Question

Here is the line:

find /localdir/ | grep '[0-9']$ | xargs -i% cp % /tftpboot

I specifically want to know what grep is looking for exactly here.
Can anyone translate it for me please ?

I am also kind of interested in what the xargs cmd is going to look like…

Thanks in advance.

Answer 1

[0-9] means any character from 0 through 9. $ means the end of a line. So your grep will find any line (i.e., filename) that ends with a digit, and xargs will copy them each to /tftpboot.

Of course, you’ll have some surprises if any of those filenames contain spaces. You can do this entirely within the shell in zsh (and I think in recent versions of bash):

cp /localdir/**/*[0-9] /tftpboot

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addendum: If you’re asking about the funny quoting, that will work, though it’s not very human-friendly.

The key is that you can have quoted strings and non-quoted strings right next to each other in shell, and they’ll become a single string; echo "fo"ob'ar' will produce foobar.

The first part is quoted because [ is special to bash. ] is also special, but since bash never saw a special (unquoted) [, it leaves the ] alone. $ would normally substitute a variable, but nothing comes after it, so bash leaves that alone too.

The string that actually gets passed to grep is still [0-9]$.