Access local shell variables in vim - bash

In vim I can access my bash environment variables such as $PWD and $PATH. I would like to know how to access my temporary shell variables in vim too.
For example, suppose I was in my terminal and define a variable foo="bar". Then I enter vim and try to access this variable with the following command :!echo $foo, but it does not recognize this variable. From my understanding, vim starts a new shell each time a bash command is invoked and then closes it immediately after. Is there a way to use the same shell in vim that my local variable foo was defined in?

No, you can't interact with the parent shell from a subprocess it spawned (without that shell's active participation, which isn't reasonably/practically available in the scenario at hand) -- but you can export your variables to make them accessible to new shells started in child processes.
Running
set -a
...will make any variable defined going forward be automatically exported to the environment, even without an explicit export command.
Since (unlike the C system() function) vim's system() honors the SHELL environment variable, if SHELL=/bin/bash (or :set shell=/bin/bash has been run in vim), you can also invoke exported functions from vim. That is, if you define the function and export it as follows:
foo() { echo "bar"; }
export -f foo
...then you can invoke it with !foo from inside vim.
Even then, however, this is running in a new, transient shell instance, not the original parent process.

Explanation
Environment variables and shell variables are two entirely different concepts, but as we manipulate them in a similar way in bash, it's easy to get confused.
Whenever a process is created (by fork), it may include an environment, given by its parent at fork-time. The child process may then access and modify its content. How this is done as a user depends on the program :
In vim, you can access an environment variable like this : :echo $foo
In bash, you can access it like this : $ echo "$foo"
In most programming languages, you can access it with a syntax coherent with the rest of the language, such as ENV['foo'] in ruby
On the other hand, a program may allocate memory for any internal use, but notably, it will quite often define and use variables. Once again, this depends on the program :
In vim, you would use the :let command to assign an internal variable
In bash, you would assign a variable with $ foo='bar', and then read it with $ echo "$foo"
In most programming languages, you have a variation of the foo='bar' syntax, sometimes with type declarations, etc
As you can see, bash uses the same syntax to read an environment variable and one of its own private variables, which can lead to some confusion.
When you execute vim from your bash shell, the environment is copied over from the parent process (bash) to the child (vim), but the private memory of bash (including the variables you may have defined) are not.
Thus, accessing them from the child process would require some inter-process communication mechanism, between parent and child. While technically doable, this option is not implemented in bash nor vim.
Solution
In order for your variable to be accessible from vim (or any forked process, for that matter), you need it to be present in the environment of your vim process.
Several options to do that :
$ export foo='bar' : This will mark your variable for export to the environment of subsequently executed commands. That's what you want in most cases.
$ foo='bar' vim : This adds your variable to the environment of this vim command. Very useful for troubleshooting, or for one-liners.
$ set -a : As you can see in bash manpage, this marks every subsequent definitions for export to the environment of subsequent commands. It's essentially equivalent to prepending every subsequent definition by export.
To go further
The question uses the :!echo $foo syntax to display the value of foo, which is yet another usecase. The ! here is actually an escape sequence that allows you to execute a shell command from vim.
However, vim cannot execute anything in the parent shell (the one you executed the vim command in), so it creates a new bash shell in a child process, executes echo in it, and displays the result.
In the current case, the result is mostly the same, but it could easily be misleading in other situations, so it's important to understand what is happening here.
There is another vim syntax, using expand, that allows one to lookup variables : :echo expand("$foo")
It however works entirely differently.
If no internal variable named foo exists, vim will invoke a shell to look it up (similarly to what ! would do).
This options is way slower than an environment lookup, and not recommended for most usecases.

If you want to use a value from your shell on the :substitute command, there's actually a way to do it.
I don't know if it solves your need but here we go.
Let's say we want to substitute Mydir by your PWD:
:s/Mydir/\=expand($PWD)/g

Related

Script runs when executed but fails when sourced

Original Title: Indirect parameter substitution breaks when the script is sourced (zsh)
zsh 5.7.1 (x86_64-apple-darwin19.0)
GNU bash, version 4.4.20(1)-release (x86_64-pc-linux-gnu)
I’m developing a shell script on a Mac and I’m trying to keep it portable between bash & zsh, so array indexing is a consideration. I know that I can set KSH_ARRAYS to get indexing to start at 0, but I decided to query the OS for the shell that’s in use and set the start index accordingly, which led to the issue described below.
It made sense (to me anyway!) to use indirect expansion, which is what led to the problem. Consider the script indirect.sh:
#! /bin/bash
declare -r ARRAY_START_BASH=0
declare -r ARRAY_START_ZSH=1
declare -r SHELL_BASH=0
declare -r SHELL_ZSH=1
# Indirect expansion is used to reference the values of the variables declared
# in this case statement e.g. ${!ARRAY_START}
case $(basename $SHELL) in
"bash" )
declare -r SHELL_ID=SHELL_BASH
declare -r ARRAY_START=ARRAY_START_BASH
;;
"zsh" )
declare -r SHELL_ID=SHELL_ZSH
declare -r ARRAY_START=ARRAY_START_ZSH
;;
* )
return 1
;;
esac
echo "Shell ID: ${!SHELL_ID} Index arrays from: ${!ARRAY_START}"
It works fine when run from the command line while in the same directory:
<my home> ~ % echo "$(./indirect.sh)"
Shell ID: 1 Index arrays from: 1
Problems arise when I source the script:
<my home> ~ % echo "$(. ~/indirect.sh)"
/Users/<me>/indirect.sh:28: bad substitution
I don’t understand why sourcing the script changes the behavior of the parameter expansion.
Is this expected behavior? If so, I’d be grateful if someone could explain it and hopefully, offer a work around.
The problem described in the original post has nothing to do with indirect expansion. The difference in behavior is a result of different shells being invoked depending on whether the script is “executed” or “sourced”. These differences reveal the basic flaw in deriving the shell from the $SHELL variable that underpins the script's design. If the shell defined in $SHELL does not match the shebang, the script will fail either when sourced or executed. An explanation follows.
Indirect expansion doesn’t offer value in the given scenario because values could just as easily be assigned directly. They’ll have to be assigned that way regardless given the different syntax used for indirect expansion between shells. In fact, other syntax differences between shells makes the entire premise for detecting the shell moot! However, putting that aside, the difference in behavior is a result of different shells being invoked based on whether the script is “executed” or “sourced”. The behavior of sourcing is well documented with numerous explanations on the web, but for context here’s how it works:
Executing a Script
Use the “./“ syntax to execute a script.
When run this way, the script executes in a sub-shell. Any changes the
script makes to it’s shell are applied to the sub-shell, not the shell
in which the script was launched, so those changes are lost when the
shell exits because the sub-shell in which it executed is destroyed as
well. For example, if the script changes the working directory, it
does so in the sub-shell. The working directory of the main shell that
launched the script is unchanged when the script terminates. If you
want to make changes to the shell in which the script was launched, it
must be sourced.
Sourcing a Script
Use the “source “ syntax to source a
script. When run this way, the script essentially becomes an argument
for the source command, which handles invoking the appropriate
execution. Some shells (e.g. ksh) use a single period “.” instead of
“source”.
When a script is executed with the “./“ syntax, the shebang at the top of the file is used to determine which shell to use. When a script is sourced, the shebang is ignored and the shell in which the script is launched is used instead. Also note that the period that appears in the “./“ command syntax used to execute a script, is not related to the period that’s occasionally used as an alias for the source command.
The script in the post uses bash in the shebang statement, so it works when executed because it’s run using bash. When it’s sourced from zsh, it encounters the incorrect indirect expansion syntax:
“${!A_VAR}"
The correct syntax is:
"${(P)A_VAR}"
However, correcting the syntax won’t help because it will then fail when executed. The shebang will invoke bash and the syntax will be wrong again. That renders indirection useless for accessing a variable designed to indicate the shell in use. More importantly, a design based on querying an environment variable for the shell is flawed due to differences in the shell that’s ultimately used depending on whether the script is executed or sourced.
To add to your answer (what I'm going to say is too long for a comment), I can not think of any application, why your script could be useful if not sourced. Actually, I came accross the need of such a script by myself in exactly one occasion:
Since I use as interactive shell not only zsh, but also sometimes bash, so I have written my .zshrc and .bashrc to set up everything (including defining variables and shell functions for interactive use). In order to safe work,
I try to put code which works under both bash and zsh into a single file (say: .commonrc), and my .zshrc and .bashrc have inside them a
source .commonrc
While many things are so different in bash and zsh, that I can't put them into .commonrc, some can, provided I do some tweaking. One reason for headache is obviously the different indexing of arrays, which you seemingly try to solve. So I have also a similar feature. However, I don't nee ca case construct for this. Instead, my .bashrc looks like this (using your naming of the variables):
...
declare -r ARRAY_START=0
source .commonrc
...
and my .zshrc looks like this:
...
declare -r ARRAY_START=1
source .commonrc
...
Since it does not happen that the .bashrc is run from a zsh and vice versa, I don't need to query what kind of shell I have.

How to overwrite environment variables in mac using Shell Script

I have a set of environment variables that need to be set on the basis of the arguments specified in the shell script.
But the problem is that those variables are already defined in the bash profile
FOR EXAMPLE:
bash_profile has a variable called "KARAN":
export KARAN=/config/1
Now on running the shell script, this is what it should do:
export KARAN=/config/2 (Changed the bash profile's KARAN value to 2)
Your question is not clear. If your script needs to set the env var to a specific value just do so using export VAR=val. What I think you're asking is how to have a script modify the environment of the current shell. And that is impossible without the cooperation of both shells. That is because environment vars are inherited by child processes. But a child process cannot directly modify the environment of its parent process (or some other random process for that matter). To do so the two processes must coordinate the exchange of data. This is typically done by using the source command if the child process is a shell script. Or by having the child process write a series of export statements to stdout and having the parent shell capture and execute those statements. For example, let's say I have a script named set_env that looks like this
#!/bin/sh
echo export KARAN=/config_2
echo export VAR2=val2
The current shell would then do
eval $(set_env)
Note, however, eval is dangerous. I prefer to do this which is slightly safer:
set_env | source /dev/stdin
That, however, only works in shells like ksh and zsh. Due to how bash handles pipelines the source is actually executed in a child shell and therefore the vars won't be set in the current shell.
You can create a new Profile with all the new definitions. and then call the line below on top of your shell script. Similarly, you can create as many profiles as you want and use it.
source bash_profile_new

Bash what bash alias actually is? [duplicate]

I'm surprised hasn't been asked before, but…
What is the difference between
alias ⇢ alias EXPORT='alias'
function ⇢ function exporter() { echo $EXPORT }
and
export ⇢ export ALIAS='export'
and for that matter...
alias export=$(function) (j/k)
in bash (zsh, et al.)
Specifically, I'd be most interested in knowing the lexical/practical difference between
alias this=that
and
export that=this
I have both forms... all over the place - and would prefer to stop arbitrarily choosing one, over the other. 😂
I'm sure there is a great reference to a "scopes and use-cases for unix shells", somewhere... but thought I'd post the question here, in the name of righteous-canonicalicism.
You're asking about two very different categories of things: aliases and functions define things that act like commands; export marks a variable to be exported to child processes. Let me go through the command-like things first:
An alias (alias ll='ls -l') defines a shorthand for a command. They're intended for interactive use (they're actually disabled by default in shell scripts), and are simple but inflexible. For example, any arguments you specify after the alias simply get tacked onto the end of the command; if you wanted something like alias findservice='grep "$1" /etc/services', you can't do it, because $1 doesn't do anything useful here.
A function is like a more flexible, more powerful version of an alias. Functions can take & process arguments, contain loops, conditionals, here-documents, etc... Basically, anything you could do with a shell script can be done in a function. Note that the standard way to define a function doesn't actually use the keyword function, just parentheses after the name. For example: findservice() { grep "$1" /etc/services; }
Ok, now on to shell variables. Before I get to export, I need to talk about unexported variables. Basically, you can define a variable to have some (text) value, and then if you refer to the variable by $variablename it'll be substituted into the command. This differs from an alias or function in two ways: an alias or function can only occur as the first word in the command (e.g. ll filename will use the alias ll, but echo ll will not), and variables must be explicitly invoked with $ (echo $foo will use the variable foo, but echo foo will not). More fundamentally, aliases and functions are intended to contain executable code (commands, shell syntax, etc), while variables are intended to store non-executable data.
(BTW, you should almost always put variable references inside double-quotes -- that is, use echo "$foo" instead of just echo $foo. Without double-quotes the variable's contents get parsed in a somewhat weird way that tends to cause bugs.)
There are also some "special" shell variables, that are automatically set by the shell (e.g. $HOME), or influence how the shell behaves (e.g. $PATH controls where it looks for executable commands), or both.
An exported variable is available both in the current shell, and also passed to any subprocesses (subshells, other commands, whatever). For example, if I do LC_ALL=en_US.UTF-8, that tells my current shell use the "en_US.UTF-8" locale settings. On the other hand, if I did export LC_ALL=en_US.UTF-8 that would tell the current shell and all subprocesses and commands it executes to use that locale setting.
Note that a shell variable can be marked as exported separately from defining it, and once exported it stays exported. For example, $PATH is (as far as I know) always exported, so PATH=/foo:/bar has the same effect as export PATH=/foo:/bar (although the latter may be preferred just in case $PATH somehow wasn't already exported).
It's also possible to export a variable to a particular command without defining it in the current shell, by using the assignment as a prefix for the command. For example LC_ALL=en_US.UTF-8 sort filename will tell the sort command to use the "en_US.UTF-8" locale settings, but not apply that to the current shell (or any other commands).
TL;DR:
The shell evaluation order (per POSIX) for the entities in your question is:
aliases --> variables --> command substitutions --> special built-ins --> functions --> regular built-ins
Aliases do not persist across subshells, but variables (and in Bash, functions) can be made to do so with the export command.
Regular built-ins can be overridden by writing functions that have the same name as the regular built-in (since functions expand before regular built-ins). (NOTE: If you're trying to add functionality to the regular built-in, call the built-in with command in your function definition so you don't accidentally create a recursive function.)
Variables can be made readonly with the (special built-in) readonly command, but aliases cannot.
USE CASES:
Export a variable if you need to use a variable across subshells.
Make a variable readonly if you don't want it changed for the life of the parent shell (once performed, this cannot be undone with unset; you must restart the parent shell).
If you want to override or add functionality to a regular built-in, use a function.
NOTE: If you want to be sure that you're using a special or regular built-in and not someone else's function, use builtin the_builtin, or if the shell doesn't support the builtin command, use the POSIX comand command -p the_builtin, where the -p switch tells command to use the $PATH that ships with the shell by default (in case the user has overriden path).
NOTE: A variable can be made to act like an alias that also persists across subshells and cannot be changed. For example,
#! /bin/sh
my_cmd='ls -al'
export my_cmd
readonly my_cmd
will act like
#! /bin/sh
alias my_cmd='ls -al'
so long as
my_cmd is used without double-quotes (i.e. ${my_cmd}, NOT "${my_cmd}") so it isn't treated as a single string, and
IFS is the standard space-tab-newline and not switched to something else so that the elements of my_cmd are globbed and each part separated by a space is evaluated as a single token (otherwise it will be evaluated as a single string).
Each shell (e.g. bash, zsh, ksh, yash, etc.) is a bit different, so be sure to review the reference manual for it (they each implement POSIX in a unique way, or sometimes not at all).

Access variable declared inside Makefile command

I'm trying to access a variable declared by previous command (inside a Makefile).
Here's the Makefile:
all:
./script1.sh
./script2.sh
Here's the script declaring the variable I want to access,script1.sh:
#!/usr/bin/env bash
myVar=1234
Here's the script trying to access the variable previously defined, script2.sh:
#!/usr/bin/env bash
echo $myVar
Unfortunately when I run make, myVar isn't accessible. Is there an other way around? Thanks.
Make will run each shell command in its own shell. And when the shell exits, its environment is lost.
If you want variables from one script to be available in the next, there are constructs which will do this. For example:
all:
( . ./script1.sh; ./script2.sh )
This causes Make to launch a single shell to handle both scripts.
Note also that you will need to export the variable in order for it to be visible in the second script; unexported variables are available only to the local script, and not to subshells that it launches.
UPDATE (per Kusalananda's comment):
If you want your shell commands to populate MAKE variables instead of merely environment variables, you may have options that depend on the version of Make that you are running. For example, in BSD make and GNU make, you can use "variable assignment modifiers" including (from the BSD make man page):
!= Expand the value and pass it to the shell for execution and
assign the result to the variable. Any newlines in the result
are replaced with spaces.
Thus, with BSD make and GNU make, you could do this:
$ cat Makefile
foo!= . ./script1.sh; ./script2.sh
all:
#echo "foo=${foo}"
$
$ cat script1.sh
export test=bar
$
$ cat script2.sh
#!/usr/bin/env bash
echo "$test"
$
$ make
foo=bar
$
Note that script1.sh does not include any shebang because it's being sourced, and is therefore running in the calling shell, whatever that is. That makes the shebang line merely a comment. If you're on a system where the default shell is POSIX but not bash (like Ubuntu, Solaris, FreeBSD, etc), this should still work because POSIX shells should all understand the concept of exporting variables.
The two separate invocations of the scripts create two separate environments. The first script sets a variable in its environment and exits (the environment is lost). The second script does not have that variable in its environment, so it outputs an empty string.
You can not have environment variables pass between environments other than between the environments of a parent shell to its child shell (not the other way around). The variables passed over into the child shell are only those that the parent shell has export-ed. So, if the first script invoked the second script, the value would be outputted (if it was export-ed in the first script).
In a shell, you would source the first file to set the variables therein in the current environment (and then export them!). However, in Makefiles it's a bit trickier since there's no convenient source command.
Instead you may want to read this StackOverflow question.
EDIT in light of #ghoti's answer: #ghoti has a good solution, but I'll leave my answer in here as it explains a bit more verbosely about environment variables and what we can do and not do with them with regards to passing them between environments.

Why are bash script variables not saving?

I have a simple bash script:
#!/bin/bash
JAVA_HOME=/usr
EC2_HOME=~/ec2-api
echo $EC2_HOME
export PATH=$PATH:$EC2_HOME/bin
I run the script like so
$ ./ec2
/Users/user/ec2-api
The script runs and produces the correct output.
However, when I now try to access the EC2_HOME variable, I get nothing out:
$ echo $EC2_HOME
I get a blank string back. What am I doing wrong?
Do either of the following instead:
source ec2
or
. ec2
(note the . notation is just a shortcut for source)
Explanation:
This is because ./ec2 actually spawns a subshell from your current shell to execute the script, and subshells cannot affect the environment of the parent shell from which it spawned.
Thus, EC2_HOME does get set to /Users/user/ec2-api correctly in the subshell (and similarly the PATH environment variable is updated and exported correctly in the subshell as well), but those changes won't propagate back to your parent shell.
Using source runs the script directly in the current shell without spawning a subshell, so the changes made will persist.
(A note on export: export is used to tell new shells spawned from the current shell to use the variables exported from the current shell. So for any variables you would only use in the current shell, they need not be exported.)
A shell script can never modify the environment of their parent.
To fix your problem, you can use the dot (.) command:
$ . ./ec2
and that should work. In cshell, it would be
% source ./ec2
To learn more about shells and scripts, my best resource is by far Unix power tools.

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