need an explanation on this script...

Shell Programming and Scripting
1

The following script will create a directory in a directory and will go on as many times as the number you will give in.

I am trying to find out how it works ... can someone please help me with that?

#!/bin/sh
#create a variable and set it to 1
n=1                     
#start a loop as long as $n is less or equal then 1 just keep on running      
while [ $n -le $1 ] 
#start the loop
do
  #set a variable so that the directory can be variable as well 
  dir=map_$n
  #create the directory "map_$n"
  #if the directory is created "cd" into it
  #otherwise exit the script
  mkdir "$dir" && cd "$dir" || exit 1
  #do a +1 on $n
  n=$(( $n + 1 ))
done

I have commented out what I understand ...
But here is the thing.
How does the script know when it needs to go to exit?

2

You misunderstood this line. It runs what's between do and done, as long as $n is lower or equal than the first argument (i.e. $1) of the script.

1 Like
3

So lets say I execure the script with the number 3
./script.sh 3
it will see that $n = 3
So when this is the case it will run what is between "do" and "done"
so it creates a directory "map_1" because $n=1
and then it moves into the directory and it is doing a +1 to $n and it creates another directory ...

But how does the script knows how to stop at 3?

4

No. $1 will equal 3. $1 is the first argument used when you invoke the command. $2 is the second. $3 is the third. And so on.

$n is a counter set to 1 which is incremented until its value is no longer less than or equal to $1 (the first argument to the script, which in your example is 3).

Regards,
Alister