CODEWITHSUNDEEP
bashquotesmkdir
shs
shs

Reputation: 23

basic question: using a variable content as directory name in Linux

I want to make a new directory with the name of every zip file currently exists in a specific directory, in another one. I have written a 'for loop' for this purpose:

files=NL*.zip
for a in $files; do 
b=echo $a; 
sudo mkdir '/home/shokufeh/Desktop/normals/T1/"${b}"';
done

but it creates a directory named ${b} for the first round and make errors for next ones. Could you please tell me what I should do?

Upvotes: 2

Views: 58

Answers (1)

Adhoc
Adhoc

Reputation: 121

You put your variable in simple quotes, there fore '${b}' will never be interpreted. Try this:

for a in NL*.zip; do 
  sudo mkdir "/home/shokufeh/Desktop/normals/T1/${a}";
done

No need for variables $files and $b.

To summarize, let's say var=3,

  • echo "$var" will display 3
  • echo '$var' will display $var
  • echo "'$var'" will display '3'
  • echo '"$var"' will display "$var"

Hopefully this makes sense. Quotes function like parentheses and brackets.

Upvotes: 3

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