awk/sed replace multiple newlines in the record except end of record - awk

Question

I have file where:
field delimiter is \x01
the record delimiter is \n
Some lines contain multiple newlines I need to remove them, however I don't want to remove the legitimate newlines at the end of each lines. I have tried this with awk:
awk -F '\x01' 'NF < 87 {getline s; $0 = $0 s} 1' infile > outfile
But this is only working when the line contains one newline in the record (except end of line newline). This does not work for multiple newlines.
Note: the record contains 87 fields.
What am I doing wrong here?
Example of file:
PL^ANov-21^A29-11-2021^A0^A00^A00^A0000000
test^A00000000
Test^A^A^A^A
PL^ANov-21^A29-11-2021^A0^A00^A00^A0000000
test^A00000000
Test^A^A^A^A
SL^ANov-21^A30-11-2021^AB^A0000^A1234567^A00000
test^A12102120^A00000^A00^A^A
NOTE: The file contains 11 fields; field separate \x01; record separator \n
Expected result:
PL^ANov-21^A29-11-2021^A0^A00^A00^A0000000test^A00000000 Test^A^A^A^A
PL^ANov-21^A29-11-2021^A0^A00^A00^A0000000test^A00000000 Test^A^A^A^A
SL^ANov-21^A30-11-2021^AB^A0000^A1234567^A00000test^A12102120^A00000^A00^A^A
Note: I need to preserve the field delimiter (\x01) and record delimiter (\n)
Thank you very much in advance for looking into this.
The file always contains 87 fields;
The fild delimiter is '\x01', but when viewing in Linux it is represented as '^A'
Some lines contain newlines - I need to remove them, but I don't want to remove the legitimate newlines at the end of each line.
The newline appears twice in the 1st and second record and once in third record - this are the newlines I want to remove.
In the examples/expected results there are 11 delimiters "x01" represented as "^A",
I expect to have 3 records and not 6, i.e.:
First record:
test^A00000000 should be joined to the previous line
Test^A^A^A^A should be joined to the first line as well
forming one record:
PL^ANov-21^A29-11-2021^A0^A00^A00^A0000000test^A00000000 Test^A^A^A^A
Second record
test^A00000000 should be joined to the previous line
Test^A^A^A^A should be joined to that previous line as well
forming one record:
PL^ANov-21^A29-11-2021^A0^A00^A00^A0000000test^A00000000 Test^A^A^A^A
Third record:
test^A12102120^A00000^A00^A^A should be joined to the previous line
forming one record:
SL^ANov-21^A30-11-2021^AB^A0000^A1234567^A00000test^A12102120^A00000^A00^A^A
Note:
The example of awk - provided works when there is one unwanted newline in the record but not when there are multiple newlines
Thank you so very much. It works perfectly. Thank you for explaining it so well to me too.

Answer 1

This might work for you (GNU sed):
sed ':a;N;s/\x01/&/87;Ta;s/\n//g' file
Gather up lines until there are 87 separators, remove any newlines and print the result.

Answer 2

What's wrong with your attempt is that you concatenate two lines, print the result and move to the next line. NF is then reset to the next fields count. As all your lines have less than 87 fields the NF < 87 condition is useless, your script would work the same without it.
Try this awk script:
$ awk -F'\x01' -vn=87 -vi=0 '
{printf("%s", $0); i+=NF; if(i==n) {i=0; print "";} else i-=1;}' file
Here, we use the real \x01 field separator and the NF fields count. Variable i counts the number of already printed fields. We first print the current line without the trailing newline (printf("%s", $0)). Then we update our i fields counter. If it is equal to n we reset it and print a newline. Else we decrement it such that we do not count the last field of this line and the first of the next as 2 separate fields.
Demo with n=12 instead of 87 and your own input file (with \x01 field separators):
$ awk -F'\x01' -vn=12 -vi=0 '
{printf("%s", $0); i+=NF; if(i==n) {i=0; print "";} else i-=1;}' file |
sed 's/\x01/|/g'
PL|Nov-21|29-11-2021|0|00|00|0000000test|00000000 Test||||
PL|Nov-21|29-11-2021|0|00|00|0000000test|00000000 Test||||
SL|Nov-21|30-11-2021|B|0000|1234567|00000test|12102120|00000|00||
The sed command shows the result with the \x01 replaced by | for easier viewing.