Shell script to filter date alone from a particular column of a .csv file and save the output in another csv file

Question

Consider a .csv file which has dates listed in the 6th column in format 7/20/2017 17:00 for example.

If I give input as 7/11/2017 then it should filter the column that has the matching input date and the output should be saved in another .csv file

I need a command to filter 6th column based on input given.

file.csv has below entries:

23,54,one,tree,56,7/20/2017 17:00,23,iii,89
23,54,one,tree,56,7/20/2017 17:00,23,iii,89
23,54,one,tree,56,7/21/2017 17:00,23,iii,89
23,54,one,tree,56,7/30/2017 17:00,23,iii,89

similarly multiple entries that date and time format in 6th column. Input given in runtime like read 7/20/2017, then it should filter the date from 6th column and save the output entries to another .csv file.

Answer 1

Simply with awk:

awk -F, -v d="7/20/2017" '$6~d' input.csv > dates.csv

The dates.csv contents:

23,54,one,tree,56,7/20/2017 17:00,23,iii,89
23,54,one,tree,56,7/20/2017 17:00,23,iii,89

---

If date value comes from shell variable:

d="7/20/2017"
awk -F, -v d="$d" '$6~d' input.csv > dates.csv

Answer 2

Assuming there is only one column with a date in that format:

$ grep -F '7/20/2017' file.csv
23,54,one,tree,56,7/20/2017 17:00,23,iii,89
23,54,one,tree,56,7/20/2017 17:00,23,iii,89

You may redirect that to a new file:

$ grep -F '7/20/2017' file.csv >new.csv

To properly parse a CSV file, use a CSV parser, like the one built into the CSVkit tools:

$ csvgrep -c 6 -m '7/20/2017' file.csv
23,54,one,tree,56,7/20/2017 17:00,23,iii,89
23,54,one,tree,56,7/20/2017 17:00,23,iii,89