copy lines where a character occurs even number of times

Question

I've received some files that need to be loaded into the database. These files have user input, and there are instances where the quote character is used an odd number of times. I'd like to filter out these records.

I would like to grep lines that contain a specific character an odd or even number of times.

Sample input:

12345|what"s wrong|20121212

Answer 1

With awk:

awk -F \" 'NF % 2' < yourfile

For even number of times (odd number of fields where fields are "-separated).

awk -F \" 'NF % 2 == 0' < yourfile

Or to split the file into two files:

awk -F \" '{if (NF%2) print > "even.txt"
            else print > "odd.txt"}' < yourfile

For odd number of times.

With grep, for even number:

grep -Ex '(([^"]*"){2})*[^"]*'

For odd number, add the -v option.

Answer 2

Alternative perl approach:

-- to output lines with odd number of " occurrences

perl -ne 'print if y/\"// % 2' yourfile

---

-- to output lines with even number of " occurrences

perl -ne 'print if y/\"// % 2 == 0' yourfile

---

• y/// - Perl transliteration operator

Answer 3

sed -ne '
   h;:a
   s/"//;T
   s/"//;ta
   g;p
' yourfile

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Working

1. Store the original since destructive process will commence.
2. We setup a loop on in which we successively scrub a quote at a time, when the first del is unsuccessful => terminate operations for this line => even number of quotes were present.
3. In 2nd scrub, if we were unsuccessful => an odd number of quotes, then we retrieve from hold the original and print.
4. Else we loop back.