[SOLVED] Linux bash: How do I replace a string on a line based on a pattern on another/different line?

Issue

I have a file that contains the following data:

GS*PO*112233*445566*20211006*155007*2010408*X*004010~

ST*850*0001~
BEG*00*DS*A-112233**20211005~
REF*K6*Drop Ship Order~
REF*ZZ*SO168219~
REF*DC*ABC~

ST*850*0002~
BEG*00*DS*A-44556**20211005~
REF*K6*Drop Ship Order~
REF*ZZ*PO54361~

ST*850*0003~
BEG*00*DS*A-12345**20211005~
REF*K6*Drop Ship Order~
REF*DC*XYZ~
REF*ZZ*SO897654~

For clarity, I have inserted blank line above each ST*850 line. Here is what I want to do:

1. Search for the pattern REF*ZZ*SO
2. If found, then replace the preceding ST*850 line with ST*850C

So the resultant file would look like this:

GS*PO*112233*445566*20211006*155007*2010408*X*004010~

ST*850C*0001~
BEG*00*DS*A-112233**20211005~
REF*K6*Drop Ship Order~
REF*ZZ*SO168219~
REF*DC*ABC~

ST*850*0002~
BEG*00*DS*A-44556**20211005~
REF*K6*Drop Ship Order~
REF*ZZ*PO54361~

ST*850C*0003~
BEG*00*DS*A-12345**20211005~
REF*K6*Drop Ship Order~
REF*DC*XYZ~
REF*ZZ*SO897654~

Here is what I have tried:

sed -i -n '/^REF\*ZZ\*SO/!{x;s/ST\*850\*/ST\*850C\*/;x};x;1!p;${x;p}' file

This replaces all the three ST*850 lines with ST*850C and not just the 1st and the 3rd. What am I doing wrong?

Solution

How about a perl solution although perl is not included in the tags.

perl -0777 -aF'(?=ST\*850)' -ne '
    print map {/REF\*ZZ\*SO/ && s/ST\*850/$&C/; $_} @F;
' file

Output:

GS*PO*112233*445566*20211006*155007*2010408*X*004010~

ST*850C*0001~
BEG*00*DS*A-112233**20211005~
REF*K6*Drop Ship Order~
REF*ZZ*SO168219~
REF*DC*ABC~

ST*850*0002~
BEG*00*DS*A-44556**20211005~
REF*K6*Drop Ship Order~
REF*ZZ*PO54361~

ST*850C*0003~
BEG*00*DS*A-12345**20211005~
REF*K6*Drop Ship Order~
REF*DC*XYZ~
REF*ZZ*SO897654~

• The -0777 option tells perl to slurp whole file at once.
• The -a option enables the auto split mode then the split fragments are stored in the array @F.
• The -F option specifies the pattern to split the input.
• The regex (?=ST\*850) is a positive lookbehind which matches at the beginning of a string ST*850.
• The -ne option is mostly equivalent to that of sed.
• The map {..} @F function converts all elements of @F according to the statement within the curly brackets.
• The statement /REF\*ZZ\*SO/ && s/ST\*850/$&C/ is translated as: "if the element of @F matches the pattern /REF*ZZ*SO/, then perform the substitution s/ST*850/$&C/ for the element."
• The final $_ is the perl's default variable similar to the pattern space of sed and will be the return values of the map function.

Answered By - tshiono