I have the following shell file that contains this:
sh
nightlyTag() {
echo $1-alpha.$(date +%Y%m%d).$(($(date +%s%N)/1000000))
}
yarnPubCanaryVersion() {
if [ -z "$1" ]
then
echo "No version argument supplied, maybe you meant v1.0.0?"
return 1
fi
version=`nightlyTag $1`
yarn version --new-version $version --no-git-tag-version
npm publish --tag canary
git reset --hard HEAD
}
I make the file executable with chmod +x canary.sh, then I run it doing ./canary.sh then my terminal changes to sh-3.2$ then I try to run the functions in the terminal like this nightlyTag and I get
sh: nightlyTag: command not found
Same for yarnPubCanaryVersion.
I was looking at this SO question
You won't be able to run functions from the terminal after you run the script.
You need to source the script to do this:
source ./canary.sh
Or add the contents of the file to the .bashrc file or its equivalent, and then source it.
The source command is used to load any function file into the current shell.
Now once you call those functions you will get the expected output.
At the top of your sh file you need to include:
#! /path/to/bash
the path to the bash that you are using.
Related
The following simple script is apparently not so simple.
The entire script appears to work properly until I get to the npm command.
I have looked at the numerous threads here, but none of the solutions fix the issue.
Each of the scripts is kicked off by a parent script.
Here is the parent:
#!/bin/bash/
authGogglesPath='/c/sandBox/amazon-sandbox/CraigMonroe/platform.shared.auth-goggles'
echo $'\nExecuting node commands for local running solution...\n'
#echo $(pwd)
# run the scripts
bash edge.sh ${edgePath} &
exec bash
I checked my path in the terminal and it's aware
I thought that it might be running as another associated profile so I tried the full path to npm, but the same results.
#!/bin/bash/
authGogglesPath='/c/sandBox/amazon-sandbox/CraigMonroe/platform.shared.auth-goggles'
echo $'\nExecuting node commands for local running solution...\n'
#echo $(pwd)
# run the scripts
bash edge.sh ${edgePath} &
exec bash
That calls edge.sh with a string path for arg (more for later)
edge.sh is another simple script
#!/bin/bash/
PATH=$1
#echo $PATH
if [ -z "${PATH}" ] ; then
"PATH is empty! Aborting"
exit 1
fi
cd "${PATH}"
echo $'\nExecuting Edge...\n'
npm run dev
Each time I run this I'm receiving:
$ bash edge.sh /c/sandBox/amazon-sandbox/CraigMonroe/platform.shared.auth-goggles/
Executing Edge...
edge.sh: line 13: npm: command not found
cmonroe#LP10-G6QD2X2 MINGW64 ~/cruxScripts
$
When in the terminal and manually navigating to the directory and running the command it works properly. Where the edge builds and starts.
Unless npm is in /c/sandBox/amazon-sandbox/CraigMonroe/platform.shared.auth-goggles/, doing PATH=$1 means your PATH only refers to that one folder.
No more /usr/bin or any other folders your bash session might need.
As commented, adding to the PATH should work
PATH="$1:${PATH}"
How is useful env in this install.sh script:
env git clone --depth=1 https://github.com/robbyrussell/oh-my-zsh.git $ZSH || {
printf "Error: git clone of oh-my-zsh repo failed\n"
exit 1
}
Is it because of the variable $ZSH?
The $ZSH variable is defined a couple of lines above so it is not the purpose here.
The env command can be used to run a command/program in a modified environment, but as there are no additional variables specified here, the command will be run in an unmodified environment.
The main purpose in this case is to get rid of aliases and functions that could shadow the command to be run.
gcloud init command doesn't offer login prompt during a bash script execution.
But it offered the login after I typed exit command manually after script ended.
vagrant#vagrant-ubuntu-trusty-64:~$ exit
logout
Welcome! This command will take you through the configuration of gcloud.
Settings from your current configuration [default] are:
Your active configuration is: [default]
Pick configuration to use:
[1] Re-initialize this configuration [default] with new settings
[2] Create a new configuration
Please enter your numeric choice: 1
Your current configuration has been set to: [default]
To continue, you must log in. Would you like to log in (Y/n)?
My bash script:
#!/usr/bin/env bash
OS=`cat /proc/version`
function setupGCE() {
curl https://sdk.cloud.google.com | bash
`exec -l $SHELL`
`gcloud init --console-only`
`chown -R $USER:$USER ~/`
}
if [[ $OS == *"Ubuntu"* || $OS == *"Debian"* ]]
then
sudo apt-get -y install build-essential python-pip python-dev curl
sudo pip install apache-libcloud
setupGCE
fi
How can I get the login prompt during the bash script execution?
There are a number of issues with the posted snippet.
The correct snippet is (probably):
function setupGCE() {
curl https://sdk.cloud.google.com | bash
gcloud init --console-only
chown -R $USER:$USER ~/
}
The first error with the original, which you discovered yourself (the what of it at least it not the why), is that exec -l $SHELL is blocking progress. It does that because you've run an interactive shell that is now waiting on you for input and the function is waiting for that process to exit before continuing.
Additionally, exec replaces the current process with the spawned process. You got lucky here actually. Had you not wrapped the call to exec in single quotes your function would have exited the shell script entirely when you exited the $SHELL it launched. As it is, however, exec just replaced the sub-shell that the backticks added and so you were left with a child process that could safely exit and return you to the parent/main script.
The second issue is that backticks run the command they surround and then replace themselves with the output. This is why
echo "bar `echo foo` baz"
outputs bar foo baz, etc. (Run set -x before running that to see what commands are actually being run.) So when you write
`gcloud init --console-only`
what you are saying is "run gcloud init --console-only then take its output and replace the command with that" which will then attempt to run the output as a command itself (which is likely not what you wanted). Similarly on the other lines.
This happens to not have been problematic here though as chown and likely gcloud init don't return anything and so the resulting command line is empty.
Somehow the exec -l $SHELL did all the mess. I changed it to source ~/.bashrc and now it works.
I'm on a Linux machine using screen, and I'm attempting to write a (fairly portable) function which runs a bash function in a new, detached screen session which automatically closes upon completion. I've had some success, but I noticed the following behavior:
If I include the definition of mail_submit() in my ~/.bashrc file, I can run
mail_submit foo
in the terminal, and also I can access the alias in a new screen session:
screen -S test
mail_submit foo
However, the following command does not work:
screen -d -m -S test sh -c 'mail_submit foo'
presumably because sh -c starts a fresh shell that has no knowledge of my ~/.bashrc profile. So, I can use the following fix:
screen -d -m -S test sh -c 'source ~/.bashrc; mail_submit foo'
which does work.
But if I want to wrap this functionality up into a bash alias (which is my ultimate goal here), this will cause a weird self-referential situation.
Question: What is an easy way to either have sh -c know the location of my ~/.bashrc profile, or use a variant of sourcing the file and creating an alias?
EDIT: I could save the shell script in my home directory, and create an alias which runs
screen -d -m -S test bash -c '~/mail_submit.sh $1'
but I'd still be curious to hear other possible fixes.
A default ~/.bashrc contains this ([[ "$-" != *i* ]] && return) little piece of code on top of it (or somewhere else in the upper part). This line will prevent the ~/.bashrc from beeing sourced if the bash shell doesn't run in interactive mode.
You could:
Remove this line
Create a new file which will only contain the alias you need and source that
Create a little bash script instead of an alias and run that
Do you mean screen -d -m -S test bash -c 'mail_submit foo'?
It looks like you're trying to run the command with the shell (sh), and not the bourne again shell (bash), which is the shell interpreter which actually reads the ~/.bashrc profile.
Edit: The .bashrc file is not being sourced by default because screen does not create the bash process as a login shell, which is when the .bashrc file is read. Creating a .screenrc file with the line defshell -bash will create the bash process as a login shell instead, which will then call the .bashrc file.
My bash script:
#!/bin/bash
cd /tmp
Before running my script:
pwd: /
After running my script:
pwd: /
After runnig my script trough sourcing it:
pwd: /tmp
How I can stay at the path from the script without sourcing it ?
You can't. Changes to the current directory only affect the current process.
Let me elaborate a little bit on this:
When you run the script, bash creates a new process for it, and changes to the current directory only affect that process.
When you source the script, the script is executed directly by the shell you are running, without creating extra processes, so changes to the current directory are visible to your main shell process.
So, as Ignacio pointed out, this can not be done
Ignacio is correct. However, as a heinous hack (totally ill advised and this really should get me at least 10 down votes) you can exec a new shell when you're done
#!/bin/bash
...
cd /
exec bash
Here's a silly idea. Use PROMPT_COMMAND. For example:
$ export PROMPT_COMMAND='test -f $CDFILE && cd $(cat $CDFILE) && rm $CDFILE'
$ export CDFILE=/tmp/cd.$$
Then, make the last line of your script be 'pwd > $CDFILE'
If you really need this behavior, you can make your script return the directory, then use it somehow. Something like:
#!/bin/bash
cd /tmp
echo $(pwd)
and then you can
cd $(./script.sh)
Ugly, but does the trick in this simple case.
You can define a function to run in the current shell to support this. E.g.
md() { mkdir -p "$1" && cd "$1"; }
I have the above in my ~/.bashrc