I have a section of code in a bash script that uses a while loop to grep a file until the string I am looking for is there, then exit. Currently, its just hanging using the following code:
hostname="test-cust-15"
VAR1=$(/bin/grep -wo -m1 "HOST ALERT: $hostname;DOWN" /var/log/logfile)
while [ ! "$VAR1" ]
do
sleep 5
done
echo $VAR1 was found
I know the part of the script responsible for inserting this string into the logfile works, as I can grep it out side of the script and find it.
One thing I have tried is to change up the variables. Like this:
hostname="test-cust-15"
VAR1="HOST ALERT: $hostname;DOWN"
while [ ! /bin/grep "$VAR1" /var/log/logfile ]
do
sleep 5
done
echo $VAR1 was found
But i get a binary operator expected message and once I got a too many arguments message when using this:
while [ ! /bin/grep -q -wo "$VAR1" /var/log/logfile ]
What do I need to do to fix this?
while/until can work off of the exit status of a program directly.
until /bin/grep "$VAR1" /var/log/logfile
do
sleep 5
done
echo "$VAR1" was found
You also mentioned that it prints out the match in an above comment. If that's not desirable, use output redirection, or grep's -q option.
until /bin/grep "$VAR1" /var/log/logfile >/dev/null
until /bin/grep -q "$VAR1" /var/log/logfile
No need to bother with command substitution or test operator there. Simply:
while ! grep -wo -m1 "HOST ALERT: $hostname;DOWN" /var/log/logfile; do
sleep 5
done
Don't waste resources, use tail!
#!/bin/bash
while read line
do
echo $line
break
done < <(tail -f /tmp/logfile | grep --line-buffered "HOST ALERT")
Related
I'm trying to understand why this very simple script is not working:
echo "SomeText" > test
if [[ "grep 'FindMe' ./test" ]] ; then
echo Why is this line written to screen?
fi
I have tried with:
"grep 'FindMe' ./test"
$(grep 'FindMe' ./test)
`grep 'FindMe' ./test`
also with a lot of grep options.
I have tried with [] and [[]] and {} and any combination.
In fact, to check the exit status, use no parentheses and no quotes:
if grep 'FindMe' ./test ; then
Using grep -q is common in such situations as we don't want the output of the script to be cluttered with random matches.
I've got the following code in my shell script:
SERVER=`ps -ef | grep -v grep | grep -c sof2ded`
if ["$SERVER" != "0"]; then
echo "Already Running, exiting"
exit
else
echo "Starting up the server..."
cd /home/sof2/
/home/sof2/crons/start.sh > /dev/null 2>&1
fi
I did chmod a+x status.sh
Now I try to run the script but it's returning this error:
./status.sh: line 5: [1: command not found
Starting up the server...
Any help would be greatly appreciated.
Could you please try changing a few things in your script as follows and let me know if that helps you?(changed back-tick to $ and changed [ to [[ in code)
SERVER=$(ps -ef | grep -v grep | grep -c sof2ded)
if [[ "$SERVER" -eq 0 ]]; then
echo "Already Running, exiting"
exit
else
echo "Starting up the server..."
cd /home/sof2/
/home/sof2/crons/start.sh > /dev/null 2>&1
fi
The problem is with the test command. "But", I hear you say, "I am not using the test command". Yes you are, it is also known as [.
if statement syntax is if command. The brackets are not part of if syntax.
Commands have arguments separated (tokenized) by whitespace, so:
[ "$SERVER" != "0" ]
The whitespace is needed because the command is [ and then there are 4 arguments passed to it (the last one must be ]).
A more robust way of comparing numerics is to use double parentheses,
(( SERVER == 0 ))
Notice that you don't need the $ or the quotes around SERVER. Also the spacing is less important, but useful for readability.
[[ is used for comparing text patterns.
As a comment, backticks ` ` are considered deprecated because they are difficult to read, they are replaced with $( ... ).
Is it possible to write a script that does not proceed till a given line appears in a particular file?
For example I want to do something like this:
CANARY_LINE='Server started'
FILE='/var/logs/deployment.log'
echo 'Waiting for server to start'
.... watch $FILE for $CANARY_LINE ...
echo 'Server started'
Basically, a shell script that watches a file for line (or regex).
tail -n0 -f path_to_my_log_file.log | sed '/particular_line/ q'
You can use the q flag while parsing the input via sed. Then sed will interrupt tail as soon as Server started appears in /var/logs/deployment.log.
tail -f /var/logs/deployment.log | sed '/Server started/ q'
Another way to do the same thing
( tail -f -n0 /var/logs/deployment.log & ) | grep -q "Server Started"
Previous answer (works but not as efficient than this one)
We have to be careful with loops.
For example if you want to check for a file to start an algorithm you've probably have to do something like that:
FILE_TO_CHECK="/var/logs/deployment.log"
LINE_TO_CONTAIN="Server started"
SLEEP_TIME=10
while [ $(cat FILE_TO_CHECK | grep "${LINE_TO_CONTAIN}") ]
do
sleep ${SLEEP_TIME}
done
# Start your algorithm here
But, in order to prevent an infinite loop you should add some bound:
FILE_TO_CHECK="/var/logs/deployment.log"
LINE_TO_CONTAIN="Server started"
SLEEP_TIME=10
COUNT=0
MAX=10
while [ $(cat FILE_TO_CHECK | grep "${LINE_TO_CONTAIN}") -a ${COUNT} -lt ${MAX} ]
do
sleep ${SLEEP_TIME}
COUNT=$(($COUNT + 1))
done
if [ ! $(cat FILE_TO_CHECK | grep "${LINE_TO_CONTAIN}") ]
then
echo "Let's go, the file is containing what we want"
# Start your algorithm here
else
echo "Timed out"
exit 10
fi
CANARY_LINE='Server started'
FILE='/var/logs/deployment.log'
echo 'Waiting for server to start'
grep -q $CANARY_LINE <(tail -f $FILE)
echo 'Server started'
Source: adapted from How to wait for message to appear in log in shell
Try this:
#!/bin/bash
canary_line='Server started'
file='/var/logs/deployment.log'
echo 'Waiting for server to start'
until grep -q "${canary_line}" "${file}"
do
sleep 1s
done
echo 'Server started'
Adjust sleep's parameter to your taste.
If the line in the file needs to match exactly, i.e. the whole line, change grep's second parameter to "^${canary_line}$".
If the line contains any characters that grep thinks are special, you're going to have to solve that... somehow.
I have following code in my build script:
if [ -z "$1" ]; then
make -j10 $1 2>&1 | tee log.txt && notify-send -u critical -t 7 "BUILD DONE"
else
make -j10 $1 2>&1 | tee log.txt | grep -i --color "Error" && notify-send -u critical -t 7 "BUILD DONE"
fi
I tried to optimize it to:
local GREP=""
[[ ! -z "$1" ]] && GREP="| grep -i --color Error" && echo "Grepping for ERRORS"
make -j10 $1 2>&1 | tee log.txt "$GREP" && notify-send -u critical -t 7 "BUILD DONE"
But error thrown in make line if $1 isn't empty. I just can't figure out how to pass command with grep pipe through the variable.
Like others have already pointed out, you cannot, in general, expect a command in a variable to work. This is a FAQ.
What you can do is execute commands conditionally. Like this, for example:
( make -j10 $1 2>&1 && notify-send -u critical -t 7 "BUILD DONE" ) |
tee log.txt |
if [ -z "$1" ]; then
grep -i --color "Error"
else
cat
fi
This has the additional unexpected benefit that the notify-send is actually conditioned on the exit code of make (which is probably what you intended) rather than tee (which I would expect to succeed unless you run out of disk or something).
(Or if you want the notification regardless of the success status, change && to just ; -- I think this probably makes more sense.)
This is one of those rare Useful Uses of cat (although I still feel the urge to try to get rid of it!)
You can't put pipes in command variables:
$ foo='| cat'
$ echo bar $foo
bar | cat
The linked article explains how to do such things very well.
As mentioned in #l0b0's answer, the | will not be interpreted as you are hoping.
If you wanted to cut down on repetition, you could do something like this:
if [ $(make -j10 "$1" 2>&1 > log.txt) ]; then
[ "$1" ] && grep -i --color "error" log.txt
notify-send -u critical -t 7 "BUILD DONE"
fi
The inside of the test is common to both branches. Instead of using tee so that the output can be piped, you can just indirect the output to log.txt. If "$1" isn't empty, grep for any errors in log.txt. Either way, do the notify-send.
This question already has answers here:
Quick-and-dirty way to ensure only one instance of a shell script is running at a time
(43 answers)
Closed 5 years ago.
now this is embarrassing. I'm writing quick script and I can't figure out why this statement don't work.
if [ $(pidof -x test.sh | wc -w) -eq 1 ]; then echo Passed; fi
I also tried using back-ticks instead of $() but it still wouldn't work.
Can you see what is wrong with it? pidof -x test.sh | wc -w returns 1 if I run it inside of script, so I don't see any reason why basically if [ 1 -eq 1 ] wouldn't pass.
Thanks a lot!
Jefromi is correct; here is the logic I think you want:
#!/bin/bash
# this is "test.sh"
if [ $(pidof -x test.sh| wc -w) -gt 2 ]; then
echo "More than 1"
exit
fi
echo "Only one; doing whatever..."
Ah, the real answer: when you use a pipeline, you force the creation of a subshell. This will always cause you to get an increased number:
#!/bin/bash
echo "subshell:"
np=$(pidof -x foo.bash | wc -w)
echo "$np processes" # two processes
echo "no subshell:"
np=$(pidof -x foo.bash)
np=$(echo $np | wc -w)
echo "$np processes" # one process
I'm honestly not sure what the shortest way is to do what you really want to. You could avoid it all by creating a lockfile - otherwise you probably have to trace back via ppid to all the top-level processes and count them.
you don't have to pass the result of pidof to wc to count how many there are..use the shell
r=$(pidof -x -o $$ test.sh)
set -- $r
if [ "${##}" -eq 1 ];then
echo "passed"
else
echo "no"
fi
If you use the -o option to omit the PID of the script ($$), then only the PID of the subshell and any other instances of the script (and any subshells they might spawn) will be considered, so the test will pass when there's only one instance:
if [ $(pidof -x -o $$ test.sh | wc -w) -eq 1 ]; then echo Passed; fi
Here's how I would do it:
if [ "`pgrep -c someprocess`" -gt "1" ]; then
echo "More than one process running"
else
echo "Multiple processes not running"
fi
If you don't want to use a lockfile ... you can try this:
#!/bin/bash
if [[ "$(ps -N -p $$ -o comm,pid)" =~ $'\n'"${0##*/}"[[:space:]] ]]; then
echo "aready running!"
exit 1
fi
PS: it might need adjustment for a weird ${0##*/}
Just check for the existence of any one (or more) process identified as test.sh, the return code will be 1 if none are found:
pidof -x test.sh >/dev/null && echo "Passed"