Shell脚本| tr -d'\\ [\\] \\ _'

Question

I am working on a project, and wondering about this command in particular.

tr -d '\[\]\_'

I am aware that the -d flag deletes all characters matching the criteria of the regular expression inside, but I am unsure of what this expression is saying. Is it deleting all special characters such as '!"[]\\ and such? Or is that this command?

tr -cd "[[:alnum:]\n ]"

Thank you in advance!

Answer 1

echo "][][_word[]_" | tr -d '\\[\\]\\_' echo "][][_word[]_" | tr -d '\\[\\]\\_' outputs word

The backslashes are escaping the brackets because they're special characters in regex expressions. So this command is just deleting all [ ] and _ characters from the string.

However, tr -d '[]_' is enough because tr does not use regex.

Answer 2

tr -d '\\[\\]\\_' removes square brackets and underscores from standard input.

Escaping [ , ] and _ is useless here as these characters have no special meaning with tr , they will always be handled as literals.

Note that tr does not use regular expressions .

From man page :

The format of the SET1 and SET2 arguments resembles the format of regular expressions; however, they are not regular expressions, only lists of characters.

[:alnum:] matches alphanumeric characters (equivalent to 0-9A-Za-z ), so nothing to do with special characters.