I have 2 scripts one is software.sh and the other timer.sh both get executed in separate terminals. I want the if statement that checks for software execution duration to be checking continuously but apparently it only gets checked once at start of execution :(
This is the timer.sh which gets the PID of software.sh and counts its running time
#!/bin/bash
while ((1)); do
PID_Ardu=$(pidof arducopter)
Time= ps -p $PID_Ardu -oetime= | tr '-' ':' | awk -F: '{ total=0; m=1; } { for (i=0; i < NF; i++) {total += $(NF-i)*m; m *= i >= 2 ? 24 : 60 }} {print total}'
if [[ $Time < 20 ]]; then **#THIS SHOULD BE EXECUTED CONTINOUSLY**
echo $Time
else
echo INSIDE
ps -ef | grep software| grep -v grep | awk '{print $2}' | xargs kill
fi
done
This is software.sh which executes various generated files
#!/bin/bash
chmod a+x *.exp
./ACCF01.exp | wait |echo "FINISHED SCRIPT 1" #Run the ACCF01 script
sed -i '/ACCF01.exp/d' ./software.sh #Delete previous line after finishing execution.
./ACCF02.exp | wait |echo "FINISHED SCRIPT 2"
sed -i '/ACCF02.exp/d' ./software.sh
./ACCF03.exp | wait |echo "FINISHED SCRIPT 3"
sed -i '/ACCF03.exp/d' ./software.sh
./ACCF04.exp | wait |echo "FINISHED SCRIPT 4"
The Time variable is assigned a string, not the return value of the pipeline, because it needs to enclose it in $(..) or backticks.
In any case the If is executed multiple times, it just has no changing numerical value in the variable.
When you try to debug loops and variable assignments you can use set -x or start with bash -x time.sh in the future. It will print out all lines execute and variables expanded.
Related
I want to run a script in the background, and when the time comes bring it back from background and call another script that will take in user input from the command-line, when I use this code, it's able to run the script effectively, but when it calls the other script, it only prints out a couple lines and It isn't able to take in user input, just straight up exits out. Been at it for a couple hours, no idea what to do from here.
I call this script with ./runTool.sh &
currTime=$(ls -lu | grep test01 | awk '{print $8}')
currHour="${currTime:0:2}"
currMin="${currTime:3:3}"
check=0
while [ true ]
do
timestamp=$(ls -lu | grep test01 | awk '{print $8}')
timeHour="${timestamp:0:2}"
timeMin="${timestamp:3:3}"
if (( $timeHour > $currHour )) || (( $timeMin > $currMin )) || (($timeHour < $currHour ))
then
check=1
set -m
fg %1
./tool.sh
break
fi
sleep 1
done
I'm capturing stdout (log) in a file using tail -f file_name to save a specific string with grep and sed (to exit the tail) :
tail -f log.txt | sed /'INFO'/q | grep 'INFO' > info_file.txt
This works fine, but I want to terminate the command in case it does not find the pattern (INFO) in the log file after some time
I want something like this (which does not work) to exit the script after a timeout (60sec):
tail -f log.txt | sed /'INFO'/q | grep 'INFO' | read -t 60
Any suggestions?
This seems to work for me...
read -t 60 < <(tail -f log.txt | sed /'INFO'/q | grep 'INFO')
Since you only want to capture one line:
#!/bin/bash
IFS= read -r -t 60 line < <(tail -f log.txt | awk '/INFO/ { print; exit; }')
printf '%s\n' "$line" >info_file.txt
For a more general case, where you want to capture more than one line, the following uses no external commands other than tail:
#!/usr/bin/env bash
end_time=$(( SECONDS + 60 ))
while (( SECONDS < end_time )); do
IFS= read -t 1 -r line && [[ $line = *INFO* ]] && printf '%s\n' "$line"
done < <(tail -f log.txt)
A few notes:
SECONDS is a built-in variable in bash which, when read, will retrieve the time in seconds since the shell was started. (It loses this behavior after being the target of any assignment -- avoiding such mishaps is part of why POSIX variable-naming conventions reserving names with lowercase characters for application use are valuable).
(( )) creates an arithmetic context; all content within is treated as integer math.
<( ) is a process substitution; it evaluates to the name of a file-like object (named pipe, /dev/fd reference, or similar) which, when read from, will contain output from the command contained therein. See BashFAQ #24 for a discussion of why this is more suitable than piping to read.
The timeout command, (part of the Debian/Ubuntu "coreutils" package), seems suitable:
timeout 1m tail -f log.txt | grep 'INFO'
Code sample :
declare -i a=1
echo "The number of NMON instances running in Performance VM"
ps -ef | grep nmon | awk '{ print $2 }' | wc -l
echo "---------------------------------------------"
num2= ps -ef | grep nmon | awk '{ print $2 }' | wc -l
#num1=${num2%%.*}
#num2 = $(ps -ef | grep nmon | awk '{ print $2 }' | wc -l)
echo "---------------------------------------------"
echo "${num2}"
while [ "$a" -lt "$num2" ]
do
kill -USR2 $(ps -ef | grep nmon | awk '{ print $2 }' | head -1)
a=`expr $a + 1`
done
In the Output i am getting the following error
[: : integer expression expected
in the debug it shows
++ '[' 1 -lt '' ']'
that num2 is empty but when i echo the num2 value i am getting the value correctly.
Output:
The number of NMON instances running in Performance VM
1
1
thanks in advance
The 1 you see in the output is not from echo "${num2}". Like the diagnostics already tell you, this variable is empty.
The general syntax of shell scripts is
[ variable=value ...] command parameters ...
which will assign value to variable for the duration of command, then restore its original value. So the pipeline you are running temporarily sets num2 to the empty string (which apparently it already contained anyway), then runs the pipeline without storing the output anywhere (such as, I imagine you expected, in num2).
Here is a fixed version of your script, with the additional change that the Awk scripts handle stuff you used grep and head and wc for. Because the functionality of these commands is easily replaced within Awk, using external utilities is doubtful (especially so for grep which really is useless when you just run it as a preprocessor for a simple Awk script).
countnmon () {
ps -ef | awk '/[n]mon/ { ++n } END { print n }'
}
declare -i a=1
echo "The number of NMON instances running in Performance VM"
countnmon
echo "---------------------------------------------"
num2=$(countnmon)
#num1=${num2%%.*}
#num2 = $(countnmon)
echo "---------------------------------------------"
echo "${num2}"
while [ "$a" -lt "$num2" ]
do
kill -USR2 $(ps -ef | awk '/[n]mon/ { print $2; exit }')
a=`expr $a + 1`
done
The repeated code could be refactored even further to avoid all code duplication but that will somewhat hamper the readability of this simple script so I have not done that.
Whitespaces matter in bash.
The syntax for command execution is:
command arg1 arg2 ...
So,
var = value # command: var, arg1: =, arg2: value
There's are two exceptions to this rule
exporting variables to an executed command (the variables vanish after the command finishes):
var1=value1 var2=value2 .. command arg1 arg2 ...
assigning a variable (you want this one):
var=value
I have this code:
total=0;
ps -u $(whoami) --no-headers | awk {'print $1'} | while read line;
do vrednost=$(pmap $line | tail -n1 | column -t | cut -d" " -f3 | tr "K" " ");
total=$(( vrednost + total ))
echo $total
done
echo total: $total
As you can see, my code sums usage of all my processes. When I echo my total every time in while, it is working ok, but at the end... When i want total to be a value (echo total: $total) it is still zero. but before (in while) has right value.
BASH FAQ #24: "I set variables in a loop that's in a pipeline. Why do they disappear after the loop terminates? Or, why can't I pipe data to read?"
#!/bin/bash
while read ...
do
...
done < <(ps ...)
Okay, pick and choose. You can either do it in BASH or AWK, but don't do both. You've seen a BASH example, here's an AWK example:
ps -e -o user -o vsz | awk -v USER="$(whoami)" '
BEGIN {TOTAL = 0}
END {print "Total is " TOTAL}
{
if ($1 == USER) {
TOTAL += $2
}
}
'
Awk is like a programming language that assumes a loop (like perl -n) and processes each line in the file. Each field (normally separated by whitespace) is given a $ variable. The first is $1, the second is $2, etc.
The -v option allows me to define an awk variable (in this case USER) before I run awk.
The BEGIN line is what I want to do before I run my awk script. In this case, initialize TOTAL to zero. (NOTE: This really isn't necessary since undefined variables automatically are given a value of zero). The END line is what I want to do afterwards. In this case, print out my total.
So, if the first field ($1) is equal to my user, I'll add the second field (the vsize) to my total.
All Awk programs are surrounded by {...} and they usually have single quotes around them to prevent shell interpolation of $1, etc.
Try this
#!/bin/bash
total=0;
for line in `ps -u $(whoami) --no-headers | awk {'print $1'}`;
do
vrednost=$(pmap $line | tail -n1 | column -t | cut -d" " -f3 | tr "K" " ");
total=$(( $vrednost + $total ))
echo $total
done
echo "total: $total"
Ignacio's answer is fine, but process substitution is not portable. And there is a simpler solution. You need to echo the total in the same subshell in which it is calculated:
... | { while read line; do ...; done; echo total: $total; }
Let's cut down on the number of extra processes you need :)
declare -i total=0
for size in $( ps -u $(whoami) --no-header -o vsz ); do
total+=$size
done
echo $total
First, use various options for ps to generate the desired list of process sizes, in kilobytes. Iterate over that list using a bash for-loop, keeping a running total in a parameter declared with the 'integer' attribute for easy arithmetic. The desired sum is now in total, ready for whatever use you need. The sum includes the memory used by the ps process itself.
Using while (Dennis' suggestion) and avoiding process substitution (William's suggestion):
ps -u $(whoami) --no-header -o vsz | {
while read -r var; do
((total+=$var))
done
echo $total
}
(For real one-liner, here's a dc command that I borrowed from https://stackoverflow.com/a/453290/1126841:
ps -u $(whoami) --no-header -o vsz | dc -f - -e '[+z1<r]srz1<rp'
This sum includes the memory used by the ps and dc commands themselves.)
Infinite loop on bash script and I want to run forever but (I guess) something goes wrong script is killed. Is there any way like try-catch, just continue to running forever, unconditionaly.
#!/bin/bash
iteration=0
for (( ; ; ))
do
process_id=`ps -ef | grep java | grep TEST | awk '{print $2}' `
kill_command='kill -3 '$process_id
time=`date | awk '{print substr($4,0,5)}' `
last_write=`ls -l /files/*.txt | awk '{print $8}' `
if [ "$time" != "$last_write" ]
then
$kill_command
sleep 1
$kill_command
sleep 1
$kill_command
sleep 1
/test/show_queue.sh
fi
let "iteration+=1"
if [ "$iteration" == "30" ]
then
let "iteration=0"
$kill_command
echo '------------' >> memory_status.log
date >> memory_status.log
prstat -n 7 1 1 >> memory_status.log
echo '------------' >> memory_status.log
/test/show_queue.sh
fi
sleep 60
done
A very simple way to do it is to use two scripts. One with the loop and one which does the killing task :
for (( ; ; ))
do
DoKillingTask
rc=$? # <- You get the return code of the script and decide what to do
done
If it continues to be killed, Mikel (in comment of your question) is right.