How to run a function with parameters with in a bash script - shell

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).

Related

Call from script A a function defined in script B - without executing the body of script B [duplicate]

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
~

How to run my bash functions in terminal using a parent name?

* the wording of the question is terrible, sorry!
I have some bash functions I create
test() {echo "hello wold"}
test2() {echo "hello wold"}
Then in my .bashrc I source the file that has the above function . ~/my_bash_scripts/testFile
In the terminal I can run test and get hello world.
is there a way for me to add parent variable that holds all my functions together. For example personal test, personal test2.
Similar to every other gem out there, I downloaded a tweeter one. All it's methods are followed by the letter t, as in t status to write a status, instead of just status
You are asking about writing a command-line program. Just a simple one here:
#!/usr/bin/env bash
if [[ $# -eq 0 ]]; then
echo "no command specified"
exit
elif [[ $# -gt 1 ]]; then
echo "only one argument expected"
exit
fi
case "$1" in
test)
echo "hello, this is test1"
;;
test2)
echo "hello, this is test2"
;;
*)
echo "unknown command: $1"
;;
esac
Then save it and make it an executable by run chmod +x script.sh, and in your .bashrc file, add alias personal="/fullpath/to/the/script.sh".
This is just very basic and simple example using bash and of course you can use any language you like, e.g. Python, Ruby, Node e.t.c.
Use arguments to determine final outputs.
You can use "$#" for number of arguments.
For example,
if [ $# -ne 2 ]; then
# TODO: print usage
exit 1
fi
Above code exits if arguments not euqal to 2.
So below bash program
echo $#
with
thatscript foo bar baz quux
will output 4.
Finally you can combine words to determine what to put stdout.
If you want to flag some functions as your personal functions; no, there is no explicit way to do that, and essentially, all shell functions belong to yourself (although some may be defined by your distro maintainer or system administrator as system-wide defaults).
What you could do is collect the output from declare -F at the very top of your personal shell startup file; any function not in that list is your personal function.
SYSFNS=$(declare -F | awk '{ a[++i] = $3 }
END { for (n=1; n<=i; n++) printf "%s%s", (n>1? ":" : ""), a[n] }')
This generates a variable SYSFNS which contains a colon-separated list of system-declared functions.
With that defined, you can check out which functions are yours:
myfns () {
local fun
declare -F |
while read -r _ _ fun; do
case :$SYSFNS: in *:"$fun":*) continue;; esac
echo "$fun"
done
}

How to modify "source" shell builtin?

I'd like to modify the "source" shell builtin so that every time source is called the command "echo ${file I am sourcing}" is called.
I'd like to do this so that I can always tell which files have been sourced when I open a new bash instance.
You can define a new function:
mysource() { echo "sourcing file: $1" && source "$#"; }
But, if you really must call it source:
source() { echo "sourcing file: $1" && builtin source "$#"; }
Note that I am using "$#" so that any arguments after the filename are passed to the builtin source command too.
Define a function in your shell:
source() { echo "$1"; . "$1"; }
This will only be valid for the shell in which the function is defined. Put it in the appropriate startup file if you want it to be defined in all new shells. (eg ~/.bashrc)

Importing functions from a shell script

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
~

Bash - How to call a function declared in a parent shell?

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

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