sed：在特定位置替换空格

Question

I'd like to replace a space character in position 4 of "abc " , other characters at that position should be left untouched. Matching only for a space doesn't seem to work:

$ echo "abc " | sed "s/\ /d/4"
abc

Matching with . seems to do it, but it's matching too broad for my case:

$ echo "abc " | sed "s/./d/4"
abcd

Am I missing a detail about sed , or its regex flavor?

Answer 1

You could use range quantifier \\{min,max\\} or \\{num\\} .

$ echo "abc " | sed 's/^\(.\{3\}\) /\1d/'
abcd

^ Asserts that we are at the start. \\(..\\) called capturing groups. .\\{3\\} matches exactly three character, those three characters are captured by the group index 1. Later we could refer those captured characters by back-referencing.

Answer 2

You're trying to replace the 4th instance of space, which in this case it isn't. echo "abc " | sed "s/ /d/"

If there were at least 4 spaces in your text, it would work.

Edit: If you need to replace the 4th character you can do this: echo "abc " | sed "s/./d/4" echo "abc " | sed "s/./d/4"

This means "replace the 4th instance of any character with d".

Answer 3

Another way of doing it with perl :

echo "abc " | perl -pe 's/...\K /d/'
abcd

Ps. \\K is positive look-behind.

Answer 4

If you want to replace it then.

$ echo "abc " |sed -r 's/ /d/1'

The above command will search for the 1st occurrence of space character and replace it with d , for your case you can replace it with any thing you want.