sed remove special characters and spaces in line

Question

Input

[*] 111, 22, 33, 44

Output

111,22,33,44

Using sed command sed 's/^\[\*\][[:space:]]*//' file were able to display the output as 111, 22, 33, 44 which means space still there.

So what is the correct sed command?

Answer 1

The short answer is that this should do what you want:

sed 's/[][*]\|[[:space:]]//g'

Test it out like so:

echo '[*] 111, 22, 33, 44' | sed 's/[][*]\|[[:space:]]//g'

Output:

111,22,33,44

A longer explanation follows.

---

The key ingredient that your expression was missing was the use of the g command to perform global substitution. Without this command only the first match on each line will be replaced.

The most counter-intuitive bit (in my opinion) of my solution is how to include the square brackets inside a character class. For that, we refer to the sed manual:

‘]’
    ends the bracket expression if it’s not the first list item.
    So, if you want to make the ‘]’ character a list item, you must put it first.

For further discussion on that topic, see the following post on SeverFault:

• Using sed to remove both an opening and closing square bracket around a string

Alternatively, you could have taken your expression and piped it into another sed command, e.g.:

echo '[*] 111, 22, 33, 44' \
| sed 's/^\[\*\]//' \
| sed 's/[[:space:]]//g'