Perl extract group with lookbehind from different line - regex

Question

I've tried web search and have read several answers on stackexchange, still cannot grasp why command does not extract anything. At the end I want to extract group with lookbehind from different line, e.g. from
Code>TEST1<Code Code2>best<Code2
Code>test2<Code
Type>false<Type
by finding needed key between Type and extracting first Code above the finding, so it case above to get test2. But I cannot succeed to extract even something from multiple lines, i.e.
perl -lne 'print $1,"_",$2 if /Code>(.*)<Code[\s\S\n]*?Type>(.*)<Type/'<test.txt prints nothing.
I've played with removing ln parameters and adding/removing greedy ? and trying just . in place of [\s\S\n].
perl -lne 'print $1,"_",$2 if /Code>(.*)<Code[\s\S\n]*?Code2>(.*)<Code2/'<test.txt
gives TEST1_best so same line extraction works.
What am I missing? Can what I want be done in one line of command?

Answer 1

The following command answers your question: it collects all values contained in a Code>...<Code pattern, if they are followed by a Type>...<Type pattern (with potentially other patterns in between, but no other occurrences of Code>...<Code in between):
perl -lne 's/^.*?(?=Code>)//s; for (split /Code>/) { print qq($1:$2\n) if /(.*?)<Code.*?Type>(.*?)<Type/s }' -0777 <test.txt
If e.g. test.txt contains the following lines,
Code>test4<Code Type>false<Type
Code>test3<Code
Type>true<Type
Code>TEST1<Code Code2>best<Code2
Code>test2<Code
Type>false<Type
then the command will collect the following value pairs:
test4:false
test3:true
test2:false
Edited on 04/08/2019, 17:38 CEST I edited the command to remove the "header part" of the file (the part before the first occurrence of Code>), as it might - by some error of the file's editor - contain a closing tag <Code which had not been opened with Code> but instead with a typo like e.g. Cde>. My assumption was that the complete file was "syntactically correct" in the sense that it consists of elements of type /(\w+)>.*?<\1/, separated by whitespace (including newlines). For files which do not conform to this syntax, the statement was not waterproof.

Answer 2

Another way to do it, using progressive matching and embedded code
perl -lne 'while (/\b(?:Code>(.*?)<Code(?{$c=$1})|Type>(.*?)<Type(?{print qq($c:$2\n) if defined $c;undef $c}))\b/g){}' -0777 <test.txt
Explanations:
Basically, the expression finds occurrences of Code>(.*?)<Code or Type>(.*)<Type. This gives the basic form of an alternation in an unnamed grouping expression: (?:Code>(.*?)<Code|Type>(.*?)<Type).
The word boundary assertions \b around the group ensure that the keywords Codeand Type are matched, but not e.g. Code2 or TType.
The modifier g ensures progressive application of the regular expression on the string. Since I want to extract the result inside of the expression itself, I place the regex in an empty loop, i.e. while (/.../g) {}.
You suppose a grammar rule Code ⟶ Type, i.e. you look for occurrences of a Type token following a Code token. For this, a Code token is memorized in a variable $c with the code expression (?{$c=$1}). If a Type token is found, it is considered a match only if formerly a Code token has been found, indicated by the fact that the variable $c is defined. In any case, if a Type token has been found, the variable $c will be undefd to prepare it for the next search. This gives the code evaluation (${print qq($c:$2\n) if defined $c;undef $c;}) in the Type branch of the regular expression.
Note that the captures of the Code>(.*?)<Code and Type>(.*?)<Type tokens may be the empty string. This is why I am working with undef $c and if defined $c instead of the simpler $c='' and if $c.

Answer 3

if your data in 'd', by gnu sed;
sed -Ez 's/.*Code>(\w+)<Code\sType>\w*<Type.*/\1/' d
Perl
perl -ne 'BEGIN{undef $/} /Code>(\w+)<Code\nType>\w*<Type/; print $1' d