Say I have a bash script and I want some variables to appear when sourced and others to only be accessible from within the script (both functions and variables). What's the convention to achieve this?
Let's say test.sh is your bash script.
What you can do is extract all the common items and put them in common.sh which can be sourced by other scripts.
The BASH_SOURCE array helps you here:
Consider this script, source.sh
#!/bin/bash
if [[ ${BASH_SOURCE[0]} == "$0" ]]; then
# this code is run when the script is _executed_
foo=bar
privFunc() { echo "running as a script"; }
main() {
privFunc
publicFunc
}
fi
# this code is run when script is executed or sourced
answer=42
publicFunc() { echo "Hello, world!"; }
echo "$0 - ${BASH_SOURCE[0]}"
[[ ${BASH_SOURCE[0]} == "$0" ]] && main
Running it:
$ bash source.sh
source.sh - source.sh
running as a script
Hello, world!
Sourcing it:
$ source source.sh
bash - source.sh
$ declare -p answer
declare -- answer="42"
$ declare -p foo
bash: declare: foo: not found
$ publicFunc
Hello, world!
$ privFunc
bash: privFunc: command not found
$ main
bash: main: command not found
Related
I have a bash script named test.sh with:
#!/bin/bash
var1=hello
sh test2.sh $var1="/var/log"
My test2.sh looks like this:
#!/bin/bash
function jumpto
{
label=$1
cmd=$(sed -n "/$label:/{:a;n;p;ba};" $0 | grep -v ':$')
eval "$cmd"
exit
}
start=${1:-"start"}
jumpto $start
start:
echo "variable -- " $var1
This does not work due to I used jumpto function. When execution, always $var1 assigned to the $1 variable in jumpto function.
Is there a different way to do this.?
We suggest to learn about environment variable and export command and their scope.
Try modify your test.sh
#!/bin/bash
export var1=hello
sh test2.sh $var1="/var/log"
I have a script that I'd like people to source, but optionally so. So they can run it with or without sourcing it, it's up to them.
e.g. The following should both work:
$ . test.sh
$ test.sh
The problem is, test.sh contains exit statements if correct args aren't passed in. If someone sources the script, then the exit commands exit the terminal!
I've done a bit of research and see from this StackOverflow post that I could detect if it's being sourced, and do something different, but what would that something different be?
The normal way to exit from a sourced script is simply to return (optionally adding the desired exit code) outside of any function. Assuming that when run as a command we have the -e flag set, this will also exit from a shell program:
#!/bin/sh -eu
if [ $# = 0 ]
then
echo "Usage $0 <argument>" >&2
return 1
fi
If we're running without -e, we might be able to return || exit instead.
There may be better ways to do this, but here's a sample script showing how I got this to work:
bparks#home
$ set | grep TESTVAR
bparks#home
$ ./test.sh
Outputs some useful information to the console. Please pass one arg.
bparks#home
$ set | grep TESTVAR
bparks#home
$ . ./test.sh
Outputs some useful information to the console. Please pass one arg.
bparks#home
$ set | grep TESTVAR
bparks#home
$ ./test.sh asdf
export TESTVAR=me
bparks#home
$ set | grep TESTVAR
bparks#home
$ . ./test.sh asdf
bparks#home
$ set | grep TESTVAR
TESTVAR=me
bparks#home
$
test.sh
#!/usr/bin/env bash
# store if we're sourced or not in a variable
(return 0 2>/dev/null) && SOURCED=1 || SOURCED=0
exitIfNotSourced(){
[[ "$SOURCED" != "0" ]] || exit;
}
showHelp(){
IT=$(cat <<EOF
Outputs some useful information to the console. Please pass one arg.
EOF
)
echo "$IT"
}
# Show help if no args supplied - works if sourced or not sourced
if [ -z "$1" ]
then
showHelp
exitIfNotSourced;
return;
fi
# your main script follows
# this sample shows exporting a variable if sourced,
# and outputting this to stdout if not sourced
if [ "$SOURCED" == "1" ]
then
export TESTVAR=me
else
echo "export TESTVAR=me"
fi
Checkout this answer for better description and porper solution.
And here is how it is used in docker-entrypoint.sh in official Mysql image:
# check to see if this file is being run or sourced from another script
_is_sourced() {
# https://unix.stackexchange.com/a/215279
[ "${#FUNCNAME[#]}" -ge 2 ] \
&& [ "${FUNCNAME[0]}" = '_is_sourced' ] \
&& [ "${FUNCNAME[1]}" = 'source' ]
}
I have a shell script that I would like to test with shUnit. The script (and all the functions) are in a single file since it makes installation much easier.
Example for script.sh
#!/bin/sh
foo () { ... }
bar () { ... }
code
I wanted to write a second file (that does not need to be distributed and installed) to test the functions defined in script.sh
Something like run_tests.sh
#!/bin/sh
. script.sh
# Unit tests
Now the problem lies in the . (or source in Bash). It does not only parse function definitions but also executes the code in the script.
Since the script with no arguments does nothing bad I could
. script.sh > /dev/null 2>&1
but I was wandering if there is a better way to achieve my goal.
Edit
My proposed workaround does not work in the case the sourced script calls exit so I have to trap the exit
#!/bin/sh
trap run_tests ERR EXIT
run_tests() {
...
}
. script.sh
The run_tests function is called but as soon as I redirect the output of the source command the functions in the script are not parsed and are not available in the trap handler
This works but I get the output of script.sh:
#!/bin/sh
trap run_tests ERR EXIT
run_tests() {
function_defined_in_script_sh
}
. script.sh
This does not print the output but I get an error that the function is not defined:
#!/bin/sh
trap run_tests ERR EXIT
run_tests() {
function_defined_in_script_sh
}
. script.sh | grep OUTPUT_THAT_DOES_NOT_EXISTS
This does not print the output and the run_tests trap handler is not called at all:
#!/bin/sh
trap run_tests ERR EXIT
run_tests() {
function_defined_in_script_sh
}
. script.sh > /dev/null
According to the “Shell Builtin Commands” section of the bash manpage, . aka source takes an optional list of arguments which are passed to the script being sourced. You could use that to introduce a do-nothing option. For example, script.sh could be:
#!/bin/sh
foo() {
echo foo $1
}
main() {
foo 1
foo 2
}
if [ "${1}" != "--source-only" ]; then
main "${#}"
fi
and unit.sh could be:
#!/bin/bash
. ./script.sh --source-only
foo 3
Then script.sh will behave normally, and unit.sh will have access to all the functions from script.sh but will not invoke the main() code.
Note that the extra arguments to source are not in POSIX, so /bin/sh might not handle it—hence the #!/bin/bash at the start of unit.sh.
Picked up this technique from Python, but the concept works just fine in bash or any other shell...
The idea is that we turn the main code section of our script into a function. Then at the very end of the script, we put an 'if' statement that will only call that function if we executed the script but not if we sourced it. Then we explicitly call the script() function from our 'runtests' script which has sourced the 'script' script and thus contains all its functions.
This relies on the fact that if we source the script, the bash-maintained environment variable $0, which is the name of the script being executed, will be the name of the calling (parent) script (runtests in this case), not the sourced script.
(I've renamed script.sh to just script cause the .sh is redundant and confuses me. :-)
Below are the two scripts. Some notes...
$# evaluates to all of the arguments passed to the function or
script as individual strings. If instead, we used $*, all the
arguments would be concatenated together into one string.
The RUNNING="$(basename $0)" is required since $0 always includes at
least the current directory prefix as in ./script.
The test if [[ "$RUNNING" == "script" ]].... is the magic that causes
script to call the script() function only if script was run directly
from the commandline.
script
#!/bin/bash
foo () { echo "foo()"; }
bar () { echo "bar()"; }
script () {
ARG1=$1
ARG2=$2
#
echo "Running '$RUNNING'..."
echo "script() - all args: $#"
echo "script() - ARG1: $ARG1"
echo "script() - ARG2: $ARG2"
#
foo
bar
}
RUNNING="$(basename $0)"
if [[ "$RUNNING" == "script" ]]
then
script "$#"
fi
runtests
#!/bin/bash
source script
# execute 'script' function in sourced file 'script'
script arg1 arg2 arg3
If you are using Bash, a similar solution to #andrewdotn's approach (but without needing an extra flag or depending on the script name) can be accomplished by using BASH_SOURCE array.
script.sh:
#!/bin/bash
foo () { ... }
bar () { ... }
main() {
code
}
if [[ "${#BASH_SOURCE[#]}" -eq 1 ]]; then
main "$#"
fi
run_tests.sh:
#!/bin/bash
. script.sh
# Unit tests
If you are using Bash, another solution may be:
#!/bin/bash
foo () { ... }
bar () { ... }
[[ "${FUNCNAME[0]}" == "source" ]] && return
code
I devised this. Let's say our shell library file is the following file, named aLib.sh:
funcs=("a" "b" "c") # File's functions' names
for((i=0;i<${#funcs[#]};i++)); # Avoid function collision with existing
do
declare -f "${funcs[$i]}" >/dev/null
[ $? -eq 0 ] && echo "!!ATTENTION!! ${funcs[$i]} is already sourced"
done
function a(){
echo function a
}
function b(){
echo function b
}
function c(){
echo function c
}
if [ "$1" == "--source-specific" ]; # Source only specific given as arg
then
for((i=0;i<${#funcs[#]};i++));
do
for((j=2;j<=$#;j++));
do
anArg=$(eval 'echo ${'$j'}')
test "${funcs[$i]}" == "$anArg" && continue 2
done
unset ${funcs[$i]}
done
fi
unset i j funcs
At the beginning it checks and warns for any function name collision detected.
At the end, bash has already sourced all functions, so it frees memory from them and keeps only the ones selected.
Can be used like this:
user#pc:~$ source aLib.sh --source-specific a c
user#pc:~$ a; b; c
function a
bash: b: command not found
function c
~
I have a shell script that I would like to test with shUnit. The script (and all the functions) are in a single file since it makes installation much easier.
Example for script.sh
#!/bin/sh
foo () { ... }
bar () { ... }
code
I wanted to write a second file (that does not need to be distributed and installed) to test the functions defined in script.sh
Something like run_tests.sh
#!/bin/sh
. script.sh
# Unit tests
Now the problem lies in the . (or source in Bash). It does not only parse function definitions but also executes the code in the script.
Since the script with no arguments does nothing bad I could
. script.sh > /dev/null 2>&1
but I was wandering if there is a better way to achieve my goal.
Edit
My proposed workaround does not work in the case the sourced script calls exit so I have to trap the exit
#!/bin/sh
trap run_tests ERR EXIT
run_tests() {
...
}
. script.sh
The run_tests function is called but as soon as I redirect the output of the source command the functions in the script are not parsed and are not available in the trap handler
This works but I get the output of script.sh:
#!/bin/sh
trap run_tests ERR EXIT
run_tests() {
function_defined_in_script_sh
}
. script.sh
This does not print the output but I get an error that the function is not defined:
#!/bin/sh
trap run_tests ERR EXIT
run_tests() {
function_defined_in_script_sh
}
. script.sh | grep OUTPUT_THAT_DOES_NOT_EXISTS
This does not print the output and the run_tests trap handler is not called at all:
#!/bin/sh
trap run_tests ERR EXIT
run_tests() {
function_defined_in_script_sh
}
. script.sh > /dev/null
According to the “Shell Builtin Commands” section of the bash manpage, . aka source takes an optional list of arguments which are passed to the script being sourced. You could use that to introduce a do-nothing option. For example, script.sh could be:
#!/bin/sh
foo() {
echo foo $1
}
main() {
foo 1
foo 2
}
if [ "${1}" != "--source-only" ]; then
main "${#}"
fi
and unit.sh could be:
#!/bin/bash
. ./script.sh --source-only
foo 3
Then script.sh will behave normally, and unit.sh will have access to all the functions from script.sh but will not invoke the main() code.
Note that the extra arguments to source are not in POSIX, so /bin/sh might not handle it—hence the #!/bin/bash at the start of unit.sh.
Picked up this technique from Python, but the concept works just fine in bash or any other shell...
The idea is that we turn the main code section of our script into a function. Then at the very end of the script, we put an 'if' statement that will only call that function if we executed the script but not if we sourced it. Then we explicitly call the script() function from our 'runtests' script which has sourced the 'script' script and thus contains all its functions.
This relies on the fact that if we source the script, the bash-maintained environment variable $0, which is the name of the script being executed, will be the name of the calling (parent) script (runtests in this case), not the sourced script.
(I've renamed script.sh to just script cause the .sh is redundant and confuses me. :-)
Below are the two scripts. Some notes...
$# evaluates to all of the arguments passed to the function or
script as individual strings. If instead, we used $*, all the
arguments would be concatenated together into one string.
The RUNNING="$(basename $0)" is required since $0 always includes at
least the current directory prefix as in ./script.
The test if [[ "$RUNNING" == "script" ]].... is the magic that causes
script to call the script() function only if script was run directly
from the commandline.
script
#!/bin/bash
foo () { echo "foo()"; }
bar () { echo "bar()"; }
script () {
ARG1=$1
ARG2=$2
#
echo "Running '$RUNNING'..."
echo "script() - all args: $#"
echo "script() - ARG1: $ARG1"
echo "script() - ARG2: $ARG2"
#
foo
bar
}
RUNNING="$(basename $0)"
if [[ "$RUNNING" == "script" ]]
then
script "$#"
fi
runtests
#!/bin/bash
source script
# execute 'script' function in sourced file 'script'
script arg1 arg2 arg3
If you are using Bash, a similar solution to #andrewdotn's approach (but without needing an extra flag or depending on the script name) can be accomplished by using BASH_SOURCE array.
script.sh:
#!/bin/bash
foo () { ... }
bar () { ... }
main() {
code
}
if [[ "${#BASH_SOURCE[#]}" -eq 1 ]]; then
main "$#"
fi
run_tests.sh:
#!/bin/bash
. script.sh
# Unit tests
If you are using Bash, another solution may be:
#!/bin/bash
foo () { ... }
bar () { ... }
[[ "${FUNCNAME[0]}" == "source" ]] && return
code
I devised this. Let's say our shell library file is the following file, named aLib.sh:
funcs=("a" "b" "c") # File's functions' names
for((i=0;i<${#funcs[#]};i++)); # Avoid function collision with existing
do
declare -f "${funcs[$i]}" >/dev/null
[ $? -eq 0 ] && echo "!!ATTENTION!! ${funcs[$i]} is already sourced"
done
function a(){
echo function a
}
function b(){
echo function b
}
function c(){
echo function c
}
if [ "$1" == "--source-specific" ]; # Source only specific given as arg
then
for((i=0;i<${#funcs[#]};i++));
do
for((j=2;j<=$#;j++));
do
anArg=$(eval 'echo ${'$j'}')
test "${funcs[$i]}" == "$anArg" && continue 2
done
unset ${funcs[$i]}
done
fi
unset i j funcs
At the beginning it checks and warns for any function name collision detected.
At the end, bash has already sourced all functions, so it frees memory from them and keeps only the ones selected.
Can be used like this:
user#pc:~$ source aLib.sh --source-specific a c
user#pc:~$ a; b; c
function a
bash: b: command not found
function c
~
Is there any variable in bash that contains the name of the .sh file executed? The line number would be great too.
I want to use it in error messages such as:
echo "ERROR: [$FILE:L$LINE] $somefile not found"
#!/bin/bash
echo $LINENO
echo `basename $0`
$LINENO for the current line number
$0 for the current file. I used basename to ensure you only get the file name and not the path.
UPDATE:
#!/bin/bash
MY_NAME=`basename $0`
function ouch {
echo "Fail # [${MY_NAME}:${1}]"
exit 1
}
ouch $LINENO
You have to pass the line as a parameter if you use the function approach else you will get the line of the function definition.
I find the "BASH_SOURCE" and "BASH_LINENO" built-in arrays very useful:
$ cat xx
#!/bin/bash
_ERR_HDR_FMT="%.23s %s[%s]: "
_ERR_MSG_FMT="${_ERR_HDR_FMT}%s\n"
error_msg() {
printf "$_ERR_MSG_FMT" $(date +%F.%T.%N) ${BASH_SOURCE[1]##*/} ${BASH_LINENO[0]} "${#}"
}
error_msg "here"
error_msg "and here"
Invoking xx yields
2010-06-16.15:33:13.069 xx[11]: here
2010-06-16.15:33:13.073 xx[14]: and here
You just need to
echo $LINENO
echo $(basename $0)
Here's how to do it in a reusable function. if the following is in a file named script:
#!/bin/bash
debug() {
echo "${BASH_SOURCE[1]##*/}:${FUNCNAME[1]}[${BASH_LINENO[0]}]" > /dev/tty
}
debug
This produces the output:
script:main[5]
Which indicates the line on which debug was called.
The following will print out the filename, function, line and an optional message.
Also works in zsh for extra goodness.
# Say the file, line number and optional message for debugging
# Inspired by bash's `caller` builtin
# Thanks to https://unix.stackexchange.com/a/453153/143394
function yelp () {
# shellcheck disable=SC2154 # undeclared zsh variables in bash
if [[ $BASH_VERSION ]]; then
local file=${BASH_SOURCE[1]##*/} func=${FUNCNAME[1]} line=${BASH_LINENO[0]}
else # zsh
emulate -L zsh # because we may be sourced by zsh `emulate bash -c`
# $funcfiletrace has format: file:line
local file=${funcfiletrace[1]%:*} line=${funcfiletrace[1]##*:}
local func=${funcstack[2]}
[[ $func =~ / ]] && func=source # $func may be filename. Use bash behaviour
fi
echo "${file##*/}:$func:$line $*" > /dev/tty
}