shell - How do i replace [] brackets using SED

Question

I have a string that i am want to remove punctuation from.

I started with

sed 's/[[:punct:]]/ /g'

But i had problems on HP-UX not liking that all the time, and some times i would get a 0 and anything after a $ in my string would dissappear. So i decided to try to do it manually.

I have the following code which works on all my punctuation that I am interested in, except I cannot seem to add square brackets "[]" to my sed with anything else, otherwise it does not replace anything, and i dont get an error, so I am not sure what to fix.

Anyways this is what i currently have and would like to add [] to.

sed 's/[-=+|~!@#$%^&*(){}:;''''"''`''.''/''\']/ /g'

BTW I am using KSH on Solaris, Redhat & HP

Answer 1

You need to place the brackets early in the expression:

sed 's/[][=+...-]/ /g'

By placing the ']' as the first character immediately after the opening bracket, it is interpreted as a member of the character set rather than a closing bracket. Placing a '[' anywhere inside the brackets makes it a member of the set.

For this particular character set, you also need to deal with - specially, since you are not trying to build a range of characters between [ and =. So put the - at the end of the class.