So here's my problem. I have an issue with a .csv file (Current.csv) in that there are commas randomly place throughout the data, so awk-ing the file doesn't give me constant column numbers for a particular type of information I'm looking for. Luckily, I realized this info is always the third instance of a date format (m,mm)/(d,dd)/yy. So I'm trying the regular expression below to display just the dates within the ith line:
awk -F',' '{if (NR==$i)print}' Current.csv | grep -o "[0-9]{1,2}/[0-9]{1-2}/[0-9]{1,2}" | echo
It doesn't display anything so far and I'm totally stuck as to why. My guess for displaying the third is to just pipe this all into:
awk {print $3}
Any ideas on the awk'd regular expression search problem?
Sample line
"lettershere",numbershere,"retardedbrokenquoteshere,mm/dd/yy,morestuff,mm/dd/yy,numbers,mm/dd/yy