accessing newly created directory in shell script - bash

I'm attempting to make a new folder, a duplicate of the input, and then tar the contents of that folder. I can't figure out why - but it seems like instead of searching the contents of my newly created directory - it is searching my entire computer... returning lines such as
/Applications/GarageBand.app/Contents/Frameworks/MAAlchemy.framework/Resources/Libraries/WaveOsc/Sine/Sine - Vocal 1.raw is a file
/Applications/GarageBand.app/Contents/Frameworks/MAAlchemy.framework/Resources/Libraries/WaveOsc/Sine/Sine - Vocal 2.raw is a file
/Applications/GarageBand.app/Contents/Frameworks/MAAlchemy.framework/Resources/Libraries/WaveOsc/Sine/Triangle - Arp.raw is a file
/Applications/GarageBand.app/Contents/Frameworks/MAAlchemy.framework/Resources/Libraries/WaveOsc/Sine/Triangle - Asym 4.raw is a file
/Applications/GarageBand.app/Contents/Frameworks/MAAlchemy.framework/Resources/Libraries/WaveOsc/Sine/Triangle - Eml.raw is a file
/Applications/GarageBand.app/Contents/Frameworks/MAAlchemy.framework/Resources/Libraries/WaveOsc/Square is a folder
/Applications/GarageBand.app/Contents/Frameworks/MAAlchemy.framework/Resources/Libraries/WaveOsc/Square/Square - Arp.raw is a file
/Applications/GarageBand.app/Contents/Frameworks/MAAlchemy.framework/Resources/Libraries/WaveOsc/Square/Square - Bl Saw.raw is a file
can you guys spot a simple error?
BTW, I know that the script to tar isn't present yet, but that will be easy once i can navigate the new folder.
#!/bin/bash
##--- deal with help args ------------------
##
print_help_message() {
printf "Usage: \n"
printf "\t./`basename $0` <input_dir> <output_dir>\n"
printf "where\n"
printf "\tinput_dir : (required) the input directory.\n"
printf "\toutput_dir : (required) the output directory.\n"
}
if [ "$1" == "help" ]; then
print_help_message
exit 1
fi
## ------ get cli args ----------------------
##
if [ $# == 2 ]; then
INPUT_DIR="$1"
OUTPUT_DIR="$2"
fi
## ------ tree traversal function -----------
##
mkdir "$2"
cp -r "$1"/* "$2"/
## ------ return output dir name ------------
##
return_output_dir() {
echo $OUTPUT_DIR/$(basename $(basename $(dirname $1)))
}
bt() {
output_dir="$1"
for filename in $output_dir/*; do
if [ -d "${filename}" ]; then
echo "$filename is a folder"
bt $filename
else
echo "$filename is a file"
fi
done
}
## ------ main ------------------------------
##
main() {
bt $return_output_dir
exit 0
}
main
}

Well, I can tell you why it's doing that, but I'm not clear on what it's supposed to be doing, so I'm not sure how to fix it. The immediate problem is that return_output_dir is a function, not a variable, so in the command bt $return_output_dir the $return_output_dir part expands to ... nothing, and bt gets run with no argument. That means that inside bt, output_dir gets set to the empty string, so for filename in $output_dir/* becomes for filename in /*, which iterates over the top-level items on your boot volume.
There are a number of other things that're confusing/weird about this code:
The function main() doesn't seem to serve any purpose -- some of the main-line code is outside it (notably, the argument parsing stuff), some inside, for no apparent reason. Having a main function is required in some languages, but in a shell script it generally makes more sense to just put the main code inline. (Also, functions shouldn't exit, they should return.)
You have variables named both OUTPUT_DIR and output_dir. Use distinct names. Also, it's generally best to stick to lowercase (or mixed-case) variable names, to avoid conflicts with the variables that're used by the shell and other programs.
You copy $1 and $2 into INPUT_DIR and OUTPUT_DIR, then continue to use $1 and $2 rather than the more-clearly-named variables you just copied them into.
output_dir is changed in the recursive function, but not declared as local; this means that inner invocations of bt will be changing the values that outer ones might try to use, leading to weirdness. Declare function-local variables as local to avoid trouble.
$(basename $(basename $(dirname $1))) doesn't make sense. Suppose $1 is "/foo/bar/baz/quux": then dirname $1 returns /foo/bar/baz, basename /foo/bar/baz returns "baz", and basename baz returns "baz" again. The second basename isn't doing anything! And in any case, I'm pretty sure the whole thing isn't doing what you expect it to.
What directory is bt supposed to be recursing through? Nothing in how you call it has any reference to either INPUT_DIR or OUTPUT_DIR.
As a rule, you should put variable references in double-quotes (e.g. for filename in "$output_dir"/* and bt "$filename"). You do this in some places, but not others.

Related

Shell script: Copy file and folder N times

I've two documents:
an .json
an folder with random content
where <transaction> is id+sequancial (id1, id2... idn)
I'd like to populate this structure (.json + folder) to n. I mean:
I'd like to have id1.json and id1 folder, an id2.json and id2 folder... idn.json and idn folder.
Is there anyway (shell script) to populate this content?
It would be something like:
for (i=0,i<n,i++) {
copy "id" file to "id+i" file
copy "id" folder to "id+i" folder
}
Any ideas?
Your shell syntax is off but after that, this should be trivial.
#!/bin/bash
for((i=0;i<$1;i++)); do
cp "id".json "id$i".json
cp -r "id" "id$i"
done
This expects the value of n as the sole argument to the script (which is visible inside the script in $1).
The C-style for((...)) loop is Bash only, and will not work with sh.
A proper production script would also check that it received the expected parameter in the expected format (a single positive number) but you will probably want to tackle such complications when you learn more.
Additionaly, here is a version working with sh:
#!/bin/sh
test -e id.json || { (>&2 echo "id.json not found") ; exit 1 ; }
{
seq 1 "$1" 2> /dev/null ||
(>&2 echo "usage: $0 transaction-count") && exit 1
} |
while read i
do
cp "id".json "id$i".json
cp -r "id" "id$i"
done

Saving function output into a variable named in an argument

I have an interesting problem that I can't seem to find the answer for. I am creating a simple app that will help my dev department auto launch docker containers with NginX and config files. My problem is, for some reason I can't get the bash script to store the name of a folder, while scanning the directory. Here is an extremely simple example of what I am talking about....
#!/bin/bash
getFolder() {
local __myResultFolder=$1
local folder
for d in */ ; do
$folder=$d
done
__myResultFolder=$folder
return $folder
}
getFolder FOLDER
echo "Using folder: $FOLDER"
I then save that simple script as folder_test.sh and put it in a folder where there is only one folder, change owner to me, and give it correct permissions. However, when I run the script I keep getting the error...
./folder_test.sh: 8 ./folder_test.sh: =test_folder/: not found
I have tried putting the $folder=$d part in different types of quotes, but nothing works. I have tried $folder="'"$d"'", $folder=`$d`, $folder="$d" but none of it works. Driving me insane, any help would be greatly appreciated. Thank you.
If you want to save your result into a named variable, what you're doing is called "indirect assignment"; it's covered in BashFAQ #6.
One way is the following:
#!/bin/bash
# ^^^^ not /bin/sh; bash is needed for printf -v
getFolder() {
local __myResultFolder=$1
local folder d
for d in */ ; do
folder=$d
done
printf -v "$__myResultFolder" %s "$folder"
}
getFolder folderName
echo "$folderName"
Other approaches include:
Using read:
IFS= read -r -d '' "$__myResultFolder" < <(printf '%s\0' "$folder")
Using eval (very, very carefully):
# note \$folder -- we're only trusting the destination variable name
# ...not trusting the content.
eval "$__myResultFolder=\$folder"
Using namevars (only if using new versions of bash):
getFolder() {
local -n __myResultFolder=$1
# ...your other logic here...
__myResultFolder=$folder
}
The culprit is the line
$folder=$d
which is treating the folder names to stored with a = sign before and tried to expand it in that name i.e. literally treats the name =test_folder/ as an executable to be run under shell but does not find a file of that name. Change it to
folder=$d
Also, bash functions' return value is only restricted to integer types and you cannot send a string to the calling function. If you wanted to send a non-zero return code to the calling function on $folder being empty you could add a line
if [ -z "$folder" ]; then return 1; else return 0; fi
(or) if you want to return a string value from the function, do not use return, just do echo of the name and use command-substitution with the function name, i.e.
getFolder() {
local __myResultFolder=$1
local folder
for d in */ ; do
folder=$d
done
__myResultFolder=$folder
echo "$folder"
}
folderName=$(getFolder FOLDER)
echo "$folderName"

Get bash function path from name

A hitchhiker, waned by the time a function is taking to complete, wishes to find where a function is located, so that he can observe the function for himself by editting the file location. He does not wish to print the function body to the shell, simply get the path of the script file containing the function. Our hitchhiker only knows the name of his function, which is answer_life.
Imagine he has a function within a file universal-questions.sh, defined like this, the path of which is not known to our hitchhiker:
function answer_life() {
sleep $(date --date='7500000 years' +%s)
echo "42"
}
Another script, called hitchhiker-helper-scripts.sh, is defined below. It has the function above source'd within it (the hitchhiker doesn't understand source either, I guess. Just play ball.):
source "/usr/bin/universal-questions.sh"
function find_life_answer_script() {
# Print the path of the script containing `answer_life`
somecommand "answer_life" # Should output the path of the script containing the function.
}
So this, my intrepid scripter, is where you come in. Can you replace the comment with code in find_life_answer_script that allows our hitchhiker to find where the function is located?
In bash operating in extended debug mode, declare -F will give you the function name, line number, and path (as sourced):
function find_life_answer_script() {
( shopt -s extdebug; declare -F answer_life )
}
Like:
$ find_life_answer_script
answer_life 3 ./universal-questions.sh
Running a sub-shell lets you set extdebug mode without affecting any prior settings.
Your hitchhiker can also try to find the answer this way:
script=$(readlink -f "$0")
sources=$(grep -oP 'source\s+\K[\w\/\.]+' $script)
for s in "${sources[#]}"
do
matches=$(grep 'function\s+answer_life' $s)
if [ -n "${matches[0]}" ]; then
echo "$s: Nothing is here ("
else
echo "$s: Congrats! Here is your answer!"
fi
done
This is for case if debug mode will be unavailable on some planet )

Are there any existing methods for importing functions from other scripts without sourcing the entire script?

I am working on a large shell program and need a way to import functions from other scripts as required without polluting the global scope with all the internal functions from that script.
UPDATE: However, those imported functions have internal dependancies. So the imported function must be executed in the context of its script.
I came up with this solution and wonder if there is any existing strategy out there and if not, perhaps this is a really bad idea?
PLEASE TAKE A LOOK AT THE POSTED SOLUTION BEFORE RESPONDING
example usage of my solution:
main.sh
import user get_name
import user set_name
echo "hello $(get_name)"
echo "Enter a new user name :"
while true; do
read user_input < /dev/tty
done
set_name $user_input
user.sh
import state
set_name () {
state save "user_name" "$1"
}
get_name () {
state get_value "user_name"
}
As one approach, you could put a comment in the script to indicate where you want to stop sourcing:
$ cat script
fn() { echo "You are running fn"; }
#STOP HERE
export var="Unwanted name space pollution"
And then, if you are using bash, source it like this:
source <(sed '/#STOP HERE/q' script)
<(...) is process substitution and our process, sed '/#STOP HERE/q' script just extracts the lines from script until the stop line is reached.
Adding more precise control
We can select particular sections from a file if we add both start and stop flags:
$ cat script
export var1="Unwanted name space pollution"
#START
fn1() { echo "You are running fn1"; }
#STOP
export var2="More unwanted name space pollution"
#START
fn2() { echo "You are running fn2"; }
#STOP
export var3="More unwanted name space pollution"
And then source the file like this:
source <(sed -n '/#START/,/#STOP/p' script)
create standalone shel script that do this
will have 2 argument the file name and the function name
it will source the input file first
it will then use declare -f function name
in your code you can include functions like this
eval "./importfunctions.sh filename functionaname"
what is happening here :
step 1 basically read the file and source it in new shell environment . then it will echo the function declaration
step 2 will eval that function into our main code
So final result is as if we wrote just that function in our main script
When the functions in the script indent untill the closing } and all start with the keyword function, you can include specific functions without changing the original files:
largeshell.sh
#!/bin/bash
function demo1 {
echo "d1"
}
function demo2 {
echo "d2"
}
function demo3 {
echo "d3"
}
function demo4 {
echo "d4"
}
echo "Main code of largeshell... "
demo2
Now show how to source demo1() and forget demo4():
source <(sed -n '/^function demo1 /,/^}/p' largeshell.sh)
source <(sed -n '/^function demo3 /,/^}/p' largeshell.sh)
demo1
demo4
Or source all functions in a loop:
for f in demo1 demo3; do
echo sourcing $f
source <(sed -n '/^function '$f' /,/^}/p' largeshell.sh)
done
demo1
demo4
You can make it more fancy when you source a special script that will:
grep all strings starting with largeshell., like largefile.demo1
generate functions like largefile.demo1 that will call demo1
and source all functions that are called.
Your new script will look like
source function_includer.sh
largeshell.demo1
largeshell.demo4
EDIT:
You might want to reconsider your requirements.
Above solution is not only slow, but it will also make it hard for the
guys and ladies who made tha largeshell.sh. As soon as they are going to refactor their code or replace it with something in another language,
they have to refactor, test and deploy your code as well.
A better path is extracting the functions from largeshell.sh into some smaller files ("modules"), and put them in a shared directory (shlib?).
With names as sqlutil.sh, datetime.sh, formatting.sh, mailstuff.sh and comm.sh you can pick the includes file you need (and largefile.sh will include them all).
It's been a while and it would appear that my original solution is the best one out there. Thanks for the feedback.

How to add/use a variable to my bashrc file?

I'm a newbie to Linux operating system
I need to do the following:-
I have multiple projects under "~/myprojects"
Think of like >ls ~/myprojects
project1 project2i newproject project_possible....
All my projects have a fixed structure see as below:-
ls ~/myprojects/
src lib inc common test_scripts
(all these are directories having some files in them
For navigating the current()
I want to do something like this in my bashrc file.
assign curr_project = "$1"
alias psrc='cd ~/myprojects/curr_project/src/'
alias plib='cd ~/myprojects/curr_project/lib/'
Thanks in advance
You can use an environment variable to specify the current project and use the variable in your aliases:
current() {
export CURR_PROJECT=$1
}
alias psrc='cd ~/myprojects/$CURR_PROJECT/src/'
alias plib='cd ~/myprojects/$CURR_PROJECT/lib/'
First you set the CURR_PROJECT by using
$ current project1
Then you call your alias to change directories:
$ psrc
Hope that helps.
I use something similar for my work environment - many projects with a common directory structures. I also use a selector to allow me choose projects quickly without typing their name. You may find it useful.
E.g.
current()
{
export PROJECT_ROOT=~/myprojects
# If you pass a project name, use it, otherwise print a list
# for the user to select
if [ -n "$1" ]; then
export CURRENT_PROJECT=$1
else
# Find subdirectories in PROJECT_ROOT
SUBDIRS=`find $PROJECT_ROOT -mindepth 1 -maxdepth 1 -type d -printf "%f "`
if [ -n "$SUBDIRS" ]; then
PS3="Select project: "
select d in $SUBDIRS; do
if [[ -n $d ]]; then
export CURRENT_PROJECT=$d
break
else
echo "Bad choice"
return
fi
done
else
echo "No projects found"
return
fi
fi
# Now we have the CURRENT_PROJECT name, set up the aliases
alias psrc='cd $PROJECT_ROOT/$CURRENT_PROJECT/src/'
alias plib='cd $PROJECT_ROOT/$CURRENT_PROJECT/lib/'
}
Then if you type "current", you will get a choice:
~$ current
1) proj1
2) proj2
3) proj3
Select project:
This is a real time-saver for me - maybe it will be for you too.

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