Split file based on a date field

Question

I have a file abc.txt. I need to grep the file on the date field which is present at 4^th position in the text file. I need to split the file based on the date from 1^st to 15^th August the data need to be in one file and from 16^th to 31^st August the data need to be in another file.

How can I do that?

a|b|c|08/01/2017|d
a|b|c|08/15/2017|d
a|b|c|08/16/2017|d
a|b|c|08/31/2017|d

Thank you.

Answer 1

This is easy job with awk:

awk -F'|' '{print >((substr($4,4,2)<=15)?"1-15Aug":"16-31Aug")}' infile

This is checking if the day of month in date was ‎≤15 in filed #4 it will write the lines into the file 1-15Aug otherwise those will redirect to 16-31Aug file using Ternary condition.

The output would be 2 files:

==> 1-15Aug <==
a|b|c|08/01/2017|d
a|b|c|08/15/2017|d

==> 16-31Aug <==
a|b|c|08/16/2017|d
a|b|c|08/31/2017|d

Answer 2

Using sed:

sed -n -e '\#08/01/2017#,\#08/15/2017#w first.out' \
       -e '\#08/16/2017#,\#08/31/2017#w second.out' file.in

This will write all lines between the line containing 08/01/2017 to the one containing 08/15/2017 (inclusive) to the file first.out. Similarly for the second range of dates, but to second.out.

This relies on the file file.in being sorted in date order.

The # character was chosen as the pattern delimiter as the pattern already contains /.

Answer 3

The shortest awk approach:

awk -F'[|/]' '{print >(($5<=15)?"1-15":"16-31")"Aug"}' file

• -F'[|/]' - complex field separator

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Viewing results:

for f in [0-9]*Aug; do echo "file: $f"; cat "$f"; echo; done

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The output:

file: 1-15Aug
a|b|c|08/01/2017|d
a|b|c|08/15/2017|d

file: 16-31Aug
a|b|c|08/16/2017|d
a|b|c|08/31/2017|d

Answer 4

You can do a simple csplit if your data is sorted by date.

split -f abc greptest.txt '/08\/16\/2017.*$/'

It will cut your file into two when finding the first Line containing 08/16/2017. They will be named abc00 and abc01