CODEWITHSUNDEEP
bashsed
Kevin H.
Kevin H.

Reputation: 11

sed - Replacing brackets with characters?

I have a string that with brackets that enclose a single character, like such:

[a]

I want to take the character within the bracket and replace the bracket with the character, so the end result would look like:

aaa

This is what I came up with, but it doesn't work:

sed 's/\[ \([a-z]\) \]/\2/g' < testfile

Can someone please help me, and explain why my command isn't working?

Upvotes: 1

Views: 1833

Answers (3)

Ren
Ren

Reputation: 2946

Try the following code:

$ echo "[a]" | sed 's/\[\([a-zA-Z]\)\]/\1\1\1/g'

or

$ echo "[a]" | sed -r 's/\[([a-zA-Z])\]/\1\1\1/g'

Output:

aaa

Upvotes: 1

Quinn
Quinn

Reputation: 4504

The issue with your patern sed 's/\[ \([a-z]\) \]/\2/g' < testfile:

1) The pattern has only one group \([a-z]\), so \2 is invalid;
2) The pattern contains space, there is no match found;
3) To replace brackets, you need to capture them in a group.

My idea is, to catch all groups in a pattern, and replace them with \2\2\2:

echo "[a]" | sed 's/\(\[\)\([a-z]\)\(\]\)/\2\2\2/g'

Or

echo "[a]" | sed 's/\(.\)\(.\)\(.\)/\2\2\2/g'

The output is:

aaa

Upvotes: 0

karakfa
karakfa

Reputation: 67567

I think you missed some basic concepts. First let's duplicate a single char

$ echo a | sed -r 's/(.)/\1\1/' 
aa

parenthesis indicates the groups and \1 refers to the first group

Now, to match a char in square brackets and triple it.

$ echo [a]b | sed -r 's/\[(.)\]/\1\1\1/'
aaab

you need to escape square bracket chars since they have special meaning in regex. The key is you have to bracket in parenthesis the regex you're interested in and refer to them in the same order with \{number} notation.

Upvotes: 0

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