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

Solution 1:

This is easy, if you follow the manual carefully: all members inside a character class lose special meaning (with a few exceptions). And ] loses its special meaning if it is placed first in the list. Try:

$ echo '[123]' | sed 's/[][]//g'
123
$

This says:

1. inside the outer [ brackets ], replace any of the included characters, namely:
   • ] and
   • [
2. replace any of them by the empty string — hence the empty replacement string //,
3. replace them everywhere (globally) — hence the final g.

Again, ] must be first in the class whenever it is included.

Solution 2:

I'm not sure why that doesn't work but this does:

echo '[123]' | sed 's/\(\[\|\]\)//g'

or this:

echo '[123]' | sed -r 's/(\[|\])//g'

You can also try a different approach and match the string inside the brackets (assuming the string can be matched easily and is not defined by the brackets):

echo '[123]' | egrep -o "[0-9]+"

I'm having the same troubles with your original regex using grep so I suspect this is not just a sed thing.

Weirdly, these produce different results but one of them matches what you want:

echo '[123]' | egrep -o '[^][]+'
123

echo '[123]' | egrep -o '[^[]]+'
3]

Applying this to your original sed (and adding the /g modifier so it removes both brackets):

echo '[123]' | sed 's/[][]//g'
123

---

Solution 3:

To remove everything before and after the brackets :

$ echo '[123]' | sed 's/.*\[//;s/\].*//;'
123

If your data is like this always meaning starting and ending with square brackets:

$ echo '[123]' | sed 's/.//;s/.$//;'
123