'How do i replace [] brackets using SED

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

shellreplacesedkshbrackets


Solution 1:[1]

Here is the final code I ended up with

`echo "$string" | sed 's/[^a-zA-Z0-9]/ /g'`

I had to put = and - at the very end.

Solution 2:[2]

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.

Solution 3:[3]

You can also specify the characters you want to keep [with inversion]:

sed 's/[^a-zA-Z0-9]/ /g'

Solution 4:[4]

You can do it manually: 

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

This remove the 32 punctuation character, the order of some characters is important:

  • - should be at the end like this -]
  • [] should be like that [][other characters]
  • ' should be escaped like that '\''
  • not begin with ^ like in [^
  • not begin with [. [= [: and end with .] =] :]
  • not end with $]

here you can have explication of why all that http://pubs.opengroup.org/onlinepubs/9699919799/basedefs/V1_chap09.html#tag_09_03_03

Solution 5:[5]

Can be handled using the regex capture technique too (Eg: here below) :

echo "narrowPeak_SP1[FLAG]" | sed -e 's/\[\([a-zA-Z0-9]*\)\]/_\1/g'
> narrowPeak_SP1_FLAG

\[ : literal match to open square bracket, since [] is a valid regex
\] : literal match to square close bracket
\(...\) : capture group
\1 : represents the capture group within the square brackets

Solution 6:[6]

If you need to replace the brackets totally along with the content then you need to escape it. For example below, replacing the whole brackets along with colon

echo "listen [::]:8080 default_server" sed -i 's|listen \[::\]:8080 default_server|listen       8080|' filename.txt

Sources

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Source: Stack Overflow

Solution Source
Solution 1
Solution 2
Solution 3 perreal
Solution 4
Solution 5
Solution 6 Saad Khan

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