For some reason, this function is working properly. The terminal is outputting
newbootstrap.sh: 2: Syntax error: "(" unexpected
Here is my code (line 2 is function MoveToTarget() {)
#!/bin/bash
function MoveToTarget() {
# This takes two arguments: source and target
cp -r -f "$1" "$2"
rm -r -f "$1"
}
function WaitForProcessToEnd() {
# This takes one argument. The PID to wait for
# Unlike the AutoIt version, this sleeps for one second
while [ $(kill -0 "$1") ]; do
sleep 1
done
}
function RunApplication() {
# This takes one application, the path to the thing to execute
exec "$1"
}
# Our main code block
pid="$1"
SourcePath="$2"
DestPath="$3"
ToExecute="$4"
WaitForProcessToEnd $pid
MoveToTarget $SourcePath, $DestPath
RunApplication $ToExecute
exit
You're using the wrong syntax to declare functions. Use this instead:
MoveToTarget() {
# Function
}
Or this:
function MoveToTarget {
# function
}
But not both.
Also, I see that later on you use commas to separate arguments (MoveToTarget $SourcePath, $DestPath). That is also a problem. Bash uses spaces to separate arguments, not commas. Remove the comma and you should be golden.
I'm also new to defining functions in Bash scripts. I'm using a Bash of version 4.3.11(1):-release (x86_64-pc-linux-gnu) on Ubuntu 14.04 (Trusty Tahr).
I don't know why, but the definition that starts with the keyword function never works for me.
A definition like the following
function check_and_start {
echo Hello
}
produces the error message:
Syntax error: "}" unexpected
If I put the { on a new line like:
function my_function
{
echo Hello.
}
It prints a Hello. when I run the script, even if I don't call this function at all, which is also what we want.
I don't know why this wouldn't work, because I also looked at many tutorials and they all put the open curly brace at the end of the first line. Maybe it's the version of Bash that we use?? Anyway, just put it here for your information.
I have to use the C-style function definition:
check_and_start() {
echo $1
}
check_and_start World!
check_and_start Hello,\ World!
and it works as expected.
If you encounter "Syntax error: "(" unexpected", then use "bash" instead of using "sh".
For example:
bash install.sh
I had the same issue. I was running scripts on Ubuntu sometimes using sh vs. Dash. It seems running scripts with sh causes the issue, but running scripts with Dash works fine.
Related
My current setup starts with a function that is ostensibly in .bashrc (.bash_it/custom/funcs.bash to be precise)
#!/usr/bin/env bash
function proset() {
. proset-core "$#";
}
proset-core does some decrypting of secrets and exports those secrets to the session, hence the need for the . instead of just running it as a script/subshell.
If something goes wrong in proset-core, I use return instead of exit since I don't want the SSH connection to be dropped.
if [ "${APP_JSON}" = "null" ] ; then
echo -e "\n${redtext}App named $NAME not found in ${APPCONF}. Aborting.${resettext}\n";
return;
fi
This makes sense in the context of the exported proset function, but precludes usage as a script since return isn't valid except from within a function.
Is there a way to detect how it's being called and return one or the other as appropriate?
Just try to return, and exit if it fails.
_retval=$?
return 2>/dev/null || exit "$_retval"
The only case where your code will still be continuing after the return was invoked at top-level (outside of a function) is if you were executed rather than sourced, and should that happen, exiting is the Right Thing.
Make the builtin variable $SHLVL part of $# args as the last arg. Then at test point:
if [ "${#: -1}" -lt $SHLVL ]; then
# SHLVL arg is less than current SHLVL
# we are in a subshell
exit
else
return
fi
Ended up using
calledBy="$(ps -o comm= $PPID)";
if [ "x${calledBy}" = "xsshd" ]; then
return 1;
else
exit 1;
fi
since it didn't require passing anything extra. Anything that might cause this to be problematic please comment. Not too worried about being bash-specific or portable.
Credit: get the name of the caller script in bash script
include.sh
#!/bin/bash -
export gg
f() {
for i in "${gg[#]}"
do
echo $i
done
}
run.sh
#!/bin/bash -
set -o nounset
. include.sh || exit 1
f
I get this error
scripts/include.sh: line 5: gg[#]: unbound variable
Isn't export keyword supposed to make gg global and available anywhere?
If not, how to make gg available everywhere from include.sh ?
UPDATE
Environment:
$ cat /etc/*-release
NAME="SLES"
VERSION="11.4"
VERSION_ID="11.4"
PRETTY_NAME="SUSE Linux Enterprise Server 11 SP4"
Legacy Unix shells don't support () arrays. You should invoke the scripts with bash and they'll run as expected.
As written in the updated question, you need to define gg before exporting it:
gg=()
export gg
I tested the patch and it works fine.
If you source a file (equivalent to .), the commands are executed in the current shell context, therefore there is no need for export. However, using unset/empty array will cause bash to terminate your script if you are using set -o nounset. Either assign some values to your array:
gg=(value1 value2) #can be assigned in both run.sh and/or include.sh
f() {
for i in "${gg[#]}"; do
echo "$i"
done
}
Or use parameter expansion to handle this:
f() {
for i in ${gg[#]+"${gg[#]}"}; do
echo "$i"
done
}
${parameter+word} will expand to word only if parameter is set, otherwise nothing is substituted. If you want to read more about how this works with arrays: wiki.bash-hackers.org.
As mentioned by #CharlesDuffy, in bash 4.4 empty array doesn't become an error. There is no error even without assignment array=(). For more information, see:
BashFAQ #112.
I'm about to learn bash scripting and wrote a little script like this for testing purposes:
#!/bin/bash
function time {
echo $(date)
}
time
However the function doesn't get executed, instead the command time is running.
So what do I have to do to execute the function instead?
I'm running bash 4.2.45
To run a function with the same name as the special keyword time, quote it e.g.:
function time {
echo "$(date)"
}
'time'
You need to add () to the function definition. I'm not sure you need the function text there
Following should work:
#!/bin/bash
get_time() {
echo $(date)
}
get_time
Edited: time seems to be a reserved keyword, so changed the function name
When I use exit command in a shell script, the script will terminate the terminal (the prompt). Is there any way to terminate a script and then staying in the terminal?
My script run.sh is expected to execute by directly being sourced, or sourced from another script.
EDIT:
To be more specific, there are two scripts run2.sh as
...
. run.sh
echo "place A"
...
and run.sh as
...
exit
...
when I run it by . run2.sh, and if it hit exit codeline in run.sh, I want it to stop to the terminal and stay there. But using exit, the whole terminal gets closed.
PS: I have tried to use return, but echo codeline will still gets executed....
The "problem" really is that you're sourcing and not executing the script. When you source a file, its contents will be executed in the current shell, instead of spawning a subshell. So everything, including exit, will affect the current shell.
Instead of using exit, you will want to use return.
Yes; you can use return instead of exit. Its main purpose is to return from a shell function, but if you use it within a source-d script, it returns from that script.
As §4.1 "Bourne Shell Builtins" of the Bash Reference Manual puts it:
return [n]
Cause a shell function to exit with the return value n.
If n is not supplied, the return value is the exit status of the
last command executed in the function.
This may also be used to terminate execution of a script being executed
with the . (or source) builtin, returning either n or
the exit status of the last command executed within the script as the exit
status of the script.
Any command associated with the RETURN trap is executed
before execution resumes after the function or script.
The return status is non-zero if return is used outside a function
and not during the execution of a script by . or source.
You can add an extra exit command after the return statement/command so that it works for both, executing the script from the command line and sourcing from the terminal.
Example exit code in the script:
if [ $# -lt 2 ]; then
echo "Needs at least two arguments"
return 1 2>/dev/null
exit 1
fi
The line with the exit command will not be called when you source the script after the return command.
When you execute the script, return command gives an error. So, we suppress the error message by forwarding it to /dev/null.
Instead of running the script using . run2.sh, you can run it using sh run2.sh or bash run2.sh
A new sub-shell will be started, to run the script then, it will be closed at the end of the script leaving the other shell opened.
Actually, I think you might be confused by how you should run a script.
If you use sh to run a script, say, sh ./run2.sh, even if the embedded script ends with exit, your terminal window will still remain.
However if you use . or source, your terminal window will exit/close as well when subscript ends.
for more detail, please refer to What is the difference between using sh and source?
This is just like you put a run function inside your script run2.sh.
You use exit code inside run while source your run2.sh file in the bash tty.
If the give the run function its power to exit your script and give the run2.sh
its power to exit the terminator.
Then of cuz the run function has power to exit your teminator.
#! /bin/sh
# use . run2.sh
run()
{
echo "this is run"
#return 0
exit 0
}
echo "this is begin"
run
echo "this is end"
Anyway, I approve with Kaz it's a design problem.
I had the same problem and from the answers above and from what I understood what worked for me ultimately was:
Have a shebang line that invokes the intended script, for example,
#!/bin/bash uses bash to execute the script
I have scripts with both kinds of shebang's. Because of this, using sh or . was not reliable, as it lead to a mis-execution (like when the script bails out having run incompletely)
The answer therefore, was
Make sure the script has a shebang, so that there is no doubt about its intended handler.
chmod the .sh file so that it can be executed. (chmod +x file.sh)
Invoke it directly without any sh or .
(./myscript.sh)
Hope this helps someone with similar question or problem.
To write a script that is secure to be run as either a shell script or sourced as an rc file, the script can check and compare $0 and $BASH_SOURCE and determine if exit can be safely used.
Here is a short code snippet for that
[ "X$(basename $0)" = "X$(basename $BASH_SOURCE)" ] && \
echo "***** executing $name_src as a shell script *****" || \
echo "..... sourcing $name_src ....."
I think that this happens because you are running it on source mode
with the dot
. myscript.sh
You should run that in a subshell:
/full/path/to/script/myscript.sh
'source' http://ss64.com/bash/source.html
It's correct that sourced vs. executed scripts use return vs. exit to keep the same session open, as others have noted.
Here's a related tip, if you ever want a script that should keep the session open, regardless of whether or not it's sourced.
The following example can be run directly like foo.sh or sourced like . foo.sh/source foo.sh. Either way it will keep the session open after "exiting". The $# string is passed so that the function has access to the outer script's arguments.
#!/bin/sh
foo(){
read -p "Would you like to XYZ? (Y/N): " response;
[ $response != 'y' ] && return 1;
echo "XYZ complete (args $#).";
return 0;
echo "This line will never execute.";
}
foo "$#";
Terminal result:
$ foo.sh
$ Would you like to XYZ? (Y/N): n
$ . foo.sh
$ Would you like to XYZ? (Y/N): n
$ |
(terminal window stays open and accepts additional input)
This can be useful for quickly testing script changes in a single terminal while keeping a bunch of scrap code underneath the main exit/return while you work. It could also make code more portable in a sense (if you have tons of scripts that may or may not be called in different ways), though it's much less clunky to just use return and exit where appropriate.
Also make sure to return with expected return value. Else if you use exit when you will encounter an exit it will exit from your base shell since source does not create another process (instance).
Improved the answer of Tzunghsing, with more clear results and error re-direction, for silent usage:
#!/usr/bin/env bash
echo -e "Testing..."
if [ "X$(basename $0 2>/dev/null)" = "X$(basename $BASH_SOURCE)" ]; then
echo "***** You are Executing $0 in a sub-shell."
exit 0
else
echo "..... You are Sourcing $BASH_SOURCE in this terminal shell."
return 0
fi
echo "This should never be seen!"
Or if you want to put this into a silent function:
function sExit() {
# Safe Exit from script, not closing shell.
[ "X$(basename $0 2>/dev/null)" = "X$(basename $BASH_SOURCE)" ] && exit 0 || return 0
}
...
# ..it have to be called with an error check, like this:
sExit && return 0
echo "This should never be seen!"
Please note that:
if you have enabled errexit in your script (set -e) and you return N with N != 0, your entire script will exit instantly. To see all your shell settings, use, set -o.
when used in a function, the 1st return 0 is exiting the function, and the 2nd return 0 is exiting the script.
if your terminal emulator doesn't have -hold you can sanitize a sourced script and hold the terminal with:
#!/bin/sh
sed "s/exit/return/g" script >/tmp/script
. /tmp/script
read
otherwise you can use $TERM -hold -e script
If a command succeeded successfully, the return value will be 0. We can check its return value afterwards.
Is there a “goto” statement in bash?
Here is some dirty workaround using trap which jumps only backwards.
#!/bin/bash
set -eu
trap 'echo "E: failed with exitcode $?" 1>&2' ERR
my_function () {
if git rev-parse --is-inside-work-tree > /dev/null 2>&1; then
echo "this is run"
return 0
else
echo "fatal: not a git repository (or any of the parent directories): .git"
goto trap 2> /dev/null
fi
}
my_function
echo "Command succeeded" # If my_function failed this line is not printed
Related:
https://stackoverflow.com/a/19091823/2402577
How to use $? and test to check function?
I couldn't find solution so for those who want to leave the nested script without leaving terminal window:
# this is just script which goes to directory if path satisfies regex
wpr(){
leave=false
pwd=$(pwd)
if [[ "$pwd" =~ ddev.*web ]]; then
# echo "your in wordpress instalation"
wpDir=$(echo "$pwd" | grep -o '.*\/web')
cd $wpDir
return
fi
echo 'please be in wordpress directory'
# to leave from outside the scope
leave=true
return
}
wpt(){
# nested function which returns $leave variable
wpr
# interupts the script if $leave is true
if $leave; then
return;
fi
echo 'here is the rest of the script, executes if leave is not defined'
}
I have no idea whether this is useful for you or not, but in zsh, you can exit a script, but only to the prompt if there is one, by using parameter expansion on a variable that does not exist, as follows.
${missing_variable_ejector:?}
Though this does create an error message in your script, you can prevent it with something like the following.
{ ${missing_variable_ejector:?} } 2>/dev/null
1) exit 0 will come out of the script if it is successful.
2) exit 1 will come out of the script if it is a failure.
You can try these above two based on ur req.
I have a script that has some functions.
Can I run one of the function directly from command line?
Something like this?
myScript.sh func()
Well, while the other answers are right - you can certainly do something else: if you have access to the bash script, you can modify it, and simply place at the end the special parameter "$#" - which will expand to the arguments of the command line you specify, and since it's "alone" the shell will try to call them verbatim; and here you could specify the function name as the first argument. Example:
$ cat test.sh
testA() {
echo "TEST A $1";
}
testB() {
echo "TEST B $2";
}
"$#"
$ bash test.sh
$ bash test.sh testA
TEST A
$ bash test.sh testA arg1 arg2
TEST A arg1
$ bash test.sh testB arg1 arg2
TEST B arg2
For polish, you can first verify that the command exists and is a function:
# Check if the function exists (bash specific)
if declare -f "$1" > /dev/null
then
# call arguments verbatim
"$#"
else
# Show a helpful error
echo "'$1' is not a known function name" >&2
exit 1
fi
If the script only defines the functions and does nothing else, you can first execute the script within the context of the current shell using the source or . command and then simply call the function. See help source for more information.
The following command first registers the function in the context, then calls it:
. ./myScript.sh && function_name
Briefly, no.
You can import all of the functions in the script into your environment with source (help source for details), which will then allow you to call them. This also has the effect of executing the script, so take care.
There is no way to call a function from a shell script as if it were a shared library.
Using case
#!/bin/bash
fun1 () {
echo "run function1"
[[ "$#" ]] && echo "options: $#"
}
fun2 () {
echo "run function2"
[[ "$#" ]] && echo "options: $#"
}
case $1 in
fun1) "$#"; exit;;
fun2) "$#"; exit;;
esac
fun1
fun2
This script will run functions fun1 and fun2 but if you start it with option
fun1 or fun2 it'll only run given function with args(if provided) and exit.
Usage
$ ./test
run function1
run function2
$ ./test fun2 a b c
run function2
options: a b c
I have a situation where I need a function from bash script which must not be executed before (e.g. by source) and the problem with #$ is that myScript.sh is then run twice, it seems... So I've come up with the idea to get the function out with sed:
sed -n "/^func ()/,/^}/p" myScript.sh
And to execute it at the time I need it, I put it in a file and use source:
sed -n "/^func ()/,/^}/p" myScript.sh > func.sh; source func.sh; rm func.sh
Edit: WARNING - seems this doesn't work in all cases, but works well on many public scripts.
If you have a bash script called "control" and inside it you have a function called "build":
function build() {
...
}
Then you can call it like this (from the directory where it is):
./control build
If it's inside another folder, that would make it:
another_folder/control build
If your file is called "control.sh", that would accordingly make the function callable like this:
./control.sh build
Solved post but I'd like to mention my preferred solution. Namely, define a generic one-liner script eval_func.sh:
#!/bin/bash
source $1 && shift && "#a"
Then call any function within any script via:
./eval_func.sh <any script> <any function> <any args>...
An issue I ran into with the accepted solution is that when sourcing my function-containing script within another script, the arguments of the latter would be evaluated by the former, causing an error.
The other answers here are nice, and much appreciated, but often I don't want to source the script in the session (which reads and executes the file in your current shell) or modify it directly.
I find it more convenient to write a one or two line 'bootstrap' file and run that. Makes testing the main script easier, doesn't have side effects on your shell session, and as a bonus you can load things that simulate other environments for testing. Example...
# breakfast.sh
make_donuts() {
echo 'donuts!'
}
make_bagels() {
echo 'bagels!'
}
# bootstrap.sh
source 'breakfast.sh'
make_donuts
Now just run ./bootstrap.sh.Same idea works with your python, ruby, or whatever scripts.
Why useful? Let's say you complicated your life for some reason, and your script may find itself in different environments with different states present. For example, either your terminal session, or a cloud provider's cool new thing. You also want to test cloud things in terminal, using simple methods. No worries, your bootstrap can load elementary state for you.
# breakfast.sh
# Now it has to do slightly different things
# depending on where the script lives!
make_donuts() {
if [[ $AWS_ENV_VAR ]]
then
echo '/donuts'
elif [[ $AZURE_ENV_VAR ]]
then
echo '\donuts'
else
echo '/keto_diet'
fi
}
If you let your bootstrap thing take an argument, you can load different state for your function to chew, still with one line in the shell session:
# bootstrap.sh
source 'breakfast.sh'
case $1 in
AWS)
AWS_ENV_VAR="arn::mumbo:jumbo:12345"
;;
AZURE)
AZURE_ENV_VAR="cloud::woo:_impress"
;;
esac
make_donuts # You could use $2 here to name the function you wanna, but careful if evaluating directly.
In terminal session you're just entering:
./bootstrap.sh AWS
Result:
# /donuts
you can call function from command line argument like below
function irfan() {
echo "Irfan khan"
date
hostname
}
function config() {
ifconfig
echo "hey"
}
$1
Once you defined the functions put $1 at the end to accept argument which function you want to call.
Lets say the above code is saved in fun.sh. Now you can call the functions like ./fun.sh irfan & ./fun.sh config in command line.