Compare and Replace inline using Awk/Sed command - awk

Question

I have a file (fixed length) in which searching for consecutive 2 lines starting with number 30 and then comparing value in position 183-187 and if both matching printing the line number. I am able to achieve the desired results up to this stage. But I would like to replace the value present in the line number with empty spaces without tampering the fixed length.
awk '/^30*/a=substr($0,183,187);getline;b=substr($0,183,187); if(a==b) print NR}' file
Explanation of the command above:
line number start with 30*
assign value to a from position between 183 to 187
get next line
assign value to b from position between 183 to 187
compare a & b - if matches it proves the value between 183 to 187 in 2 consecutive lines which starts with 30.
print line number (this is the 2nd match line number)
Above command is working as expected and printing the line number.
Example Record (just for explanation purpose hence not used fixed length)
10 ABC
20 XXX
30 XYZ
30 XYZ
30 XYZ
30 XYZ
40 YYY
10 ABC
20 XXX
30 XYZ
30 XYZ
40 YYY
With above command I could able to get line number 3 and 4 but unable to replace the 4th line output with empty spaces (inline replace) so that fixed width will not get compromised
Expected Output
10 ABC
20 XXX
30 XYZ
30
30
30
40 YYY
10 ABC
20 XXX
30 XYZ
30
40 YYY
Length of all the above lines should be 255 chars - when replace happens it has to be inline without adding it as new spaces.
Any help will be highly appreciated. Thanks.

Answer 1

I would use GNU AWK and treat every character as field, consider following example, let file.txt content be
10 ABC
20 XXX
30 XYZ
30 XYZ
40 YYY
then
awk 'BEGIN{FPAT=".";OFS=""}prev~/^30*/{a=substr(prev,4,3);b=substr($0,4,3);if(a==b){$4=$5=$6=" "}}{print}{prev=$0}' file.txt
output
10 ABC
20 XXX
30 XYZ
30
40 YYY
Explanation: I elected to use storing whole line in variable called prev rather than using getline, thus I do {prev=$0} as last action. I set FPAT to . indicating that any single character should be treated as field and OFS (output field separator) to empty string so no unwanted character will be added during line preparing. If prev (previous line or empty string for first line) starts with 3 I get substring with characters 4,5,6 from previous line (prev) and store it in variable a and get substring with characters 4,5,6 from current line ($0) and store it in variable b, if a and b are equal I change 4th, 5th and 6th character to space each. No matter it was changed or not I print line. Disclaimer: this assume that you want to deal with at most 2 subsequent lines having equal substring. Note /^30*/ does not check if string starts with 30 but rather if it does start with 3 e.g. it will match 312, you should probably use /^30/ instead, I elected to use your pattern unchanged as you imply it does work as intended for your data.
(tested in gawk 4.2.1)

Answer 2

This might work for you (GNU sed):
sed -E '/^30/{N;s/^(.{182}(.{5}).*\n.{182})\2/\1 /}' file
Match on a line beginning 30 and append the following line.
Using pattern matching, if the 5 characters from 183-187 for both lines are the same, replace the second group of 5 characters with 5 spaces.
For multiple adjacent lines use:
sed -E '/^30/{:a;N;s/^(.{182}(.{5}).*\n.{182})\2/\1 /;ta}' file
Or alternative:
sed -E ':a;$!N;s/^(30.{180}(\S{5}).*\n.{182})\2/\1 /;ta;P;D' file

Answer 3

It sounds like this is what you want, using any awk in any shell on every Unix box:
$ awk -v tgt=30 -v beg=11 -v end=13 '
($1==tgt) && seen[$1]++ { $0=substr($0,1,beg-1) sprintf("%*s",end-beg,"") substr($0,end+1) }
1' file
10 ABC
20 XXX
30 XYZ
30
40 YYY
Just change -v beg=11 -v end=13 to beg=183 -v end=187 for your real data.
If you're ever again tempted to use getline make sure to read awk.freeshell.org/AllAboutGetline first as it's usually the wrong approach.