I can't believe I couldn't find my question anywhere else in the bash/linux community. I simply want to replace the first column of my csv file with a variable...
0,8,9,10
0,8,9,10
0,8,9,10
0,8,9,10
0,8,9,10
0,8,9,10
2/24/14,8,9,10
2/24/14,8,9,10
2/24/14,8,9,10
2/24/14,8,9,10
2/24/14,8,9,10
2/24/14,8,9,10
I simply just want to replace the first column with $date, but all the awk commands don't allow a variable as a the replacement.
This should work:
awk -v dt="$date" 'BEGIN{FS=OFS=","}{$1=dt}1' inputFile
Explaination:
-v option to set an awk variable and assign it your shell variable. , (since it is a csv)$1 to your awk variable. 1 is to print the line with modified $1. 
1 after the curly brackets or parentheses (round brackets.) That helps me understand some other commands.sed 's#^[^,]*#2/24/14#' filename
should do it. If you want to work with the current date, you should do:
DATE=$(date +'%d/%m/%y')
sed "s#^[^,]*#$DATE#" filename
This is looking for the first item before the comma and replace it with the date. The use of # as delimiter allows you to type in the date without having to escape the / character.
sed "s|[^,]*|$( date +'%d/%m/%y' )|" YourFile do all at onceEvery time you deal with complex data format you should use proper parser and CSV is complex data format (and without any proper definition and full of quirks. No RFC 4180 is not the definition. It is rather joke. There is a lot of data violating RFC 4180 and almost every decent CSV parser out there is ignoring this RFC because it would not work.)
perl -mText::CSV_XS -e'$d=shift@ARGV;$csv=Text::CSV_XS->new({ binary => 1, eol => $/ });while(my$row=$csv->getline(*ARGV)){$$row[0]=$d;$csv->print(*STDOUT,$row)}' "$date"
Changing to current date using awk
echo "0,8,9,10" | awk '{sub(/[^,]+/,d)}8' d="$(date +'%d/%m/%y')"
25/02/14,8,9,10
[^,]+ everything that is not a , will be replaced with d date
