How to exclude a file from a command with ZSH?

Question

Given this directory content :

one.file
two.file
three.file

in bash, when I enter

rm *.file !(two)

only one.file and three.file are deleted. How can I do this in ZSH?

Answer 1

I was interested in the answer too, and a quick search turned up this article on globbing in zsh. The highlights:

• ^ Acts as a negation. For example, ls ^two.file will only list the one.file and three.file.
• You can combine ^ and *. For example, ls ^two* will list anything that doesn't start with "two"
• You can use parentheses to make more complex matches. For example, ls (^two).file will list anything that doesn't start with "two", and does end in "file".

Answer 2

If you want to use the ksh syntax ls !(two).file, simply turn on the KSH_GLOB option in zsh:

$ setopt KSH_GLOB
$ ls -1 !(two).file
one.file
three.file

But zsh provides other powerful globbing techniques, activated by the EXTENDED_GLOB option. For a complete list, please read the section FILENAME GENERATION in man zshexpn. Most relevant to the question are these operators:

• ^x matches anything except the pattern x, so in your case ls -1 ^two.file

• x~y is more powerful as it matches anything that matches the pattern x but does not match y, so ls -1 *~two.file. The special thing is, that you have can use another globbing pattern for x, e.g.

  $ ls -1 *.file~two*
  one.file
  three.file
  

  This is not possible with the ^ operator, which is in this case equivalent to *~:

  $ ls -1 *.file^two*
  one.file
  three.file
  two.file