[SOLVED] Getting last X fields from a specific line in a CSV file using bash

Issue

I'm trying to get as bash variable list of users which are in my csv file. Problem is that number of users is random and can be from 1-5.

Example CSV file:

"record1_data1","record1_data2","record1_data3","user1","user2"
"record2_data1","record2_data2","record2_data3","user1","user2","user3","user4"
"record3_data1","record3_data2","record3_data3","user1"

I would like to get something like

list_of_users="cat file.csv | grep "record2_data2" | <something> "
echo $list_of_users
user1,user2,user3,user4

I'm trying this:

cat file.csv | grep "record2_data2" |  awk -F, -v OFS=',' '{print $4,$5,$6,$7,$8 }' | sed 's/"//g'
    

My result is:

user2,user3,user4,,

Question: How to remove all "," from the end of my result? Sometimes it is just one but sometimes can be user1,,,, Can I do it in better way? Users always starts after 3rd column in my file.

Solution

This will do what your code seems to be trying to do (print the users for a given string record2_data2 which only exists in the 2nd field):

$ awk -F',' '{gsub(/"/,"")} $2=="record2_data2"{sub(/([^,]*,){3}/,""); print}' file.csv
user1,user2,user3,user4

but I don't see how that's related to your question subject of Getting last X records from CSV file using bash so idk if it's what you really want or not.

Answered By - Ed Morton

Answer Checked By - David Goodson (WPSolving Volunteer)