[SOLVED] How to capture one digit with sed and replace the other?

Issue

In a text file test.txt are many lines of text, of which I want to extract a single line matching:

blabla 28.40.00 blabla

I would like to replace the first digit of the middle number (in this case 4) by three. That is, no matter what the middle number is (40, 41, 52, 63 etc), I would like it to be replaced by a number starting with 3 (40 becomes 30, 41 becomes 31, 52 becomes 32, 63 becomes 33 etc).

The following line matches the middle number and replaces it with the alphabet a:

cat test.txt |awk '/blabla/'|sed -E s_[[:digit:]][[:digit:]]_3_2

output: blabla 28.3.00 blabla

But when I want to replace only the first digit, sed doesn't work:

cat test.txt |awk '/blabla/'|sed -E s_[[:digit:]]\([[:digit:]]\)1_3\1_2

output: blabla 28.40.00 blabla

What am I doing wrong?

Solution

To always replace the third digit in the line:

$ echo 'blabla 28.40.00 blabla' | sed 's/[0-9]/3/3'
blabla 28.30.00 blabla

If the number before the . can have more than 2 digits, here's one workaround:

$ echo 'blabla 528.40.00 blabla' | sed 's/\.[0-9]/.3/'
blabla 528.30.00 blabla

---

In your second attempt, you should quote the command and use unescaped parenthesis. You also seem to have an extra 1 after the capture group.

$ echo 'blabla 28.40.00 blabla' | sed -E 's_[[:digit:]]([[:digit:]])_3\1_2'
blabla 28.30.00 blabla

Also, you don't need awk in this case. You can add filter in sed as well:

$ echo 'blabla 28.40.00 blabla' | sed -E '/blabla/ s_[[:digit:]]([[:digit:]])_3\1_2'
blabla 28.30.00 blabla

Answered By - Sundeep

Answer Checked By - Terry (WPSolving Volunteer)