Excluding enclosed delimiters with cut

Question

Let's say I have a CSV file:

"col1","col2","col3"
"col4","col5,subtext","col6

The problem I have is as follows :

cut -d, -f1,2 test.txt
"coll1","col2"
"col4","col5

The desired output is :

"col1","col2"
"col4","col5,subtext"

Answer 1

The ParseWords module, which ships with Perl, covers this quite elegantly. Example below.

$ perl -MText::ParseWords -nE '@a=quotewords ",",1,$_;say $a[0],",",$a[1]' <test.txt
"col1","col2"
"col4","col5,subtext"
$

Answer 2

If you have gawk v4 available, a fine solution exists at Parse a csv using awk and ignoring commas inside a field

Example:

gawk -vFPAT='[^,]*|"[^"]*"' '{print $1 "," $2}' test.txt

Answer 3

Another perl solution, assuming all fields are quoted

$ perl -F'/"\K,(?=")/' -lane 'print "$F[0],$F[1]"' test.txt 
"col1","col2"
"col4","col5,subtext"

• -F'/"\K,(?=")/' field separator is comma only if it is preceded by " and followed by "
• print "$F[0],$F[1]" print first two fields separated by ,

Can be done with grep as well

$ grep -oE '^"[^"]*","[^"]*"' test.txt 
"col1","col2"
"col4","col5,subtext"

If N number of fields are needed, use grep -oE '^("[^"]*",){1}"[^"]*"' where the number inside {} is N-1

Answer 4

you can also try this with awk as below;

awk -F'","'  '{printf "%s\",\"%s\"\n", $1, $2 }' test.txt 

Eg;

user@host$ awk -F'","'  '{printf "%s\",\"%s\"\n", $1, $2 }' test.txt 
"col1","col2"
"col4","col5,subtext"