I have a working command which I would like to make an alias for, but can't seem to figure out how to do. The working command is
docker-compose logs -f | less -S -R +F
This works fine, essentially removing word wrap from the logs.
The caveat is that when using this I also want to be able to optionally add arguments to the middle, to specify the particular service(s) to tail. Making something that end up looking like this
docker-compose logs -f service1 service2 etc... | less -S -R +F
To be able to do this I was thinking I could use xargs -I to pass in a variable number of arguments and inject them into the middle of the command. But my skill with bash is very limited, and so no matter how much I fiddle with it I can't seem to get it to work, and am sure there is something/some concept I'm missing.
The last iteration of the alias I tried before posting this is
alias logsx="xargs -I{} bash -c 'docker-compose logs -f \"{}\" | less -S -R +F'"
which when you run it seems to start less but without the docker logs being piped there.
Any help would be greatly appreciated. Thanks!
All you need is a function instead.
logsx() {
docker-compose logs -f "$#" | less -S -R +F
}
"$#" will expand the arguments given. So you write logsx service1 service2.
Related
#!/usr/bin/env bash
# usage: wttr [location], e.g. wttr Berlin, wttr New\ York
# Standard location if no parameters were passed
location=''
language=''
time=`date`
# Expand terminal display
if [ -z "$language" ]; then
language=${LANG%_*}
fi
curl \
-H -x "Accept-Language: ${language}" \
-x wttr.in/"${1:-${location}}" |
head -n 7 |
tee /home/of/weather.txt |
tee -a /home/of/weather.log |
tee /home/of/BACKUP/weather_"$time".txt
#cp weather.txt /home/of/BACKUP
#mv -f /home/of/BACKUP/weather.txt /home/of/BACKUP/weather_"$time".txt
I'm very new to Linux Bash and Shell scripting and can't figure out the following.
I have a problem with the shell script above.
It works fine so far (curling ASCII data from website and writing it to weather.txt and .log).
It is also in set in crontab to run every 5 minutes.
Now I need to make a backup of weather.txt under /home/of/, in /home/of/BACKUP with the filename weather_<timestamp>.txt.
I tried to delete (rm weather*.txt) the old timestamped files in /home/of/BACKUP and then copy and rename the file everytime the cronjob is running.
I tried piping cp and mv and so on but somehow I end up with producing many duplicates as due to the timestamp the filenames are different or nothing at all when I try to delete the content of the folder first.
All I need is ONE backup file of weather.txt as weather_<timestamp>.txt which gets updated every 5 minutes with the actual timestamp bit I can't figure it out.
If I understand your question at all, then simply
rm -f /home/of/BACKUP/weather_*.txt
cp /home/of/weather.txt /home/of/BACKUP/weather_"$time".txt
cp lets you rename the file you are copying to; it doesn't make sense to separately cp and then mv.
For convenience, you might want to cd /home/of so you don't have to spell out the full paths, or put them in a variable.
dir=/home/of
rm -f "$dir"/BACKUP/weather_*.txt
cp "$dir"/weather.txt "$dir"/BACKUP/weather_"$time".txt
If you are running out of the cron of the user named of then your current working directory will be /home/of (though if you need to be able to run the script manually from anywhere, that cannot be guaranteed).
Obviously, make sure the wildcard doesn't match any files you actually want to keep.
As an aside, you can simplify the tee commands slightly. If this should only update the files and not print anything to the terminal, you could even go with
curl \
-H -x "Accept-Language: ${language}" \
-x wttr.in/"${1:-${location}}" |
head -n 7 |
tee /home/of/weather.txt \
>>/home/of/weather.log
I took out the tee to the backup file since you are deleting it immediately after anyway. You could alternatively empty the backup directory first, but then you will have no backups if the curl fails.
If you want to keep printing to the terminal, too, probably run the script with redirection to /dev/null in the cron job to avoid having your email inbox fill up with unread copies of the output.
I'm a bash noob, and I am trying to set up a sort of "hot reload" functionality for a project I'm working on using inotifywait. Ubuntu 20.04 if that matters.
Here is what I hoped would have worked:
inotifywait -m -r ../.. -e modify,create,delete |
while read line; do
custom_command
done
I'm having two problems:
Issue #1 is that custom_command takes some time to work, and so if I make more changes to the directory in the meantime, custom command appears to "queue up" custom_command, where really I just want it to keep the most recent one and drop the others.
Issue #2 is that I'm getting some sort of "double output." So for example if I bash auto-exec.sh and auto-exec.sh looks like this:
inotifywait -m -r . -q -e modify,create,delete
Then each time a change registers, I get this as output (not a mistake that it's doubled -- I get two identical lines each time there is a modification):
./ MODIFY auto-exec-testfile.txt
./ MODIFY auto-exec-testfile.txt
I should note I've tried making changes both with Visual Code Studio and gedit, with the same results.
If I modify the bash file like so:
inotifywait -m -r . -q -e modify,create,delete |
while read line; do
echo "$line"
echo "..."
done
I get the following output each time there is a change:
./ MODIFY auto-exec-testfile.txt
...
./ MODIFY auto-exec-testfile.txt
...
If I modify bash_test.sh to the following:
inotifywait -m -r . -q -e modify,create,delete |
while read line; do
echo "help me..."
done
Then I get the following each time a change is made:
help me...
help me...
What happened to the the ./ MODIFY ... line?? Presumably there's something I don't understand about bash, stdout or similar /related concepts here?
And finally, if I change the .sh file to the following:
inotifywait -m -r . -q -q -e modify,create,delete |
while read _; do
echo "help me..."
done
Then I get no output at all. This one I think I understand, because the -q -q means that inotifywait is in "super silent" mode, so there is no log and therefore nothing to trigger the while.
What I'd love to do is just trigger the code once when something changes, and drop all but the most recent execution. I'm not sure doing this using a while is entirely necessary, but I tried inotifywait -m -r . -q -q -e modify,create,delete | echo "help me..", and the script printed "help me..." once at startup, then exited on modification.
Assistance very much appreciated.
EDIT - 20201-Mar-23
I removed -m and create from the inotifywait line, and it appears to work as expected, except that it doesn't stay "up" in monitor mode. So this at least only gives me one entry from inotifywait:
notifywait -r .. -q -e modify,delete |
while read line1; do
echo ${line1}
done
Related:
inotifywait - pause monitoring while executing command
https://unix.stackexchange.com/questions/140679/using-inotify-to-monitor-a-directory-but-not-working-100
inotifywait not performing the while loop in bash script
while inotifywait -e close_write,delete .; do
pkill custom_command
custom_command&
done
I have the following shell script running a inotifywait command. I want to print the output echo to the console upon every modify event.
The script:
#!/bin/sh
while inotifywait -e modify -r -m ./ --exclude '\.sh$'; do
echo test
done
When I change one file in the specified directory, i get the standard output from inotifywait:
Setting up watches. Beware: since -r was given, this may take a while!
Watches established.
./postgres/ MODIFY postgres_test.go
./postgres/ MODIFY postgres_test.go
I have two questions:
Why is the modified event registered twice? I only updated the file once.
Why is "test" not being printed to the console in which I'm running the script?
I had a similar issue. I resolved the second part by restructuring my while:
inotifywait -e modify -r -m ./ --exclude '\.sh$' |
while read E; do
echo "----------------hello $E"
done
I have been monitoring the performance of my Linux server with ioping (had some performance degradation last year). For this purpose I created a simple script:
echo $(date) | tee -a ../sb-output.log | tee -a ../iotest.txt
./ioping -c 10 . 2>&1 | tee -a ../sb-output.log | grep "requests completed in\|ioping" | grep -v "ioping statistics" | sed "s/^/IOPing I\/O\: /" | tee -a ../iotest.txt
./ioping -RD . 2>&1 | tee -a ../sb-output.log | grep "requests completed in\|ioping" | grep -v "ioping statistics" | sed "s/^/IOPing seek rate\: /" | tee -a ../iotest.txt
etc
The script calls ioping in the folder /home/bench/ioping-0.6. Then it saves the output in readable form in /home/bench/iotest.txt. It also adds the date so I can compare points in time.
Unfortunately I am no experienced programmer and this version of the script only works if you first enter the right directory (/home/bench/ioping-0.6).
I would like to call this script from anywhere. For example by calling
sh /home/bench/ioping.sh
Googling this and reading about path variables was a bit over my head. I kept up ending up with different version of
line 3: ./ioping: No such file or directory
Any thoughts on how to upgrade my scripts so that it works anywhere?
The trick is the shell's $0 variable. This is set to the path of the script.
#!/bin/sh
set -x
cd $(dirname $0)
pwd
cd ${0%/*}
pwd
If dirname isn't available for some reason, like some limited busybox distributions, you can try using shell parameter expansion tricks like the second one in my example.
Isn't it obvious? ioping is not in . so you can't use ./ioping.
Easiest solution is to set PATH to include the directory where ioping is. perhaps more robust - figure out the path to $0 and use that path as the location for ioping (assing your script sits next to ioping).
If iopinf itself depend on being ruin in a certain directory, you might have to make your script cd to the ioping directory before running.
I'm testing a Bash script I created on GitHub for behavioral correctness (e.g. that it parses options correctly). I want to do this without having to clone the repository locally, so here is how I'm doing it:
curl -sSL https://github.com/jamesqo/gid/raw/master/gid | xargs -0 bash -c
My question is, how can I pass arguments to the script in question? I tried bash -c --help, but that didn't work since it got interpreted as part of the script.
Thanks!
You’re actually over-complicating things by using xargs with Bash’s -c option.
Download the script directly
You don’t need to clone the repository to run the script. Just download it directly:
curl -o gid https://raw.githubusercontent.com/jamesqo/gid/master/gid
Now that it’s downloaded as gid, you can run it as a Bash script, e.g.,
bash gid --help
You can also make the downloaded script executable in order to run it as a regular Unix script file (using its shebang, #!/bin/bash):
chmod +x gid
./gid --help
Use process substitution
If you wanted to run the script without actually saving it to a file, you could use Bash process substitution:
bash <(curl -sSL https://github.com/jamesqo/gid/raw/master/gid) --help
I'll echo Anthony's comments - it makes a lot more sense to download the script and execute it directly, but if you're really set on using the -c option for bash, it's a little bit complicated, the problem is that when you do:
something | xargs -0 bash -c
there's no opportunity to pass any arguments. They all get swallowed as the argument to -c - it essentially gets turned into:
bash -c "$(something)"
so if you place something after the -c in the xargs, it gets before the something. There is no opportunity to put anything after something, as xargs doesn't let you.
If you want to pass arguments, you have to use the substitution position option for xargs, which allows you to place where the argument goes, The option is -J <item>, and the next thing to realize is that the first argument will be $0, so you have to do:
something | xargs -0 -I # bash -c # something <arg1> <arg2>…
I can emulate this with:
echo 'echo hi: ~$0~ ~$1~ ~$2~ ~$3~' | xargs -0 -I # bash -c # something one two three four
which yields:
hi: ~something~ ~one~ ~two~ ~three~