Trying to replace a pattern with another one - linux

Question

this is my first question on this website.(glad i found out about this community)
I am trying to replace a specific pattern in a file(multiple lines) that looks somehow like this:
Bla bla bla bla |SMTH AWESOME INSIDE >>> LOL| bla bla bla | let's do it again >>> AWESOME |
Into a format that looks like this
Bla bla bla bla ( LOL | SMTH AWESOME INSIDE ) bla bla bla ( AWESOME | let's do it again )
I tried doing this by using a code that parses the line word by word and if it finds out the "|" character starts creating a string that contains the first word,then, after it finds the >>> character it starts creating the second string till it finds the "|" last character, but it didn't work.
I also tried afterwards using AWK(but since i am new to linux i failed as well.
awk -F 'BEGIN { FS=OFS="|" } { sub(/.*<<</,"", $2); }1' $1 }'
and then parse the output with sed(removing the ) and ( characters from both strings. But it didn't work.
Thank you for reading.

Answer 1

It looks like this is just a simple substitution within each line so all you need is sed:
$ sed 's/| *\([^|]*\) >>> \([^|]*\) *|/( \2 | \1 )/g' file
Bla bla bla bla ( LOL | SMTH AWESOME INSIDE ) bla bla bla ( AWESOME | let's do it again )
You can do the same in GNU awk with gensub() or other awks with match() and substr().

Answer 2

With extended regexp in sed:
sed -r 's/\|([^|]+)[[:space:]]*>>>[[:space:]]*([^|]+)\|/( \2 | \1 )/g' File
Logic:
We look for a pattern which starts with | followed by a sequence of non-| characters followed by >>> followed by a sequence of non-| characters again. See the groupings done with ( and ). Then we substitute these patterns according to our need. ( \2 | \1 ) is the replacement pattern where \1 and \2 are the first and second groupings respectively.
With basic regexp in sed:
sed 's/|\([^|]*\)[[:space:]]*>>>[[:space:]]*\([^|]*\)|/( \2 | \1 )/g' File

Answer 3

Perl's regular expressions have a "non-greedy" matching feature that awk's do not:
perl -pe '
s/ \| # the first delimiter
(.*?) # capture up to ...
>>> # the middle delimiter
(.*?) # capture up to ...
\| # the last delimiter
/($2 | $1)/gx
' file
Bla bla bla bla ( LOL | SMTH AWESOME INSIDE ) bla bla bla ( AWESOME | let's do it again )

Answer 4

Let's try with awk:
awk 'NR%2{ printf("%s", $0) } NR%2==0{ printf("( %s %s",$NF,RS); gsub(/>>>.*$/,")"); printf("%s",$0) }' RS='|' file
Bla bla bla bla ( LOL | SMTH AWESOME INSIDE ) bla bla bla ( AWESOME | let's do it again )
The RS defines | as record separator. So when the input record number (NR) isn't module of 2 (NR%2 return 1) then print that record itself. If the NR is module of 2 (NR%2==0 means if record is module of 2), then print a single open parentheses followed by printing last field from it and print record separator (printf("( %s %s",$NF,RS)), then replace >>>.*$ with close parentheses and print the rest of record (gsub(/>>>.*$/,")"); printf("%s",$0))