bash method to remove last 4 columns from csv file

Question

Is there a way to use bash to remove the last four columns for some input CSV file? The last four columns can have fields that vary in length from line to line so it is not sufficient to just delete a certain number of characters from the end of each row.

Answer 1

Cut can do this if all lines have the same number of fields or awk if you don't.

cut -d, -f1-6 # assuming 10 fields

Will print out the first 6 fields if you want to control the output seperater use --output-delimiter=string

awk -F , -v OFS=, '{ for (i=1;i<=NF-4;i++){ printf $i, }; printf "\n"}'

Loops over fields up to th number of fields -4 and prints them out.

Answer 2

cat data.csv | rev | cut -d, -f-5 | rev

rev reverses the lines, so it doesn't matter if all the rows have the same number of columns, it will always remove the last 4. This only works if the last 4 columns don't contain any commas themselves.