[SOLVED] sed command to locate a string containing alphanumeric characters and spaces and perform positional replacement in that string

Issue

i have a file with below line

aadz(mdb awetyi),Lll,Dweuio: mdb 1:00.0  ARF = 0x000000006e1d9000

i want to locate ARF = 0x000000006e1d9000 and perform positional replacement of certain positions like 12,13,14 etc with x. the final result should look like

aadz(mdb awetyi),Lll,Dweuio: mdb 1:00.0  ARF = 0x000000006xxx000

i tried this command

sed -i -e 's/\(.\{12\}\)./\1x/' -e  's/\(.\{13\}\)./\1x/'  -e  's/\(.\{74\}\)./\1x/' file

but it replaces the characters from start of the line containing the text which is not desired I'm unable to locate only ARF = 0x000000006xxx000

Solution

You can do it with the following command

sed -E 's/(ARF = .{11}).{3}/\1xxx/g' file

It matches ARF = , 11 following symbols and replaces following three symbols with xxx

Option -i is not included in example: first test to see that replacement is happening like you want.

Answered By - markalex

Answer Checked By - Katrina (WPSolving Volunteer)