Bash script to convert a date and time column to unix timestamp in .csv

Question

I am trying to create a script to convert two columns in a .csv file which are date and time into unix timestamps. So i need to get the date and time column from each row, c

Answer 1

You don't provide an exerpt from your csv-file, so I'm using this one:

[foo.csv]
2011/11/25;12:00:00
2010/11/25;13:00:00
2009/11/25;19:00:00

Here's one way to solve your problem:

$ cat foo.csv | while read line ; do echo $line\;$(date -d "${line//;/ }" "+%s") ; done
2011/11/25;12:00:00;1322218800
2010/11/25;13:00:00;1290686400
2009/11/25;19:00:00;1259172000

(EDIT: Removed an uneccessary variable.)

(EDIT2: Altered the date command so the script actually works.)

Answer 2

Now two imporvements:

First: No need for cat foo.csv, just stream that via < foo.csv into the while loop.

Second: No need for echo & tr to create the date stringformat. Just use bash internal pattern and substitute and do it inplace

while read line ; do echo ${line}\;$(date -d "${line//;/ }" +'%s'); done < foo.csv

Answer 3

this should do the job:

 awk  'BEGIN{FS=OFS=", "}{t=$1" "$2; "date -d \""t"\"  +%s"|getline d; print $1,$2,d}' yourCSV.csv

note

you didn't give any example. and you mentioned csv, so I assume that the column separator in your file should be "comma".

test

kent$  echo "2011/11/25, 10:00:00"|awk  'BEGIN{FS=OFS=", "}{t=$1" "$2; "date -d \""t"\"  +%s"|getline d; print $1,$2,d}'
2011/11/25, 10:00:00, 1322211600