Here is my bash code:
(
flock -n -e 200 || (echo "This script is currently being run" && exit 1)
sleep 10
...Call some functions which is written in another script...
sleep 5
) 200>/tmp/blah.lockfile
I'm running the script from two shells successively and as long as the first one is at "sleep 5" all goes good, meaning that the other one doesn't start. But when the first turns to perform the code from another script (other file) the second run starts to execute.
So I have two questions here:
What should I do to prevent this script and all its "children" from run while the script OR its "child" is still running.
(I didn't find a more appropriate expression for running another script other than a "child", sorry for that :) ).
According to man page, -n causes the process to exit when it fails to gain the lock, but as far as I can see it just wait until it can run. What am I missing ?
Your problem may be fairly mundane. Namely,
false || ( exit 1 )
Does not cause the script to exit. Rather, the exit instructs the subshell to exit. So change your first line to:
flock -n -e 200 || { echo "This script is currently being run"; exit 1; } >&2
Related
I have a freestyle Jenkins job that has a simple bash script for a build.
In the bash script, there is a for loop which sometimes returns a non-zero return code.
Thing is, Jenkins quits immediately when it happens. However, if during the run it gets a non-zero return code, the job needs to continue, and in the end mark the job as failed. (In other words, don't stop on failure but show the job failed when it's finished). (that's why I can't just append || true)
Is it possible to do? Thanks ahead!
If you set the bash option -e, (ie: #!/bin/bash -e) then the shell script will exit upon error. This is typically desired to avoid lots of testing for failures and still catching them, so many scripts have that configuration. Also, typically, this is NOT applicable inside loops if it-then-else constructs, so you should investigate what inner command you are invoking and what it is doing .. it may have an exit $? that is propagating.
If running a Jenkins execute shell step, you can add the first line #!/bin/bash +e and that may override the fail. Otherwise, try:
#!/bin/bash
FAILURE=false
set +e
for ENTRY in 1 2 3 4 5 .. N
do
command ${ENTRY}
[[ $? == 0 ]] || FAILURE=true
done
set -e
[[ ${FAILURE} == 'true' ]] && return 1
return 0
Similar to these questions (1) (2), I'm wanting to run a command in a background process, carry on processing, then later use the return value from that command.
I have one function in my script that takes particularly long, so I would like to run it first before the rest of the setup so that there is less of a delay when the return value of that script is given, but currently the return value doesn't get captured.
What I've tried:
if [ $LAZY_LOAD -eq 0 ]; then
echo "INFO - Getting least loaded server in background. Can take up to 30s."
local leastLoaded=$( getLeastLoaded ) &
fi
# Other setup stuff that doesn't use leastLoaded...
# setup setup setup....
if [ $LAZY_LOAD -eq 0 ]; then
echo "INFO - Waiting for least loaded server to be retrieved before continuing"
wait
fi
echo "INFO - Doing stuff with $leastLoaded."
doThingWithLeastLoaded $leastLoaded
getLeastLoaded definitely works without the &, so I'm sure this is a concurrency issue.
Thanks!
According to bash manual:
If a command is terminated by the control operator &, the shell executes the command in the background in a subshell.
So your local command would not affect the current shell.
I'd suggest like this:
do-something > /some/file &
... ...
wait
var=$( cat /some/file )
Is there a way to terminate a shell function non-interactively without killing the shell that's running it?
I know that the shell can be told how to respond to a signal (e.g. USR1), but I can't figure out how the signal handler would terminate the function.
If necessary you may assume that the function to be terminate has been written in such a way that it is "terminable" (i.e. by declaring some suitable options).
(My immediate interest is in how to do this for zsh, but I'm also interested in knowing how to do it for bash and for /bin/sh.)
EDIT: In response to Rob Watt's suggestion:
% donothing () { echo $$; sleep 1000000 }
% donothing
47139
If at this point I hit Ctrl-C at the same terminal that is running the shell, then the function donothing does indeed terminate, and I get the command prompt back. But if instead, from a different shell session, I run
% kill -s INT 47139
...the donothing function does not terminate.
Maybe I'm not fully understand what you want, but maybe something like this?
trap "stopme=1" 2
function longcycle() {
last=$1
for i in 1 2 3 4 5
do
[ ! -z "$stopme" ] && return
echo $i
sleep 1
done
}
stopme=""
echo "Start 1st cycle"
longcycle
echo "1st cycle end"
echo "2nd cycle"
stopme=""
longcycle
echo "2nd cycle end"
The above is for bash. Run it, and try press CTRL-C.
Or for not interactively, Save the above as for example my_command, then try:
$ ./my_command & #into background
$ kill -2 $! #send CTRL-C to the bg process
EDIT:
Solution for your sleep example in the bash:
$ donothing() { trap '[[ $mypid ]] && trap - 2 && kill $mypid' 0 2; sleep 1000000 & mypid=$!;wait; }
$ donothing
when you send a signal from another terminal will terminate it. Remeber, signal '0' je "normal end of the process". Semantic name: 0=EXIT, 2=INT... etc.
and remeber too, than signals are sending to processes not to the functions. In your example, the process is the current (interactive shell), so must use the wait trick to get something interrupt-able... Not a nice solution - but the only way when want interrupt something what is running in interactive shell (not a forked one) from the another terminal...
I am writing a script that executes around 10 back-end processes in sequence, depending on if the previous process was executed without any errors.
Now let's assume the scenario, in which lets say 5th process failed and script came out. But I want to code it in a way such that, when next time user runs it(after removing the error because of which script exited last time), he should be able to run from 5th process onwards and not again from 1st process.
To be more specific, assume following is the script:
Script Starts
Process1
if [ $? -eq 0 ] then
Process2
if [ $? -eq 0 ] then
Process3
if [ $? -eq 0 ] then
..
..
..
..
if [ $? -eq 0 ] then
Process10
else
exit
So here the script will exit anytime if any one of the process fails to complete with status 0. So again, if process5 fails, and user corrects the problem and restarts script, the script should start with process5 again and not process1 or at least there should be an option to user if he wants to resume the script or start it back from beginning i.e. process1.
What all possible ways we can code this kind of script, also please bear in mind, I am not allowed to use a temporary db, where I can store the status of each process.
I need to code in sh (shell script) in unix.
A simple solution would be to write stamp files:
#/bin/sh
set -e # Automatically abort if any simple command fails
if ! test -f cmd1-stamp; cmd1; fi
touch cmd1-stamp
if ! test -f cmd2-stamp; cmd2; fi
touch cmd2-stamp
When the script executes, if cmd1-stamp exists, cmd1 is not executed. Otherwise, cmd1 is executed. The script will abort if it fails. Note that it is very tempting to write test -f cmd1-stamp || cmd1, and this seems to work ( in bash ) but the shell specs state that the shell shall abort if the simple command that fails is not a part of an AND or OR list, and I suspect this is (yet another) instance of bash not conforming to the spec. (Although it doesn't seem to specify that the shell shall not abort if the failing command is part of an AND or OR list.)
I use SSH Secure Shell client to connect to a server and run my scripts.
I want to stop a script on some conditions, so when I use exit, not only the script stops, but all the client disconnects from the server!, Here is the code:
if [[ `echo $#` -eq 0 ]]; then
echo "Missing argument- must to get a friend list";
exit
fi
for user in $*; do
if [[ !(-f `echo ${user}.user`) ]]; then
echo "The user name ${user} doesn't exist.";
exit
fi
done
A picture of the client:
Why is this happening?
You use source to run the script, this runs it in the current shell. That means that exit terminates the current shell and with that the ssh session.
replace source with bash and it should work, or better put
#!/bin/bash
on to of the file and make it executable.
exit returns from the current shell - If you've started a script by running it directly, this will exit the shell that the script is running in.
return returns from a function or sourced file (TY Dennis Williamson) - Same thing, but it doesn't terminate your current shell.
break returns from a loop - Similar to return, but can be used anywhere within a loop to stop processing more items. This is probably what you want.
if you are running from the current shell, exit will obviously exit from the shell and disconnect you. try running it in a new shell ( use a . before the script) or else use 'return' instead of exit