Quick run down of what I'm doing, I'm making python development environments on my computer using virtualenv. I'm not one to remember all these commands, so I like to build simple(noob-ish) script that will help me.
Problem
As you see on step 2 in the Basic Usage section of the documentation. I need to run a command:
$ source venv/bin/activate
This activates my python environment. So I've tried a mixture of functions and eval. As mentioned, I am a bit of a noob when it comes to bash scripts.
My Code
File: fpyenv
#!/bin/bash
# $ cd ~/scripts/
# $ chmod u+x <filename>.sh
activateSrc(){
eval '$(source "${1}"/bin/activate)'
}
if [[ $1 == '' ]];
then
printf "ERROR: Missing Filename\n"
else
printf "Creating Virtual Environment $1\n"
# This creates the environment
virtualenv $1
printf "Do you want to activate $1 as your virtual environment?(y/n)\n"
# Get answer from user
read answer
if [[ $answer != 'y' ]];
# Said No
then
printf "Did not set $1 as your virtual environment\n"
else
# Said Yes
activateSrc $1
printf "Set $1 as your virtual environment\n"
fi
fi
This is what it should look like:
Step 1
myComputer $ fpyenv venv
returns
Creating Virtual Environment venv
Do you want to activate venv as your virtual environment?(y/n)
Step 2(user inputs y/n)
y
returns
Set venv as your virtual environment
(venv)myComputer $
But what I'm getting is:
Set venv as your virtual environment
myComputer $
So it doesn't run the source venv/bin/activate. Does anyone know what I'm doing wrong? I've looked through many answers, but given that source is commonly used in a different way in bash scripts, the answers I'm getting are no help. Thank you in advance!
FIX:
Change activateSrc to:
activateSrc(){
source $1/bin/activate
}
And execute script like this:
myComputer $ . fpyenv venv
It runs source just as written. The thing is that you need to source this new script as well, otherwise it just runs in a subshell and any changes made by the sourced script are lost when the subshell exits.
I have one script with common functions that is included in other my scripts with:
. ~/bin/fns
Since my ~/bin path is on the PATH, is there a way to prevent users to execute fns from command line (by returning from the script with a message), but to allow other scripts to include this file?
(Bash >= 4)
Just remove the executable bit with chmod -x . ~/bin/fns. It will still work when sourced, but you can't call it (accidentally) by its name anymore.
Some scripts at my workplace use a special shebang
#!/bin/echo Run:.
which returns
Run:. <pathname>
when you use it as a command.
Add the following at the beginning of the script you want to be only allowed to be sourced:
if [ ${0##*/} == ${BASH_SOURCE[0]##*/} ]; then
echo "WARNING"
echo "This script is not meant to be executed directly!"
echo "Use this script only by sourcing it."
echo
exit 1
fi
This will check if the current script and executed shell script file basenames match. If they match, then obviously you are executing it directly so we print a message and exit with status 1.
if (return 0 2>/dev/null) ; then
:
else
echo "Error: script was executed."
exit 1
fi
I want to run a executable from my shell script. The executable is located at /usr/bin/to_run.
My shell script(which is calling the above executable) is in the /usr/bin folder.
The shell script is :
#!/bin/bash
#kill all existing instances of synergy
killall synergys
sh "/usr/bin/synergys"
if [ $? -eq 1 ]; then
echo "synergy server started"
else
echo "error in starting"
fi
I am getting an error saying : "synergys : no process found".
When I run the same thing - /usr/bin/synergys directly from the terminal it runs fine, but from within a script there are problems. I don't understand why.
Thank you in advance.
That error is from the killall command, it's saying there are no candidate processes matching your argument.
If you don't want to be notified where no processes match, just use the quiet option:
killall -q synergys
From the killall man page:
-q, --quiet
Do not complain if no processes were killed.
If /usr/bin/synergys is an executable and not a shell script, you will run it directly, not via the shell:
/usr/bin/synergys
Or, since /usr/bin is on the $PATH of most people, you could simply write:
synergys
If /usr/bin/synergys is actually a shell script, it should be executable (for example, 555 or -r-xr-xr-x permissions), and you can still write just synergys to execute it. You only need to use an explicit sh if the file /usr/bin/synergys is not executable and is a shell script.
I have two shell scripts, a.sh and b.sh.
How can I call b.sh from within the shell script a.sh?
There are a couple of different ways you can do this:
Make the other script executable with chmod a+x /path/to/file(Nathan Lilienthal's comment), add the #!/bin/bash line (called shebang) at the top, and the path where the file is to the $PATH environment variable. Then you can call it as a normal command;
Or call it with the source command (which is an alias for .), like this:
source /path/to/script
Or use the bash command to execute it, like:
/bin/bash /path/to/script
The first and third approaches execute the script as another process, so variables and functions in the other script will not be accessible.
The second approach executes the script in the first script's process, and pulls in variables and functions from the other script (so they are usable from the calling script).
In the second method, if you are using exit in second script, it will exit the first script as well. Which will not happen in first and third methods.
Check this out.
#!/bin/bash
echo "This script is about to run another script."
sh ./script.sh
echo "This script has just run another script."
There are a couple of ways you can do this. Terminal to execute the script:
#!/bin/bash
SCRIPT_PATH="/path/to/script.sh"
# Here you execute your script
"$SCRIPT_PATH"
# or
. "$SCRIPT_PATH"
# or
source "$SCRIPT_PATH"
# or
bash "$SCRIPT_PATH"
# or
eval '"$SCRIPT_PATH"'
# or
OUTPUT=$("$SCRIPT_PATH")
echo $OUTPUT
# or
OUTPUT=`"$SCRIPT_PATH"`
echo $OUTPUT
# or
("$SCRIPT_PATH")
# or
(exec "$SCRIPT_PATH")
All this is correct for the path with spaces!!!
The answer which I was looking for:
( exec "path/to/script" )
As mentioned, exec replaces the shell without creating a new process. However, we can put it in a subshell, which is done using the parantheses.
EDIT:
Actually ( "path/to/script" ) is enough.
If you have another file in same directory, you can either do:
bash another_script.sh
or
source another_script.sh
or
. another_script.sh
When you use bash instead of source, the script cannot alter environment of the parent script. The . command is POSIX standard while source command is a more readable bash synonym for . (I prefer source over .). If your script resides elsewhere just provide path to that script. Both relative as well as full path should work.
Depends on.
Briefly...
If you want load variables on current console and execute you may use source myshellfile.sh on your code. Example:
#!/bin/bash
set -x
echo "This is an example of run another INTO this session."
source my_lib_of_variables_and_functions.sh
echo "The function internal_function() is defined into my lib."
returned_value=internal_function()
echo $this_is_an_internal_variable
set +x
If you just want to execute a file and the only thing intersting for you is the result, you can do:
#!/bin/bash
set -x
./executing_only.sh
bash i_can_execute_this_way_too.sh
bash or_this_way.sh
set +x
You can use /bin/sh to call or execute another script (via your actual script):
# cat showdate.sh
#!/bin/bash
echo "Date is: `date`"
# cat mainscript.sh
#!/bin/bash
echo "You are login as: `whoami`"
echo "`/bin/sh ./showdate.sh`" # exact path for the script file
The output would be:
# ./mainscript.sh
You are login as: root
Date is: Thu Oct 17 02:56:36 EDT 2013
First you have to include the file you call:
#!/bin/bash
. includes/included_file.sh
then you call your function like this:
#!/bin/bash
my_called_function
Simple source will help you.
For Ex.
#!/bin/bash
echo "My shell_1"
source my_script1.sh
echo "Back in shell_1"
Just add in a line whatever you would have typed in a terminal to execute the script!
e.g.:
#!bin/bash
./myscript.sh &
if the script to be executed is not in same directory, just use the complete path of the script.
e.g.:`/home/user/script-directory/./myscript.sh &
This was what worked for me, this is the content of the main sh script that executes the other one.
#!/bin/bash
source /path/to/other.sh
The top answer suggests adding #!/bin/bash line to the first line of the sub-script being called. But even if you add the shebang, it is much faster* to run a script in a sub-shell and capture the output:
$(source SCRIPT_NAME)
This works when you want to keep running the same interpreter (e.g. from bash to another bash script) and ensures that the shebang line of the sub-script is not executed.
For example:
#!/bin/bash
SUB_SCRIPT=$(mktemp)
echo "#!/bin/bash" > $SUB_SCRIPT
echo 'echo $1' >> $SUB_SCRIPT
chmod +x $SUB_SCRIPT
if [[ $1 == "--source" ]]; then
for X in $(seq 100); do
MODE=$(source $SUB_SCRIPT "source on")
done
else
for X in $(seq 100); do
MODE=$($SUB_SCRIPT "source off")
done
fi
echo $MODE
rm $SUB_SCRIPT
Output:
~ ❯❯❯ time ./test.sh
source off
./test.sh 0.15s user 0.16s system 87% cpu 0.360 total
~ ❯❯❯ time ./test.sh --source
source on
./test.sh --source 0.05s user 0.06s system 95% cpu 0.114 total
* For example when virus or security tools are running on a device it might take an extra 100ms to exec a new process.
pathToShell="/home/praveen/"
chmod a+x $pathToShell"myShell.sh"
sh $pathToShell"myShell.sh"
#!/bin/bash
# Here you define the absolute path of your script
scriptPath="/home/user/pathScript/"
# Name of your script
scriptName="myscript.sh"
# Here you execute your script
$scriptPath/$scriptName
# Result of script execution
result=$?
chmod a+x /path/to/file-to-be-executed
That was the only thing I needed. Once the script to be executed is made executable like this, you (at least in my case) don't need any other extra operation like sh or ./ while you are calling the script.
Thanks to the comment of #Nathan Lilienthal
Assume the new file is "/home/satya/app/app_specific_env" and the file contents are as follows
#!bin/bash
export FAV_NUMBER="2211"
Append this file reference to ~/.bashrc file
source /home/satya/app/app_specific_env
When ever you restart the machine or relogin, try echo $FAV_NUMBER in the terminal. It will output the value.
Just in case if you want to see the effect right away, source ~/.bashrc in the command line.
There are some problems to import functions from other file.
First: You needn't to do this file executable. Better not to do so!
just add
. file
to import all functions. And all of them will be as if they are defined in your file.
Second: You may be define the function with the same name. It will be overwritten. It's bad. You may declare like that
declare -f new_function_name=old_function_name
and only after that do import.
So you may call old function by new name.
Third: You may import only full list of functions defined in file.
If some not needed you may unset them. But if you rewrite your functions after unset they will be lost. But if you set reference to it as described above you may restore after unset with the same name.
Finally In common procedure of import is dangerous and not so simple. Be careful! You may write script to do this more easier and safe.
If you use only part of functions(not all) better split them in different files. Unfortunately this technique not made well in bash. In python for example and some other script languages it's easy and safe. Possible to make partial import only needed functions with its own names. We all want that in next bush versions will be done the same functionality. But now We must write many additional cod so as to do what you want.
Use backticks.
$ ./script-that-consumes-argument.sh `sh script-that-produces-argument.sh`
Then fetch the output of the producer script as an argument on the consumer script.
I have a bash script which calls another bash script, like so:
#!/bin/bash
echo "Hi"
./script-two.sh
echo "Hello!"
The problem that I have is that it never makes it to printing "Hello!"
I think this is because ./script-two.sh (Which I did not write) is somehow exiting or changing the shell. I have included this script at the end of this post.
Is there a way I can gurentee that my execution will continue after script-two.sh executes?
I have looked into using the trap command, but I don't fully understand its use properly.
Thanks,
Casey
Here is the contents of what would be script-two.sh
#!/bin/sh
# This file is part of the DITA Open Toolkit project hosted on
# Sourceforge.net. See the accompanying license.txt file for
# applicable licenses.
# (c) Copyright IBM Corp. 2006 All Rights Reserved.
export DITA_HOME=cwd
if [ "${DITA_HOME:+1}" != "1" ]; then
echo "DITA_HOME environment variable is empty or not set";
exit 127;
fi
echo $DITA_HOME
cd "$DITA_HOME"
# Get the absolute path of DITAOT's home directory
DITA_DIR="`pwd`"
echo $DITA_DIR
if [ -f "$DITA_DIR"/tools/ant/bin/ant ] && [ ! -x "$DITA_DIR"/tools/ant/bin/ant ]; then
chmod +x "$DITA_DIR"/tools/ant/bin/ant
fi
export ANT_OPTS="-Xmx512m $ANT_OPTS"
export ANT_OPTS="$ANT_OPTS -Djavax.xml.transform.TransformerFactory=net.sf.saxon.TransformerFactoryImpl"
export ANT_HOME="$DITA_DIR"/tools/ant
export PATH="$DITA_DIR"/tools/ant/bin:"$PATH"
NEW_CLASSPATH="$DITA_DIR/lib:$DITA_DIR/lib/dost.jar:$DITA_DIR/lib/commons-codec-1.4.jar:$DITA_DIR/lib/resolver.jar:$DITA_DIR/lib/icu4j.jar"
NEW_CLASSPATH="$DITA_DIR/lib/saxon/saxon9.jar:$DITA_DIR/lib/saxon/saxon9-dom.jar:$NEW_CLASSPATH"
NEW_CLASSPATH="$DITA_DIR/lib/saxon/saxon9-dom4j.jar:$DITA_DIR/lib/saxon/saxon9-jdom.jar:$NEW_CLASSPATH"
NEW_CLASSPATH="$DITA_DIR/lib/saxon/saxon9-s9api.jar:$DITA_DIR/lib/saxon/saxon9-sql.jar:$NEW_CLASSPATH"
NEW_CLASSPATH="$DITA_DIR/lib/saxon/saxon9-xom.jar:$DITA_DIR/lib/saxon/saxon9-xpath.jar:$DITA_DIR/lib/saxon/saxon9-xqj.jar:$NEW_CLASSPATH"
if test -n "$CLASSPATH"
then
export CLASSPATH="$NEW_CLASSPATH":"$CLASSPATH"
else
export CLASSPATH="$NEW_CLASSPATH"
fi
"$SHELL"
It looks like script-two.sh is setting up an ant build environment.
I think the author intended that it sets up the build environment, then you type your build commands in manually, then type exit to leave the build environment.
I say this because the bottom line of script-two.sh is:
"$SHELL"
which starts a new shell.
Try running your script, then type exit. I think you will see it print Hello! after you type exit.
I'm guessing you're trying to do something like:
#!/bin/bash
echo "Hi"
./script-two.sh
ant <some args>
To do that, what you really want to do is source it, by changing:
./script-two.sh
to
. script-two.sh
e.g.
#!/bin/bash
echo "Hi"
. script-two.sh
ant <some args>
But, you will need to edit script-two.sh and change:
"$SHELL"
to:
case $0 in *script-two.sh)
# executed, start a new shell with the new environment
"$SHELL"
;;
*)
# sourced, don't start a new shell
;;
esac
so that it only starts a shell if the script is being run like ./script-two.sh, but not if it is being sourced like . script-two.sh.
Or if you absolutely can't change script-two.sh, then you could do:
#!/bin/bash
echo "Hi"
. script-two.sh </dev/null
ant <some args>
which will trick "$SHELL" into exiting because it has no input.
Also
export DITA_HOME=cwd
doesn't seem right to me.
It should probably be
export DITA_HOME=$(pwd)
or
export DITA_HOME=`pwd`
(both are equivalent)
I had a similar problem today, up on digging I finally found the answer.
The script I was calling (from within my script) actually had an exit 0 in the end. Removing that fixed my issues.
Just leaving this here as someone may find it useful.
Well for starters, you can execute your bash script with the -x switch to see where it is failing:
bash -x script-one.sh
Secondly, if you call the second script like this:
#!/bin/bash
echo "Hi"
var=$(bash script-two.sh)
echo "Hello!"
It will continue, as long as script-two.sh exits cleanly. Again, you can run the -x script against that script find any problems.
And as Mikel mentioned, always make sure to have exit at the bottom of your scripts.