Bash script for retrieving server stats - bash

#!/bin/bash
printf "Date\tMemory"
end=$((SECONDS+3600))
while[$SECONDS -lt $end];do
Memory=$(free -m | awk 'NR==2 {printf "%.2f%%\t",$3*200/$2')
Date=$(date"+%Y-%m-%d %H:%M:%S")
echo "$Date $Memory"
sleep 15
done
Q1: I have written above code and saved in perfMemstats.sh file and executed it using below command:
bash perfMemstats.sh >> perfStats.dat
Problem :
If I execute this .sh file its executed only for 10 mins. I want to run this file in background so that I can monitor other stats. Please let me know what mistake I'm doing. I even tried giving '&' it running in background but only for 10 min.
If I run this command without saving it in an output file (bash perfMemstats.sh) it is executing for 1 hour.

Related

Script executes but fails to increment

So I have this shell script that I think should run a given number of times, sleep then resume, and output the results to a log file
#!/bin/bash
log=/path/to/file/info.log
a=$(COMMAND1 | cut -d : -f 2)
b=$(COMMAND2 | grep VALUE| cut -c 7,8)
for i in {1..4}
do
echo "Test" $i >> $log
date >> $log
echo $a >> $log
echo "$((-113 + (($b * 2)))) VALUE" >> $log
sleep 60
done
When I run ps -ef | grep scriptname.sh it seems the script does run. Executes once then the PID is gone as if the run has completed.
I have tested the script and know that it is running and capturing the data I want. But I do not understand why its not incrementing and not sure why its ending earlier than expected.
info.log output sample
Test {1..4}
DATE IN UTC
EXPECTED VALUE OF a
EXPECTED VALUE OF b
Note that the output is literally "Test {1..4}" not "Test 1" "Test 2" Test 3" and so on, as I would expect.
I have run the script as ./scriptname.sh & and as /path/to/file/scriptname.sh &
I have read that there is a difference in running the script with sh and bash though I dont fully understand what effect that would have on the script. I am not a software person at all.
I have tried to run the script with nohup to keep it running in the background if I close the terminal. I also thought the & in the command was supposed to keep the script running in the background. Still it seems the script does not continue to run
I previously asked this question and it was closed, citing that it was similar to a post about the difference between sh and bash...but thats not my main question.
also echo "$BASH_VERSION" returns nothing, a blank line. echo "$-" returns smi, and I have no idea what that means. but bash --version returns:
BusyBox v1.17.1 (2019-11-26 10:41:00 PST) built-in shell (ash)
Enter 'help' for a list of built-in commands.
So my questions are:
If running the script with sh - is that done with ./scriptname.sh & and running the script with bash is /path/to/file/scriptname.sh &...and if so what effect does that have on how the script code is processed? that is - is using sh or bash. I do not fully understand the difference between the two
why does the script not continue to run when I close the terminal? This is my big concern. I would like to run this script hourly for a set period of time. Every time I try something and come back I get one instance in the log.
Neither brace expansion nor seq are part of the POSIX specification. Use a while loop.
log=/path/to/file/info.log
a=$(COMMAND1 | cut -d : -f 2)
b=$(COMMAND2 | grep VALUE| cut -c 7,8)
i=1
while [ "$i" -le 4 ]; do
printf 'Test %s\n' "$i"
date
printf '%s\n' "$a"
printf '%s\n' "$((-113 + (($b * 2)))) VALUE"
sleep 60
i=$((i+1))
done >> "$log"
(I suspect that you want to move the assignments to a and b inside the loop as well; right now, you are simply writing identical files to the log at each iteration.)

launch multiple processes in bash with for loop

On ubuntu, I'm trying to launch multiple instances of filebeat on the same host using a for loop but I encounter a weird behavior
The command I'm trying to launch
for i in `seq 1 2` ; do /etc/filebeat-test/filebeat-$i/filebeat & done
And the error message I get and then when I exit the command:
Exiting: error loading config file: stat filebeat.yml: no such file or directory
^C
[1]- Termine 1 /etc/filebeat-test/filebeat-$i/filebeat
[2]+ Termine 1 /etc/filebeat-test/filebeat-$i/filebeat
It looks like $i is not interpreted..
The weird thing is that I had this command working a couple of times, but 2 seconds later it won't work. I also tried to launch it from a file and to replace $i with "$i" with the same result.
Any idea what is going on?
Thanks for your help
Does this work for you:
# set -x # uncomment this to see what actually happens while the lopp cycles
for i in `seq 1 2` ; do (/etc/filebeat-test/filebeat-$i/filebeat &); done

Using while read, do Loop in bash script, to parse command line output

So I am trying to create a script that will wait for a certain string in the output from the command that's starting another script.
I am running into a problem where my script will not move past this line of code
$(source path/to/script/LOOPER >> /tmp/looplogger.txt)
I have tried almost every variation I can think of for this line
ie. (./LOOPER& >> /tmp/looplogger.txt)
bash /path/to/script/LOOPER 2>1& /tmp/looplogger.txt etc.
For Some Reason I cannot get it to run in a subshell and have the rest of the script go about its day.
I am trying to run a script from another script and access it's output then parse line by line until a certain string is found
Then once that string is found my script would kill said script (which I am aware if it is sourced then then the parent script would terminate as well).
The script that is starting looper then trying to kill it-
#!/bin/bash
# deleting contents of .txt
echo "" > /tmp/looplogger.txt
#Code cannot get past this command
$(source "/usr/bin/gcti/LOOPER" >> /tmp/ifstester.txt)
while [[ $(tail -1 /tmp/looplogger.txt) != "Kill me" ]]; do
sleep 1
echo ' in loop ' >> /tmp/looplogger.txt
done >> /tmp/looplogger.txt
echo 'Out of loop' >> looplogger.txt
#This kill command works as intended
kill -9 $(ps -ef | grep LOOPER | grep -v grep | awk '{print $2}')
echo "Looper was killed" > /tmp/looplogger.txt
I have tried using while IFS= read -r as well. for the above script. But I find it's syntax alittle confusing.
Looper Script -
./LOOPER
#!/bin/bash
# Script to test with scripts that kill & start processes
let i=0
# Infinite While Loop
while :
do
i=$((i+1))
until [ $i -gt 10 ]
do
echo "I am looping :)"
sleep 1
((i=i+1))
done
echo "Kill me"
sleep 1
done
Sorry for my very wordy question.

Cron job won't start again after I stopped it?

I wrote a script to run constantly on startup. If for whatever reason the script were to fail, I wrote a second script to check if it has failed, and if so, run the first script again. I then set this second script as a cronjob to run every minute so that it is constantly checking if the first script is alive.
So to test this, I reboot my system. I can see in htop that the first script is running from start up as expected. Good. I kill the process to test the second script. Sure enough, the second script starts the first script again. Still good. I then kill this process, but the second script won't run again now. It still updates a txt file when I manually start the first script, but the second script just doesn't start the first script like it's supposed to. Is it because I killed the cronjob? Restarting the cron service doesn't fix anything though, so I don't know why my second script isn't running again at all.
First script:
#!/bin/bash
stamp=$(date +%Y%m%d-%H%M)
timeout 10d tcpdump -i eth0 -s 96 -z gzip -C 10 -w /home/user/Documents/${stamp}
Second script:
#!/bin/bash
echo "not running" > /home/working.txt
if (( $(ps -ef | grep -v grep | grep tcpdump.sh | wc -l) > 0 ))
then
echo "tcpdump is running!!!" > /home/working.txt
else
/usr/local/bin/tcpdump.sh start
fi
Any help?
You would probably be better off running a simple for loop as the main script, and that kicks off the tcpdump script in the background, so something like:
#!/bin/bash
while true; do
if ps -ef | grep -v grep | grep -q tcpdump; then
: tcpdump running OK
else
# tcpdump not running - start it off
nohup /usr/local/bin/firstscript.sh start &
fi
sleep 30
done
This checks that "tcpdump.sh" is in the output of the "ps -ef" command - if it is, then do nothing (note that you must have an actual command between the "then" and "else" - the ":" command, which just takes it s arguments and ignores them, is sufficient). If it isn't running, start the first script in the background. Then sleep 30 seconds and check again. (Yes, I could have inverted the test so that I didn't need an empty "then" arm, but it would have made the code less obvious)
You put this script as the one which starts at boot time.
Edit: Do you really want to check for "tcpdump.sh"? Is that what the first script is actually called? Assuming that you actually want to check for the tcpdump program, you could use:
if pgrep tcpdump; then

Restart a process when cpu gets high

I've got a cron job checking for webserver (seeing if its active), which is handy..
http://pastebin.com/raw.php?i=KW8crfzh
I'm wanting after something similar for cpu usage. I'm running java backend which occasionally gets 70%+ cpu. I'm after a cron script to automatically kill/restart java if cpu load gets too high, how is this possible?
You could use top in batch mode coupled with some code to parse its output. For example:
top -p 1234 -n 1 -b
Will output a snapshot of the state of process 1234.
I use this script and it is pretty cool
#!/bin/bash
# author = Jaysunn
# Log
LOGFILE=/var/log/load_kill_log
# log the process causing the load at the time.
PSFILE=/var/log/ps_log
# Obtain the server load
loadavg=`uptime |cut -d , -f 4|cut -d : -f 2`
thisloadavg=`echo $loadavg|awk -F \. '{print $1}'`
if [ "$thisloadavg" -ge "10" ]; then
ps auxfww >> $PSFILE
date >> $LOGFILE
# Issue the command of choice. This can be any shell command.
## Put the command which restarts ..
fi
give executable permissions and add this to crontab with proper path to this script.

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