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
~
Related
I declare functions in one shell file,
# a.sh
foo() { ... }
function bar() { ... }
and imported in another shell file by source:
# b.sh
source ./a.sh
# invoke foo and bar
foo
bar
Now in the shell, I can use foo/bar after executing b.sh
$ source b.sh
...
# I can call foo or bar now in the shell (undesirable)
$ foo
...
How can I make the functions be local variables in the scope of the importing file, and avoid them to contaminate global/environmental variables?
There's no such thing as "file scope" in shell -- just global scope and function scope. The closest you can come is running b.sh in another shell:
$ b.sh # run b.sh rather than reading it into the current shell
then everything in in b.sh will just be in that other shell and will "go away" when it exits. But that applies to everything defined in b.sh -- all functions, aliases, environment and other variables.
It is possible to isolates private shell functions this way.
# sourced a.sh
# a_main is exposed public
my_public_a() (
private_a() {
echo "I am private_a only visible to my_public_a"
}
private_b() {
echo "I am get_b only visible to my_public_a"
}
case "$1" in
a) private_a;;
b) private_b;;
*) exit;;
esac
)
# b.sh
source a.sh
my_public_a a
my_public_a b
private_a # command not found
private_b # command not found
Even though bash does not provide direct support, what you need is still achievable:
#!/usr/bin/env bash
# b.sh
if [[ "${BASH_SOURCE[0]}" = "$0" ]] ;then
source ./a.sh
# invoke foo and bar
foo
bar
else
echo "b.sh is being sourced. foo/bar will not be available."
fi
Above is not 100% reliable, but should cover most cases.
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 one shell script that sources a second one within the context of a function taking parameters:
#!/bin/bash
# bar.sh
function f()
{
source foo.sh
echo "Do something else with $1, after foo.sh is sourced."
}
f bar
And:
#!/bin/bash
# foo.sh
x=${1:-"default"}
echo $x
Execution output is the following:
$ ./bar.sh
bar
Do something else with bar, after foo.sh is sourced.
I was expecting to get default as first line output instead of bar. So it turns out that even though I'm not passing any arguments to foo.sh, it is taking $1 from the context of function f. I can understand this behavior from reading bash documentation, but what would be the best way to override it?
EDIT: Based on your comment and edited question:
#!/bin/bash
# bar.sh
function f()
{
# save $1
arg1="$1"
# unset $1
shift
# source your script; prints default
source ./foo.sh
# restore $1
set -- $arg1
# should print bar
echo $1
echo "Do something else with $1, after foo.sh is sourced."
}
f bar
I trying to run a function outside of my script. Example test.sh:
DAYS=10
IP=1.2.3.4
Main {
functionName ${DAYS} ${IP}
}
functionName() {
echo $1
echo "$2"
}
from command line I'm trying to run the scripts function with different parameters
./test.sh functionName 4 "1.3.4.5"
Having trouble getting it to work so any ideas would be great thanks
Inside the function, $1 is the argument passed to the function, not the argument passed to the script. Just do:
DAYS=${1-10} # set DAYS to first argument, defaulting to "10"
IP=${2-1.2.3.4} # set IP to 2nd argument, defaulting to "1.2.3.4"
Main() {
functionName ${DAYS} ${IP}
}
functionName() {
echo $1
echo "$2"
}
Main
If you source your script, then the functions will be available in your current shell:
. ./test.sh
functionName 4 "1.3.4.5"
Downside is that any code in the sourced script which is not in a function will be run. You can avoid that by (in the sourced script) doing a test like:
if [[ $0 == test.sh ]]
then
Main
fi
which might be why you have a Main? By the way, why are you using global variables? Why not declare them inside Main (using local).
I am writing a bash script that calls functions declared in the parent shell, but it doesn't work.
For example:
$ function myfunc() { echo "Here in myfunc" ; }
$ myfunc
Here in myfunc
$ cat test.sh
#! /bin/bash
echo "Here in the script"
myfunc
$ ./test.sh
Here in the script
./test.sh: line 4: myfunc: command not found
$ myfunc
Here in myfunc
As you can see the script ./test.sh is unable to call the function myfunc, is there some way to make that function visible to the script?
Try
$ export -f myfunc
in the parent shell, to export the function.
#OP, normally you would put your function that every script uses in a file, then you source it in your script. example, save
function myfunc() { echo "Here in myfunc" ; }
in a file called /path/library. Then in your script, source it like this:
#!/bin/bash
. /path/library
myfunc
This also works but I noticed ${0} takes parent's value:
Maybe more useful if you don't want to have a bunch of export calls in your scripts.
script1:
#!/bin/bash
func()
{
echo func "${1}"
}
func "1"
$(. ./script2)
script2:
#!/bin/bash
func "2"
Output:
[mymachine]# ./script1
func 1
func 2