在grep中使用的Linux终端命令行变量

Question

I want to display the amount of words that have exactly 14, 15 and 16 unique letters. I want to use a for loop. (It has to be a one liner.)

This is what I have so far:
for i in {14..16}; do echo "There are $(cat /usr/share/dict/dutch | grep -P '^.{"$i"}$' | grep -vP -c '(.).*\\1') words with exactly $i unique letters"; done

Result:

There are 0 words with exactly 14 unique letters
There are 0 words with exactly 15 unique letters
There are 0 words with exactly 16 unique letters

This means the loop works and when I run it like this:

echo "There are $(cat /usr/share/dict/dutch | grep -P '^.{14}$' | grep -vP -c '(.).*\\1') words with exactly 14 unique letters" && echo "There are $(cat /usr/share/dict/dutch | grep -P '^.{15}$' | grep -vP -c '(.).*\\1') words with exactly 15 unique letters" && echo "There are $(cat /usr/share/dict/dutch | grep -P '^.{16}$' | grep -vP -c '(.).*\\1') words with exactly 16 unique letters"

The results are:
There are 13 words with exactly 14 unique letters
There are 2 words with exactly 15 unique letters
There are 0 words with exactly 16 unique letters

This shows that I am doing something wrong with the variable ($i) inside the grep-command. I don't know how I should do it or solve this problem.

Thanks in advance

Answer 1

It looks like you need to use single quotes instead of double quotes around the variable in your first regex:

for i in {14..16}; do 
  echo "there are $(cat /usr/share/dict/dutch | grep -P '^.{'$i'}$' | grep -vP -c '(.).*\1') words with exactly $i unique letters"; 
done

Because your first regex was wrapped in single quotes, it was being used literally, without any variable expansion. By putting the variable in single quotes, you're actually specifying two literal strings wrapped tightly around your actual variable (put another way, your variable isn't actually wrapped in any quotes at all).

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Edit: This is what I get on my system:

$ for i in {14..16}; do echo "there are $(cat /usr/share/dict/dutch | grep -P '^.{'$i'}$' | grep -vP -c '(.).*\1') words with exactly $i unique letters"; done
there are 13 words with exactly 14 unique letters
there are 2 words with exactly 15 unique letters
there are 0 words with exactly 16 unique letters

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Edit 2: This might demonstrate the issue more clearly:

$ i=12
$ echo '^.{"$i"}$'
^.{"$i"}$
$ echo '^.{'$i'}$'
^.{12}$