Extract the value of nth column in the last line

Question

I have a .csv file with the following data:

Name,Age,Gender x,10,F y,20,M z,30,F . . .

The no.of lines in the file always changes when I run something. How can I extract the Age for Name in the last line of the file. So far, I tried this:

 #!/bin/sh #extracting the headings and the last line values from the data.csv file (head -1 data.csv && tail -1 data.csv) > dummy.txt #this part doesn't work (sed -n "Name" data.csv && sed -n "Age" data.csv)>>dummy.txt

The above code errors out this:

sed: -e expression #1, char 13: unterminated `s' command*

Cyrus · Accepted Answer · 2017-06-13 19:09:02Z

2

With awk:

awk -F "," 'END{print $2}' file

With GNU sed:

sed -rn '$s/.*,(.*),.*/\1/p' file

Output:

answered Jun 13, 2017 at 19:09

Cyrus

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Comments

Harini · Accepted Answer · 2017-06-14 08:15:39Z

1

`tail -1 file | cut -d, -f2`

tail -1 gets the last line,
cut -d specifies the delimeter,
-f2 specifies the field

edited Jun 14, 2017 at 8:15

Harini

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answered Jun 13, 2017 at 19:11

Jakub Jun

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