Can i use regex in the Record Separator in awk in linux

Question

I have the test file like this

fdsf fdsf fdsfds fdsf
fdsfdsfsdf fdsfsf
fsdfsdf var12=1343243432

fdsf fdsf fdsfds fdsf
fdsfsdfdsfsdf
fsdfsdf var12=13432434432

fdsf fdsf fdsfds fdsf
fsdfsdf fdsfsf var12=13443432432

Now i want to use var12=\d+ as the record separator. Is this possible in awk

Answer 1

Yes, however you should use [0-9] instead of \d:

awk '1' RS="var12=[0-9]+" file

IIRC, only GNU awk can use multi-character record separators.

Results:

fdsf fdsf fdsfds fdsf
fdsfdsfsdf fdsfsf
fsdfsdf 


fdsf fdsf fdsfds fdsf
fdsfsdfdsfsdf
fsdfsdf 


fdsf fdsf fdsfds fdsf
fsdfsdf fdsfsf 

Please post your desired output if you need further assistance.

Answer 2

Assuming GNU awk (a.k.a. gawk) on Linux, yes.

RS

This is awk's input record separator. Its default value is a string containing a single newline character, which means that an input record consists of a single line of text. It can also be the null string, in which case records are separated by runs of blank lines. If it is a regexp, records are separated by matches of the regexp in the input text.

Source: 7.5.1 Built-in Variables That Control awk, The GNU Awk User's Guide.

As @steve says, \d is not in the list of Regular Expression Operators or gawk-Specific Regexp Operators, so you need to use a bracket expression such as [0-9] or [[:digit:]] in place of your \d.

However, it's not clear from your question as to what your intention here is. I've answered your question but I doubt I've solved your underlying problem. See also What is the XY problem?