匹配 grep 正则表达式中的 X 或 Y

Question

I'm trying to run a fairly simple regular expression to clear out some home directories. For background: I'm trying to ask users on my system to clear out their unnecessary files to clear up space on their home directories, so I want to inform users with scripts such as Anaconda / Miniconda installation scripts that they can clear that out.

To generate a list of users who might need such an email, I'm trying to run a simple regular expression to list all homedirs that contain such an installation script. So my assumption would be that the follwing should suffice:

for d in $(ls -d /home/); do
    if $(ls $d | grep -q "(Ana|Mini)conda[23].*\.sh"); then
        echo $d;
    fi;
done;

But after running this, it resulted in nothing at all, sadly. After a while looking, I noticed that grep does not interpret regular expressions as I would expect it to. The following:

echo "Lorem ipsum dolor sit amet" | grep "(Lorem|Ipsum) ipsum"

results in no matches at all. Which would then explain why the above forloop wouldn't work either.

My question then is: is it possible to match the specified regular expression (Ana|Mini)conda[23].*\.sh , in the same way it matches strings in https://regex101.com/r/yxN61p/1 ? Or is there some other way to find all users who have such a file in their homedir using a simple for-loop in bash?

Answer 1

Short answer: grep defaults to Basic Regular Expressions (BRE), but unescaped () and | are part of Extended Regular Expressions (ERE). GNU grep, as an extension, supports alternation (which isn't technically part of BRE), but you have to escape \ :

grep -q "\(Ana\|Mini\)conda[23].*\.sh"

Or you can indicate that you want to use ERE:

grep -Eq "(Ana|Mini)conda[23].*\.sh"

Longer answer: this all being said, you don't need grep, and parsing the output of ls comes with a lot of pitfalls . Instead, you can use globs:

printf '%s\n' /home/*/*{Ana,Mini}conda[23]*.sh

should do it, if I understand the intention correctly.

This uses the fact that printf just repeats its formatting string if supplied with more parameters than formatting directives, printing each file on a separate line.

/home/*/*{Ana,Mini}conda[23]*.sh uses brace expansion , ie, it first expands to

/home/*/*Anaconda[23]*.sh /home/*/*Miniconda[23]*.sh

and each of those is then expanded with filename expansion . [23] works the same way as in a regular expression; * is "zero or more of any character except / ".

If you don't know how deep in the directory tree the files you're looking for are, you could use globstar and ** :

shopt -s globstar
printf '%s\n' /home/**/*{Ana,Mini}conda[23]*.sh

** matches all files and zero or more subdirectories.

Finally, if you want to handle the case where nothing matches, you could set either shopt -s nullglob (expand to nothing if nothing matches) or shopt -s failglob (error if nothing matches).

Shell patterns are described here .

Answer 2

You don't need ls or grep at all for this:

shopt -s extglob

for f in /home/*/@(Ana|Mini)conda[23].*.sh; do
  echo "$f"
done

With extglob enabled, @(Ana|Mini) matches either Ana or Mini .