Search for Double Quotes (") in the file and copy the whole line in different file - bash

Question

I have a requirement to read through all the files and look for <double quotes> (") and copy the whole line to a different file. The challenge is here that to identify the whole line when there is a new character in the line.
The file format is like this - values are separated with delimiter |*| and end with |##|.
In the attached (image), the highlighted in green should go to new file, Logic would be check for " and if it finds read line starting from (line after |##| to until next |##| )
10338|*|BVL-O-G-01020-R4|*||*|BVL|*||*|Y|*|Y|*||*|CFC6E82284990A7AE040800AA5644B19|*|jmorlan|*|2011.12.21 15:52:01|##|
10358|*|BI-MED-CDMA-MCS-90-118-EXAM|*|Exam for 001-MCS-90-118:
Planning, Conducting and Reporting Post Marketing Surveillance "Studies and Safety Reporting from Non Trial Activities |*|GLOBAL_MEDICAL|*||*|Y|*|N|*||*|CFC6E822849A0A7AE040800AA5644B19|*|finke|*|2012.04.30 04:23:27|##|
10342|*|BVL-O-4-01020-R7|*||*|DVL|*||*|Y|*|Y|*||*|RRFC6E82284990A7AE040800AA5644B19|*|sppa|*|2011.12.21 15:52:01|##|

Answer 1

Assuming you mean that the sections between |##| should be considered as newline, next question is does you file contain any real newlines? If not, grep is probably not going to be very efficient as it works on a line-by-line basis. If any real newlines are supposed to be considered part of the text, then definitely, grep is going to be unhappy.
If you really want to do it in 1 go in grep:
grep -Eoz '(^|\|##\|)([^|]|\|[^#]|\|#[^#]|\|##[^|])"([^|]|\|[^#]|\|#[^#]|\|##[^|])(\|##\||$)'
This is looking for any sequence that starts with |##| (or is the start of the file) is followed by some characters, a quote, and some more characters, then ends with |##| (or end of file). By using -z grep will ignore any newlines in the file.
The complex "any characters" ([^|]|\|[^#]|\|#[^#]|\|##[^|])* expression is because grep is greedy. It basically looks for repeating sequences that are not |##|. Perhaps turning off greed is good, but that will depend on the power of the regexp engine in your version of grep.
But much easier, and probably faster, to use sed to break up the records and inject "NULL" line-breaks:
sed 's/\|##\|/\x00/g' | grep -z '"'
This is simply replacing your end of line pattern |##| with the null character, then asking grep to find quote while treating null character as end of line.

Answer 2

This answer provides two solutions a Gnu Awk solution and a POSIX version.
POSIX awk
awk '{r=r ? r "\n" $0 : $0}
/\|##\|$/ { if (r ~ /"/) print r; r=""}' inputfile > outputfile
GNU awk 1
awk 'BEGIN{RS="\\|##\\|\n?";ORS="|##|\n"}/"/' inputfile > outputfile
GNU awk 2
awk 'BEGIN{RS="\\|##\\|\n?"}/"/{printf $0 RT}' inputfile > outputfile
On the sample data provided in the question, all provided solutions give the following output:
10358|*|BI-MED-CDMA-MCS-90-118-EXAM|*|Exam for 001-MCS-90-118:
Planning, Conducting and Reporting Post Marketing Surveillance "Studies and Safety Reporting from Non Trial Activities |*|GLOBAL_MEDICAL|*||*|Y|*|N|*||*|CFC6E822849A0A7AE040800AA5644B19|*|finke|*|2012.04.30 04:23:27|##|
note: It is possible that you are suffering from the Carriage Return problem if the file comes from a Windows machine. Pleas run dos2unix on the file before using it with these tools.
How does this work? (POSIX)
Using a POSIX version of awk we can do
awk '{r=r ? r "\n" $0 : $0}
/\|##\|$/ { if (r ~ /"/) print r; r=""}' inputfile > outputfile
The idea is to build a record r by appending every line to r. If the current line ends with "|##|", then we check if the record r contains a <double quote> ". If this is the case, we print the record r and reset the record r to an empty string. If it does not contain the <double quote>, we just reset it.
How does this work? (GNU)
Using GNU awk you can do this directly using the record separator RS
awk 'BEGIN{RS="\\|##\\|\n?";ORS="|##|\n"}/"/' inputfile > outputfile
The idea here is that the file contains various records. The OP clearly stated that the information of a record is split in fields separated by |*|, but more importantly, the records themselves are separated by |##|. So in the presented example of the OP, the first record is line1 while the second record is spread over line 2 and line 3.
In awk, you can define a record separator by means of the variable RS. In its default state, RS is the <newline> character \n which makes each line a separate record which can be referenced by $0. In POSIX, the record separator can only be a single character which separates the records, while in Gnu awk, this can be a regular expression (see addendum below).
Since the record separator of the OP is the string "|##|" followed all or not by a <newline> character \n, we need to define RS=\\|##\\|\n?. Why so complicated?
the <pipe> | symbol is the OR operation (alternation operator) in a regular expression, so we need to escape it. But since string literals that are used as regular expressions are parsed twice, we also need to escape it twice. So | &rightarrow; \\| (see here)
the \n? is because it seems that the actual record separator is the string "|##|\n", but maybe some records do not have a newline character, especially the last record.
When you print records, using the print statement it automatically appends the output record separator ORS after each line. By default this is again a <newline> character \n. Since the record separator RS is not a part of the record $0 you need to update the value ORS to ORS="|##|\n". This time, not a regex, so you do not need to escape at all.
The statement /"/ is a shorthand for /"/{print $0} which means If the current record $0 contains a <double quote> ", then print the current record $0 followed by the output record separator ORS.
Note: since we actually already use Gnu awk, we can actually reduce the whole thing even further to:
awk 'BEGIN{RS="\\|##\\|\n?"}/"/{printf $0 RT}' inputfile > outputfile
Which makes use of the matched record separator RT that corresponds to the text found by RS. By replacing the print statement by a printf statement, we do not need to ORS anymore and just manually add RT to the record $0.
RS: The 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.
The ability for RS to be a regular expression is a gawk extension. In most other AWK implementations, or if gawk is in compatibility mode (see Options), just the first character of RS’s value is used.
ORS: The output record separator. It is output at the end of every print statement. Its default value is "\n", the newline character.
RT: (GNU AWK specific) The input text that matched the text denoted by RS, the record separator. It is set every time a record is read.
source: GNU AWK manual