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

Answer 1

awk one-liner:

awk -F, '{for(i=0;++i<=NF-5;)printf $i", ";print $(NF-4)}'  file.csv

the advantage of using awk over cut is, you don't have to count how many columns do you have, and how many columns you want to keep. Since what you want is removing last 4 columns.

see the test:

kent$  seq 40|xargs -n10|sed 's/ /, /g'           
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
11, 12, 13, 14, 15, 16, 17, 18, 19, 20
21, 22, 23, 24, 25, 26, 27, 28, 29, 30
31, 32, 33, 34, 35, 36, 37, 38, 39, 40

kent$  seq 40|xargs -n10|sed 's/ /, /g' |awk -F, '{for(i=0;++i<=NF-5;)printf $i", ";print $(NF-4)}'
1,  2,  3,  4,  5,  6
11,  12,  13,  14,  15,  16
21,  22,  23,  24,  25,  26
31,  32,  33,  34,  35,  36