How to get the first column of every line from a CSV file?

Question

How do get the first column of every line in an input CSV file and output to a new file? I am thinking using awk but not sure how.

Answer 1

Try this:

 awk -F"," '{print $1}' data.txt

It will split each input line in the file data.txt into different fields based on , character (as specified with the -F) and print the first field (column) to stdout.

Answer 2

Can be done:

$ cut -d, -f1 data.txt

Answer 3

echo "a,b,c" | cut -d',' -f1 > newFile

Answer 4

Input

a,12,34
b,23,56

Code

awk -F "," '{print $1}' Input

Format

awk -F <delimiter> '{print $<column_number>}' Input

Answer 5

This can be achieved using grep:

$ grep -o '^[^,]\+' file.csv

Answer 6

Using Perl:

perl -F, -lane 'print $F[0]' data.txt > data2.txt

These command-line options are used:

• -n loop around every line of the input file
• -l removes newlines before processing, and adds them back in afterwards
• -a autosplit mode – split input lines into the @F array. Defaults to splitting on whitespace.
• -e execute the perl code
• -F autosplit modifier, in this case splits on ,

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If you want to modify your original file in-place, use the -i option:

perl -i -lane 'print $F[0]' data.txt

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If you want to modify your original file in-place and make a backup copy:

perl -i.bak -lane 'print $F[0]' data.txt

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If your data is whitespace separated rather than comma-separated:

perl -lane 'print $F[0]' data.txt