使用 Bash 脚本编辑 CSV 文件

Question

I have a CSV file that looks like this:

English, "Hawkin, Jason" ,JHAWKIN110@exampleemail.com
English, "McDonald, Matt" ,MMCDONALD114@exampleemail.com
English, "Campbell, Josh" ,JCAMPBELL111@exampleemail.com

My intention is to make a Bash Script to make a second CSV file with the following format:

ID, Email, FNAME, LNAME
jhawkin110, JHAWKIN110@exampleemail.com, Jason, Hawkin

After using "cut" to remove the first column I did not need, I am at a loss on how to create the "ID" by extracting the first part of the "Email".

Answer 1

Using awk

$ awk 'BEGIN {FS=OFS=","; print "ID, Email, FNAME, LNAME"}{gsub(/"/,"");split($NF,a,"@"); print tolower(a[1])," " $NF, $3, $2}' input_file
ID, Email, FNAME, LNAME
jhawkin110, JHAWKIN110@exampleemail.com, Jason , Hawkin
mmcdonald114, MMCDONALD114@exampleemail.com, Matt , McDonald
jcampbell111, JCAMPBELL111@exampleemail.com, Josh , Campbell

Using sed

$ sed -E 's/.*"([^,]*),([^"]*)..,(([^@]*)@.*)/\3,\2, \1,\L\4/;s/(.*),(.*)/\2, \1/;1iID, Email, FNAME, LNAME' input_file
ID, Email, FNAME, LNAME
jhawkin110, JHAWKIN110@exampleemail.com, Jason, Hawkin
mmcdonald114, MMCDONALD114@exampleemail.com, Matt, McDonald
jcampbell111, JCAMPBELL111@exampleemail.com, Josh, Campbell

Answer 2

echo 'ID, Email, FNAME, LNAME' awk -v FS=', *"?|"? *,' \ -v OFS=', ' \ '{split($4,a,"@"); print tolower(a[1]),$4,$3,$2}' file.csv