Simply bash script

Shell Programming and Scripting
1

I have a bash script that has 13 variables and runs commands on them.
I've copied the commands 13 times but I'd like to simplify it to simply add 1 to the previous and run 13 times.

The variables are
dir01=/path/to/dir01
dir01=/path/to/dir02
... up to 13. (more may be added later)

...
# executable stuff
mkdir "$dir01/".LogFiles
cp -R "$dir01"/*.aep\ Logs "$dir01"/.LogFiles/
rm -Rf "$dir01"/*.aep\ Logs "$dir01"/*.log 
 echo 01

mkdir "$dir02/".LogFiles 
cp -R "$dir02"/*.aep\ Logs "$dir02"/.LogFiles/
rm -Rf "$dir02"/*.aep\ Logs "$dir02"/*.log 
 echo 02

# repeated 13 times for each dir.

I'd like to simplify it to something like this:

n=1 # Amount to add each time
num=00 # Base number to start with


while [ x = 0 ]
do
    workon=$dir(($num += $n))
    mkdir "$workon/".LogFiles 
    cp -R "$workon"/*.aep\ Logs "$workon"/.LogFiles/
    rm -Rf "$workon"/*.aep\ Logs "$workon"/*.log 
    echo "$workon"
    x=1
done

I don't know how to get it to stop when it's done 13 or whatever number of variables I have.

BTW I'm using cp and rm because mv won't clobber a directory.
These are render log files that are annoying and mostly useless so I don't mind overwriting files.

Any help with this code is very much appreciated.

2

You are already using bash's shell arithmetics, why don't you use it for your loop:

$ while (( x-- )); do echo $x; done
12
11
10
9
8
7
6
5
4
3
2
1
0

BTW, don't cp and rm without error checking - if th cp fails, your log file is gone; use mv instead.

3

I defined paths for dir1 through dir13 as variables but I can't get the loop to add the number to the "dir" to create "dir1" and then use it as the defined variable.
This is my test code:

######################################
# dir1 through dir13 defined previously
n=1
num=1


run=13 # How many directories are we working on?
(( run += n )) # Add one to the number of directories

################################
while (( $num < $run ))
do
      echo $dir$num
    (( num += n )) # When num = run exit loop
    
done

Instead of echoing the defined path for variable "dir1, dir2 ..." it echos "dir1" and "dir2," literally.
If I echo "$dir1" by adding it manually, I get the defined path as desired.
Why won't $dir$num echo as the path?

4

Variables don't work that way. Dynamic variable names are discouraged in general -- that's the sort of thing you should be using an array or even just a space-separated string for.

You can dereference a variable name with ${!...} however. (This is a BASH-only feature.)

VAR01="asdf"
A=0
B=1
VARNAME="VAR$A$B"

echo "${!VARNAME}"

You must assign the complete variable name to a string first, you can't put substitutions directly inside the ${}.

5

Thank you. That works well.

How can I set a variable with a space in it?
"dir 1" and "dir\ 1" don't work.
I'm interested if having spaces in the variable definition would also work.

6

You can't. That's not a valid variable name.

7

They designed it like that because it's a very bad idea to want spaces in the middle of a variable name. Use the underscore.