running process - shell

crontab looks as below:
SHELL=/bin/bash
* * * * * /home/zaks/influx_rabbitmq_restorer.sh > /home/zaks/crontab.txt 2>&1
influx_rabbitmq_restorer.sh:
ps -aux | grep -v grep | grep rabbitmq_restorer
RESULT=$?
echo $RESULT
if [ $RESULT -eq 0 ]; then
echo "running"
else
echo "not running"
(nohup python -u /home/zaks/rabbitmq_restorer.py )
fi
When I manually run the script /home/zaks/influx_rabbitmq_restorer.shthe result is 1 ("not running").
But cronjob log at /home/zaks/crontab.txt shows:
zaks 11612 0.0 0.0 12504 2936 ? Ss 03:15 0:00 /bin/bash -c /home/zaks/influx_rabbitmq_restorer.sh > /home/zaks/crontab.txt 2>&1
zaks 11613 0.0 0.0 12516 2388 ? S 03:15 0:00 /bin/bash -c /home/zaks/influx_rabbitmq_restorer.sh > /home/zaks/crontab.txt 2>&1
0
running
cronjob is unable to detect rabbitmq_restorer process.

Edit first line of the script to be like:
ps -aux | grep -v grep | grep -v influx_rabbitmq_restorer.sh| grep rabbitmq_restorer

Related

ps -ef | grep * command not working properly

I have simple unix shell script as follows. Which is giving different count for a service
#!/bin/bash
service=$1
ps -ef | grep $service | grep -v "grep" | wc -l
PROCESS_NUM=$(ps -ef | grep $service | grep -v "grep"| wc -l)
echo $PROCESS_NUM
In the above code below line gives output of 2.
ps -ef | grep $service | grep -v "grep" | wc -l
But when same line is assigned to variable as code below its giving output as 3.
PROCESS_NUM=$(ps -ef | grep $service | grep -v "grep"| wc -l)
echo $PROCESS_NUM
Why this is getting increased by 1 and how to tackle it.
You can see what is happening if the script tees the output to a file before counting the lines and then displaying the output after:
#!/bin/bash
service=$1
echo Directly in Script:
ps -ef | grep $service | grep -v grep | tee test.txt | wc -l
cat test.txt
echo Inside Subshell:
RESULT=$(ps -ef | grep $service | grep -v grep | tee test.txt | wc -l)
echo $RESULT
cat test.txt
When the output of a command is captured, bash starts another shell to run the command - but that subshell also shows up in the process list.
When I run that script I get:
$ ./test.sh lca
Directly in Script:
2
gcti 4268 1 0 2018 ? 21:59:03 ./lca 4999
t816826 9159 7009 0 09:22 pts/1 00:00:00 /bin/bash ./test.sh lca
Inside Subshell:
3
gcti 4268 1 0 2018 ? 21:59:03 ./lca 4999
t816826 9159 7009 0 09:22 pts/1 00:00:00 /bin/bash ./test.sh lca
t816826 9166 9159 0 09:22 pts/1 00:00:00 /bin/bash ./test.sh lca

Bash if -gt also triggered when values equal?

I want my script to check if it's already running in another instance:
$ cat test.sh
#!/bin/bash
ps -ef | grep -v grep | grep -i "test.sh" | grep bash
ps -ef | grep -v grep | grep -i "test.sh" | grep -c bash
if [ `ps -ef | grep -v grep | grep -i "test.sh" | grep -c bash` -gt 1 ]; then echo "There's another instance running."
else echo "Only this instance is running."
fi
However the output is
$ ./test.sh
noes 9503 7494 0 09:32 pts/1 00:00:00 /bin/bash ./test.sh
1
There's another instance running.
Clearly, 1 is not greater than 1, so why is the if condition triggered?
Thanks
From test man page:
INTEGER1 -gt INTEGER2
INTEGER1 is greater than INTEGER2
So the answer is no, -gt is not triggered when values are equal. In fact, as you can see if you modify the script in this way:
$ cat test.sh
#!/bin/bash
ps -ef | grep -v grep | grep -i "test.sh" | grep bash
ps -ef | grep -v grep | grep -i "test.sh" | grep -c bash
STRINGS=`ps -ef | grep -v grep | grep -i "test.sh"`
echo "$STRINGS"
COUNT=`ps -ef | grep -v grep | grep -i "test.sh" | grep -c bash`
echo $COUNT
if [ `ps -ef | grep -v grep | grep -i "test.sh" | grep -c bash` -gt 1 ]; then echo "There's another instance running."
else echo "Only this instance is running."
fi
You get this:
$ ./test.sh
lucio 5097 4736 0 10:10 pts/2 00:00:00 /bin/bash ./test.sh
1
lucio 5097 4736 0 10:10 pts/2 00:00:00 /bin/bash ./test.sh
lucio 5106 5097 0 10:10 pts/2 00:00:00 /bin/bash ./test.sh
2
There's another instance running.
If you modify the script in this way, it will work:
#!/bin/bash
ps -ef | grep -v grep | grep -i "test.sh" | grep bash
pgrep -c test.sh
if [ $(pgrep -c test.sh) -gt 1 ]; then
echo "There's another instance running."
else
echo "Only this instance is running."
fi
This is the output:
$ ./test.sh
lucio 5197 4736 0 10:17 pts/2 00:00:00 /bin/bash ./test.sh
1
Only this instance is running.
Note the use $() instead of backticks. Check this answer for this change.

different count using ps and wc to check if service is running in bash

I have a small script to check if a service is running in bash
#!/bin/bash
countlines=""$(ps -ef | grep netdata | grep -v grep | wc -l)
echo $countlines >> mylog.log
if [ ${countlines} -lt 3 ];then
sudo /bin/systemctl restart netdata
fi
The problem is when I issue a ps -ef | grep netdata | grep -v grep | wc -l at the command line the result is always 3 but mylog.log is:
6
[update: added filtered ps -ef results]
forge#reportserver:~ ps -ef | grep netdata
netdata 22308 1 0 08:38 ? 00:00:37 /usr/sbin/netdata -D
netdata 22386 22308 0 08:38 ? 00:00:58 /usr/libexec/netdata/plugins.d/apps.plugin 1
netdata 47045 22308 0 11:38 ? 00:00:02 bash /usr/libexec/netdata/plugins.d/tc-qos-helper.sh 1
forge 52028 27902 0 12:34 pts/8 00:00:00 grep --color=auto netdata
why such a discordance?
split into 2 commands to see why:
ps_grep_output=$(ps -ef | grep netdata | grep -v grep)
echo "$ps_grep_output" >> mylog.log
countlines=$(wc -l <<< "$ps_grep_output")
echo "$countlines" >> mylog.log

echo wc result not right in shell script

this is my shell script
#!/bin/sh
echo "===="
echo $1
echo "===="
ps -ef | grep -w $1 | grep -v -e "grep"
echo "===="
echo $(ps -ef | grep -w $1 | grep -v -e "grep" | wc -l)
echo "===="
exit 0
then,i execute the shell script at command line.
./test.sh php-fpm
the result is:
====
php-fpm
====
0 986 984 0 4:43PM ?? 0:05.53 php-fpm
70 988 986 0 4:43PM ?? 0:00.00 php-fpm
70 989 986 0 4:43PM ?? 0:00.00 php-fpm
70 990 986 0 4:43PM ?? 0:00.00 php-fpm
0 984 1 0 4:43PM ttys000 0:00.01 sudo php-fpm
501 4098 827 0 10:24AM ttys001 0:00.00 /bin/sh ./test.sh php-fpm
====
7
====
so,my question is:why last output is 7 not 6?
thanks.
You can use command
ps -C $1 --no-headers
Updated code is
#!/bin/sh
echo "===="
echo $1
echo "===="
ps -C $1 --no-headers
#ps -lfC $1 --no-headers
echo "===="
COUNT=$(ps -C $1 --no-headers | wc -l)
echo $COUNT
echo "===="
exit 0
Run
sh /tmp/test.sh java
O/P
====
java
====
4969 ? 00:01:00 java
6884 ? 00:00:34 java
10200 ? 00:00:18 java
====
3
====

how to see which users are executing a command (Shell Scripting)

I'm wondering if you could help me with this issue.
I need to write a shell script to see if other users are executing a watch command in the same computer in which an x command is beeing executed.
Could you help me guys?
Thanks a lot.
To solve the question is possible to use ps -aux | grep <prgname> as root user.
E.G.: ps -aux | grep firefox executing this command (as root), it returns the following output:
sergio 3252 24.1 6.7 1840936 540264 ? Sl 09:48 123:36 /usr/lib/firefox/firefox
root 23059 0.0 0.0 15944 948 pts/7 S+ 18:20 0:00 grep --color=auto firefox
The last line is the command I've executed!
Using ps a way to solve your problem may be to use a script like the following. I think that is possible to create better solutions, but this seems to run good on my Ubuntu 14.
#!/bin/bash
i=0
search="watch"
tmp=`mktemp`
ps -aux | tr -s ' ' | grep "$search" > $tmp
while read fileline
do
user=`echo $fileline | cut -f1 -d\ `
prg=`echo $fileline | cut -f11 -d\ `
prg=`basename $prg`
if [ $prg == $search ]; then
echo "$user - $prg"
i=`expr $i + 1`
fi
done < $tmp
if [ $i == 0 ]; then
echo User not found
fi
rm $tmp
while [ 1 ]; do
if [ -n "`ssh $hostname pgrep -f 'Pino_special_command'`" ]; then
ssh $hostname "ps -aux | grep watch" | grep -v "grep"
fi
done
Use the pgrep command
pgrep watch

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