If there a way to affect the '?' variable in bash?
For a regular variable, it is possible to just use FOO=bar, thus 'echo $FOO' will output bar, but for some reason, it is not working with the '?' variable.
I have found two workaround but they are quite unsatisfactory.
First, it is possible to use true and false that will set $? to respectively 0 and 1.
#! /bin/bash
echo $?
true
echo $?
false
echo $?
This will output xxx, 0, 1. This workaround is limited, because it only allow to affect the values 0 and 1.
Then, it is possible to write some code in C (or other) that will just return the value in parameter via exit and then call this function. Example :
#! /bin/bash
rm foo.c
touch foo.c
echo "#include <stdio.h>" >> foo.c
echo "#include <stdlib.h>" >> foo.c
echo "int main(int argc, char **argv)" >> foo.c
echo "{" >> foo.c
echo " return atoi(argv[1]);" >> foo.c
echo "}" >> foo.c
gcc -o foo foo.c
./foo 42
echo $?
That will output 42. Even though it works, it is pretty nasty for doing something so simple, not to mention that this is only a simplified version without all the checkings that would have to be done in order to be sure not to overwrite anything. In addition, this require gcc to be present on the system.
There's no need to involve C at all if you just want to set the most recent return code (although I'm wondering why you're trying to do this). A simple shell function is sufficient:
ret() { return $1; }
Call it like ret 42.
Of course you know that '$?' holds the exit code of the last executed command. If the command is your own written executable, then '$?' holds its exit code.
The default value is 0 (success).
To sum up: use exit in your own scripts.
$ ls
$ echo $?
0
$ echo '#!/bin.bash' > script1.sh && chmod u+x script1.sh
$ ./script1.sh
$ echo $?
0
$ echo -e '#!/bin.bash\nexit 42' > script2.sh && chmod u+x script2.sh
$ ./script2.sh
$ echo $?
42
Also, have a look at Return value in bash script
Related
I have a simple bash function, hle (highlight errors), that highlights important text when building. It is implemented as follows:
hle() (
set -o pipefail
"$#" 2>&1 |
sed -f $hldir/red.sed -f $hldir/blue.sed -f $hldir/orange.sed
)
so you could run hle make target, and all errors would appear to be red, warnings orange, and debug would be blue
Now, I have a script that does something like this:
foo.sh:
eval hle $command
[ $? ] && echo "SUCCESS"
But this doesn't work as $? represents the return code of the eval (which I believe is the return code of hle itself...) How would I preserve the return code from $command in my script?
As you found out, your implementation of hle should already result int the correct exit code. However, the test [ $? ] is flawed. With [ $? = 0 ] it should work as expected.
Nevertheless, here is an alternative implementation of hle that doesn't require changing set -o options and therefore doesn't need a sub shell too:
hle() {
"$#" 2>&1 |
sed -f "$hldir/red.sed" -f "$hldir/blue.sed" -f "$hldir/orange.sed"
return "${PIPESTATUS[0]}"
}
For eval, the exit status is that of the evaluated command, see bash's manual:
eval [arguments]
The arguments are concatenated together into a single command, which is then read and executed, and its exit status returned as the exit status of eval
By the way: Depending on $command you don't need eval at all. Right know I can only think of one case where eval would be important, namely pipes (command="cmd1 | cmd2"). However, in such a pipe only the first command would be executed by hle. If you really need the eval then put it inside the function ({ eval "$#"; } 2>&1 |).
Sorry, so after my conversatio nwith #Socowi, I discovered the problem wasn't with the eval or function, but rather with my test.
In the script, the return value of eval hle $command is in fact the same as the return value from command. My problem was with the [ $? ] && echo SUCCESS... $? resolves to either 1 or 0, which is always considered true within test. The proper line should have been:
eval hle $command
[ $? == 0 ] && echo "SUCCESS"
While you can inline output of a program as parameters
$ echo $(ls)
cpp python bash
or as a temporary file
$ echo <(ls)
/proc/self/fd/63
I wonder how you can inline the return value with a similar syntax, so that it echoes the return-value of ls that it works like this:
$ ls
$ echo $?
0
ls_retval=$(ls >/dev/null 2>&1; echo "$?")
If you want to encapsulate that:
# define a function...
retval_of() { "$#" >/dev/null 2>&1; echo "$?"; }
# and use it
ls_retval=$(retval_of ls)
As for "with a similar syntax", though -- the shell has the syntax that it has; there doesn't exist "retval substitution" (as of bash 4.4, or POSIX sh as standardized in POSIX Issue 7).
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 [[ ${!__I} == '::' ]]; then
...
I want to create a multiline file with Make, having exact content:
#!/bin/bash
if [ "$JAVA_HOME" = "" ]; then echo "Please set JAVA_HOME"; exit 1; fi
export CONFIG_VARS=$( cat <<EOF
-Dmapred.job.tracker=$JT
EOF
)
${HADOOP_HOME}/bin/hadoop $1 $HADOOP_CONFIG_VARS ${*:2} 2>&1 | grep -v SLF4J
How can I tell make to output a file with this exact content somewhere?
I tried this:
define SCRIPT_CONTENT
#!/bin/bash
if [ "$JAVA_HOME" = "" ]; then echo "Please set JAVA_HOME"; exit 1; fi
export CONFIG_VARS=$( cat <<EOF
-Dmapred.job.tracker=$JT
EOF
)
${HADOOP_HOME}/bin/hadoop $1 $HADOOP_CONFIG_VARS ${*:2} 2>&1 | grep -v SLF4J
endef
export SCRIPT_CONTENT
bin/script:
#echo "$$SCRIPT_CONTENT" > bin/script
This paricular solution 1) wipes $ and first char after $-es and 2) is ugly because the definition should happen outside of the particular target where it's needed :(
I also tried this:
bin/script:
#echo '
#!/bin/bash
if [ "$JAVA_HOME" = "" ]; then echo "Please set JAVA_HOME"; exit 1; fi
export CONFIG_VARS=$( cat <<EOF
-Dmapred.job.tracker=$JT
EOF
)
${HADOOP_HOME}/bin/hadoop $1 $HADOOP_CONFIG_VARS ${*:2} 2>&1 | grep -v SLF4J
' > bin/script
This returns error when in make, works outside of make...
Any suggestion is very welcome!
Make wants any $ characters that should be reproduced literally to be escaped by inserting another $ in front of them.
More broadly, though, it seems like you're trying to use Make as a shell-script replacement. The more idomatic way to do this would be to put that content in a source file that you can copy to the destination, or to put it in a script that will write it into a specified destination. The Makefile then just has to invoke the copy command or the script.
With the help from this magnificent answer, I cooked up the following.
# From https://stackoverflow.com/a/8316519/874188
define \n
endef
define SCRIPT_CONTENT
#!/bin/bash
if [ "$$JAVA_HOME" = "" ]; then echo "Please set JAVA_HOME"; exit 1; fi
export CONFIG_VARS=$$( cat <<EOF
-Dmapred.job.tracker=$$JT
EOF
)
$${HADOOP_HOME}/bin/hadoop $$1 $$HADOOP_CONFIG_VARS $${*:2} 2>&1 | grep -v SLF4J
endef
bin/script:
echo '$(subst $(\n),\n,$(SCRIPT_CONTENT))' >$#
When testing, I found that I needed to have a semicolon at the end of the echo line when it didn't have any redirection. I can speculate that there is a built-in echo which gets invoked when there are no shell metacharacters in the command line?
Also, notice that the definition cannot contain any single quotes, and that all dollar signs have to be doubled. Maybe one or the other of these restrictions could be removed; I was unsuccessful, but then I didn't spend too much time or effort.
You cannot do this in make. Beyond what Novelocrat says regarding $, there's the fact that make is line-oriented and does not have any ability to generate a command that contains a newline character. All newlines that appear unescaped (without a backslash before them) are parsed by make as ending that recipe line, and each recipe line is sent to a different invocation of the shell. If you want the entire command to be sent as a single string to the same shell, then you must escape the newlines.
However, make will remove all backslash/newline pairs before it runs the command.
The only possible way to do this completely within make is to generate the file one line at a time, like this:
bin/script:
#echo '#!/bin/bash' > $#
#echo 'if [ "$$JAVA_HOME" = "" ]; then echo "Please set JAVA_HOME"; exit 1; fi' >> $#
#echo 'export CONFIG_VARS=$$( cat <<EOF' >> $#
#echo ' -Dmapred.job.tracker=$$JT' >> $#
#echo 'EOF' >> $#
#echo ')' >> $#
#echo '$${HADOOP_HOME}/bin/hadoop $$1 $$HADOOP_CONFIG_VARS $${*:2} 2>&1 | grep -v SLF4J' >> $#
As Novelocrat says, the typical way this is done is to have the script file as a separate file and copy it where you want it, rather than generating it.
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.