编译没有特定子字符串的文件

Question

I'm new to bash, I was wondering how I would go about compiling files in a certain directory that don't contain certain strings in name. So say I have gumbo.java jackson.java roosevelt.java roar.java as my files in directory. How do I compile files that don't contain roar or gumbo in the name?

Answer 1

In bash, you can use an extended glob pattern like this:

shopt -s extglob nullglob
printf '%s\n' !(roar|gumbo).java

This matches anything other than roar or gumbo , that ends in .java . You can think of it as *.java , minus roar.java and gumbo.java .

Substitute the printf for whatever you're using to compile, if you're happy with the output.

If you want to negate anything that contains those substrings, you can add some asterisks:

printf '%s\n' !(*@(roar|gumbo)*).java

The @() matches any one of the pipe-separated options. As before, the whole thing is negated with !() .

As suggested in the comments, I also enabled nullglob so that an empty match expands to nothing.

Answer 2

Try using the find command:

find . -type f -not -name '*roar*' -and -not -name '*gumbo*' -exec <command> {} \;

For example:

find . -type f -not -name '*roar*' -and -not -name '*gumbo*' -exec javac {} \;

Answer 3

I would use a standard loop:

for file in *.java; do
  case $file in
  *roar*|*gumbo*) continue;;
  *) javac "$file";;
  esac
done