How to make a code bellow as a general function to be used entire script in bash:
if [[ $? = 0 ]]; then
echo "success " >> $log
else echo "failed" >> $log
fi
You might write a wrapper for command execution:
function exec_cmd {
$#
if [[ $? = 0 ]]; then
echo "success " >> $log
else
echo "failed" >> $log
fi
}
And then execute commands in your script using the function:
exec_cmd command1 arg1 arg2 ...
exec_cmd command2 arg1 arg2 ...
...
If you don't want to wrap the original calls you could use an explicit call, like the following
function check_success {
if [[ $? = 0 ]]; then
echo "success " >> $log
else echo "failed" >> $log
fi
}
ls && check_success
ls non-existant
check_success
There's no really clean way to do that. This is clean and might be good enough?
PS4='($?)[$LINENO]'
exec 2>>"$log"
That will show every command run in the log, and each entry will start with the exit code of the previous command...
You could put this in .bashrc and call it whenever
function log_status { [ $? == 0 ] && echo success>>/tmp/status || echo fail>>/tmp/status }
If you want it after every command you could make the prompt write to the log (note the original PS1 value is appended).
export PS1="\$([ \$? == 0 ] && echo success>>/tmp/status || echo fail>>/tmp/status)$PS1"
(I'm not experienced with this, perhaps PROMPT_COMMAND is a more appropriate place to put it)
Or even get more fancy and see the result with colours.
I guess you could also play with getting the last executed command:
How do I get "previous executed command" in a bash script?
Get name of last run program in Bash
BASH: echoing the last command run
Related
I have a bash script which is only meant to used be when sourced.
I want to return from it automatically on any error, similar to what set -e does.
However setting set -e doesn't work for me because it will also exit the users shell.
Right now I'm handling returning manually like this command || return 1, for each command.
You can also use command || true or command || return.
If your requirement is something different, please update more precisely.
You can use trap. E.g.:
// foo.sh
function func() {
trap 'if [ $? -ne 0 ]; then echo "Trapped!"; return ; fi' DEBUG
echo 'foo'
find -name "foo" . 2> /dev/null
echo 'bar'
}
func
Two notes. First, the trap needs to be inside the function as shown. It won't work if it's just inside the script.
Two, there is a significant limitation. Even if you set the return to the trap (e.g., return 1), while func exists after the bad find command, $? is still zero, no matter what. I'm not sure if there's a way around that, so if it's important to preserve the exit value of the failed command, this may not work.
E.g., if you had:
func
func_return=$?
echo "return value is: $func_return"
func_return will always be zero. I've played around with trying to get the exit value of the failed command to pass out of the function trap and into the function exit value, but have not found a way to do it.
If you need to preserve the return value, you could update a global variable inside the debug trap.
If I understand well, you can set -e locally in each function.
cat sourced
f1 () {
local -
set -e
[ "$1" -eq "$1" ] 2> /dev/null && echo "$1"
}
cat script.sh
. sourced
param='bad'
ret=$(f1 "$param")
[ $? -eq 0 ] && echo "result = $ret" || \
echo "error in sourced file with param $param"
param=3
ret=$(f1 "$param")
[ $? -eq 0 ] && echo "result = $ret" || \
echo "error in sourced file with param $param"
I am running club.sh
inside club.sh script i am running below scripts.
test1.sh
test2.sh
test3.sh
my concern is it should run one by one and if test1 fails it will not run test2.sh and if test2.sh fails it willnot run test3.sh
how can we check? can any one suggest any idea it would be very helpful.
Thanks,
Two approaches -
First, you can examine the exit code of each of your inner scripts (test1.sh, test2.sh, ...) and decide whether to continue accordingly -
$? Will return the exit code of the previous command. It will be 0 (zero) if the script exited without an error. Anything that is not 0 can be considered a failure. So you could so something like this -
./test1.sh # execute script
if [[ $? != 0 ]]; then exit; fi # check return value, exit if not 0
Alternatively, you could use the && bash operator which will only execute subsequent commands if the previous one passed -
./test1.sh && ./test2.sh && test3.sh
Only if test1.sh returns an exit code of 0 (zero) will test2.sh execute and the same goes for test3.sh.
The first approach is good if you need to do some logging or cleanup between executing your scripts, but if you are only concerned that the execution should not continue if there was a failure then the && method would be they way I recommend.
Here is a related post dealing with the meaning behind &&
The returned value of the execution of the first command/script is stored in $? so using this value you can check if your command was successfully executed.
Try this:
bash test1.sh
if [ $? -eq 0 ]; then # if script succeeded
bash test2.sh
else
echo "script failed"
fi
If you want to exit your script whenever a command fails, you just add at the beginning of your script set -e.
#!/bin/bash
set -e
echo hello
ls /root/lalala
echo world
Otherwise, you have two options.
The first one is to use &&. For instance:
echo hello && ls /some_inexistant_directory && echo world
The second one is to check the return value after each command:
#!/bin/bash
echo toto
if [ "$?" != "0" ]; then
exit 1
fi
ls /root
if [ "$?" != "0" ]; then
exit 1
fi
echo world
if [ "$?" != "0" ]; then
exit 1
fi
You just have to put the below at the begging of the script:
#!/bin/bash -e
I generally have -e set in my Bash scripts, but occasionally I would like to run a command and get the return value.
Without doing the set +e; some-command; res=$?; set -e dance, how can I do that?
From the bash manual:
The shell does not exit if the command that fails is [...] part of any command executed in a && or || list [...].
So, just do:
#!/bin/bash
set -eu
foo() {
# exit code will be 0, 1, or 2
return $(( RANDOM % 3 ))
}
ret=0
foo || ret=$?
echo "foo() exited with: $ret"
Example runs:
$ ./foo.sh
foo() exited with: 1
$ ./foo.sh
foo() exited with: 0
$ ./foo.sh
foo() exited with: 2
This is the canonical way of doing it.
as an alternative
ans=0
some-command || ans=$?
Maybe try running the commands in question in a subshell, like this?
res=$(some-command > /dev/null; echo $?)
Given behavior of shell described at this question it's possible to use following construct:
#!/bin/sh
set -e
{ custom_command; rc=$?; } || :
echo $rc
Another option is to use simple if. It is a bit longer, but fully supported by bash, i.e. that the command can return non-zero value, but the script doesn't exit even with set -e. See it in this simple script:
#! /bin/bash -eu
f () {
return 2
}
if f;then
echo Command succeeded
else
echo Command failed, returned: $?
fi
echo Script still continues.
When we run it, we can see that script still continues after non-zero return code:
$ ./test.sh
Command failed, returned: 2
Script still continues.
Use a wrapper function to execute your commands:
function __e {
set +e
"$#"
__r=$?
set -e
}
__e yourcommand arg1 arg2
And use $__r instead of $?:
if [[ __r -eq 0 ]]; then
echo "success"
else
echo "failed"
fi
Another method to call commands in a pipe, only that you have to quote the pipe. This does a safe eval.
function __p {
set +e
local __A=() __I
for (( __I = 1; __I <= $#; ++__I )); do
if [[ "${!__I}" == '|' ]]; then
__A+=('|')
else
__A+=("\"\$$__I\"")
fi
done
eval "${__A[#]}"
__r=$?
set -e
}
Example:
__p echo abc '|' grep abc
And I actually prefer this syntax:
__p echo abc :: grep abc
Which I could do with
...
if [[ ${!__I} == '::' ]]; then
...
In a loop in shell script, I am connecting to various servers and running some commands. For example
#!/bin/bash
FILENAME=$1
cat $FILENAME | while read HOST
do
0</dev/null ssh $HOST 'echo password| sudo -S
echo $HOST
echo $?
pwd
echo $?'
done
Here I am running "echo $HOST" and "pwd" commands and I am getting exit status via "echo $?".
My question is that I want to be able to store the exit status of the commands I run remotely in some variable and then ( based on if the command was success or not) , write a log to a local file.
Any help and code is appreciated.
ssh will exit with the exit code of the remote command. For example:
$ ssh localhost exit 10
$ echo $?
10
So after your ssh command exits, you can simply check $?. You need to make sure that you don't mask your return value. For example, your ssh command finishes up with:
echo $?
This will always return 0. What you probably want is something more like this:
while read HOST; do
echo $HOST
if ssh $HOST 'somecommand' < /dev/null; then
echo SUCCESS
else
echo FAIL
done
You could also write it like this:
while read HOST; do
echo $HOST
if ssh $HOST 'somecommand' < /dev/null
if [ $? -eq 0 ]; then
echo SUCCESS
else
echo FAIL
done
You can assign the exit status to a variable as simple as doing:
variable=$?
Right after the command you are trying to inspect. Do not echo $? before or the new value of $? will be the exit code of echo (usually 0).
An interesting approach would be to retrieve the whole output of each ssh command set in a local variable using backticks, or even seperate with a special charachter (for simplicity say ":") something like:
export MYVAR=`ssh $HOST 'echo -n ${HOSTNAME}\:;pwd'`
after this you can use awk to split MYVAR into your results and continue bash testing.
Perhaps prepare the log file on the other side and pipe it to stdout, like this:
ssh -n user#example.com 'x() { local ret; "$#" >&2; ret=$?; echo "[`date +%Y%m%d-%H%M%S` $ret] $*"; return $ret; };
x true
x false
x sh -c "exit 77";' > local-logfile
Basically just prefix everything on the remote you want to invoke with this x wrapper. It works for conditionals, too, as it does not alter the exit code of a command.
You can easily loop this command.
This example writes into the log something like:
[20141218-174611 0] true
[20141218-174611 1] false
[20141218-174611 77] sh -c exit 77
Of course you can make it better parsable or adapt it to your whishes how the logfile shall look like. Note that the uncatched normal stdout of the remote programs is written to stderr (see the redirection in x()).
If you need a recipe to catch and prepare output of a command for the logfile, here is a copy of such a catcher from https://gist.github.com/hilbix/c53d525f113df77e323d - but yes, this is a bit bigger boilerplate to "Run something in current context of shell, postprocessing stdout+stderr without disturbing return code":
# Redirect lines of stdin/stdout to some other function
# outfn and errfn get following arguments
# "cmd args.." "one line full of output"
: catch outfn errfn cmd args..
catch()
{
local ret o1 o2 tmp
tmp=$(mktemp "catch_XXXXXXX.tmp")
mkfifo "$tmp.out"
mkfifo "$tmp.err"
pipestdinto "$1" "${*:3}" <"$tmp.out" &
o1=$!
pipestdinto "$2" "${*:3}" <"$tmp.err" &
o2=$!
"${#:3}" >"$tmp.out" 2>"$tmp.err"
ret=$?
rm -f "$tmp.out" "$tmp.err" "$tmp"
wait $o1
wait $o2
return $ret
}
: pipestdinto cmd args..
pipestdinto()
{
local x
while read -r x; do "$#" "$x" </dev/null; done
}
STAMP()
{
date +%Y%m%d-%H%M%S
}
# example output function
NOTE()
{
echo "NOTE `STAMP`: $*"
}
ERR()
{
echo "ERR `STAMP`: $*" >&2
}
catch_example()
{
# Example use
catch NOTE ERR find /proc -ls
}
See the second last line for an example (scroll down)
I want to write code like this:
command="some command"
safeRunCommand $command
safeRunCommand() {
cmnd=$1
$($cmnd)
if [ $? != 0 ]; then
printf "Error when executing command: '$command'"
exit $ERROR_CODE
fi
}
But this code does not work the way I want. Where did I make the mistake?
Below is the fixed code:
#!/bin/ksh
safeRunCommand() {
typeset cmnd="$*"
typeset ret_code
echo cmnd=$cmnd
eval $cmnd
ret_code=$?
if [ $ret_code != 0 ]; then
printf "Error: [%d] when executing command: '$cmnd'" $ret_code
exit $ret_code
fi
}
command="ls -l | grep p"
safeRunCommand "$command"
Now if you look into this code, the few things that I changed are:
use of typeset is not necessary, but it is a good practice. It makes cmnd and ret_code local to safeRunCommand
use of ret_code is not necessary, but it is a good practice to store the return code in some variable (and store it ASAP), so that you can use it later like I did in printf "Error: [%d] when executing command: '$command'" $ret_code
pass the command with quotes surrounding the command like safeRunCommand "$command". If you don’t then cmnd will get only the value ls and not ls -l. And it is even more important if your command contains pipes.
you can use typeset cmnd="$*" instead of typeset cmnd="$1" if you want to keep the spaces. You can try with both depending upon how complex is your command argument.
'eval' is used to evaluate so that a command containing pipes can work fine
Note: Do remember some commands give 1 as the return code even though there isn't any error like grep. If grep found something it will return 0, else 1.
I had tested with KornShell and Bash. And it worked fine. Let me know if you face issues running this.
Try
safeRunCommand() {
"$#"
if [ $? != 0 ]; then
printf "Error when executing command: '$1'"
exit $ERROR_CODE
fi
}
It should be $cmd instead of $($cmd). It works fine with that on my box.
Your script works only for one-word commands, like ls. It will not work for "ls cpp". For this to work, replace cmd="$1"; $cmd with "$#". And, do not run your script as command="some cmd"; safeRun command. Run it as safeRun some cmd.
Also, when you have to debug your Bash scripts, execute with '-x' flag. [bash -x s.sh].
There are several things wrong with your script.
Functions (subroutines) should be declared before attempting to call them. You probably want to return() but not exit() from your subroutine to allow the calling block to test the success or failure of a particular command. That aside, you don't capture 'ERROR_CODE' so that is always zero (undefined).
It's good practice to surround your variable references with curly braces, too. Your code might look like:
#!/bin/sh
command="/bin/date -u" #...Example Only
safeRunCommand() {
cmnd="$#" #...insure whitespace passed and preserved
$cmnd
ERROR_CODE=$? #...so we have it for the command we want
if [ ${ERROR_CODE} != 0 ]; then
printf "Error when executing command: '${command}'\n"
exit ${ERROR_CODE} #...consider 'return()' here
fi
}
safeRunCommand $command
command="cp"
safeRunCommand $command
The normal idea would be to run the command and then use $? to get the exit code. However, sometimes you have multiple cases in which you need to get the exit code. For example, you might need to hide its output, but still return the exit code, or print both the exit code and the output.
ec() { [[ "$1" == "-h" ]] && { shift && eval $* > /dev/null 2>&1; ec=$?; echo $ec; } || eval $*; ec=$?; }
This will give you the option to suppress the output of the command you want the exit code for. When the output is suppressed for the command, the exit code will directly be returned by the function.
I personally like to put this function in my .bashrc file.
Below I demonstrate a few ways in which you can use this:
# In this example, the output for the command will be
# normally displayed, and the exit code will be stored
# in the variable $ec.
$ ec echo test
test
$ echo $ec
0
# In this example, the exit code is output
# and the output of the command passed
# to the `ec` function is suppressed.
$ echo "Exit Code: $(ec -h echo test)"
Exit Code: 0
# In this example, the output of the command
# passed to the `ec` function is suppressed
# and the exit code is stored in `$ec`
$ ec -h echo test
$ echo $ec
0
Solution to your code using this function
#!/bin/bash
if [[ "$(ec -h 'ls -l | grep p')" != "0" ]]; then
echo "Error when executing command: 'grep p' [$ec]"
exit $ec;
fi
You should also note that the exit code you will be seeing will be for the grep command that's being run, as it is the last command being executed. Not the ls.