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How to extract one column from csv file in perl?

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# 1  
Old 11-26-2010
How to extract one column from csv file in perl?

Hi everyone,
i am new to perl programming, i have a problem in extracting single column from csv file. the column is the 20th column,
please help me..

at present i use this code

use warnings;
use strict;
my $file1 = $ARGV[0];
open FILE1, "<$file1"
    or die "Can't open $file1: $!\n";
    my #found;
    my $period;
    while ( <FILE1> ) {
            ### Remove the newline that acts as a record delimiter
            ### Break up the record data into separate fields
            ( $period) = /^[.*,.*{19,}].*,/  
            #( $name, $location, $mapref, $type, $description ) =

# 2  
Old 11-26-2010
Please provide an example of your input as well as the output format you expect .
What is your field delimiter ?

awk -F<your_delimiter> '{print$20}' input

# 3  
Old 11-26-2010
perl -ne 'print $_ if ($.==20)' inputfile

# 4  
Old 11-26-2010
Without knowing more about your CSV file structure, there is little anybody can do to give you a definitive answer.

However here is a generic solution which uses the Text::CSV module.
use strict;
use Text::CSV;

my $file = $ARGV[0];
my $csv = Text::CSV->new();

open (CSV, "<", $file) or die "Can't open $file: $!\n";

while (<CSV>) {
   if ($csv->parse($_)) {
      my @column = $csv->fields();
      print "$column[19]\n";
   } else {
      my $err = $csv->error_input;
      print "ERROR: Failed to parse line: $err";

close CSV;

# 5  
Old 11-29-2010
hi every one!! thanks a lot for your answers..

my file has \t(tab) as delimiter, and the input file looks like this

col1  col2 col3....col20 col21 col22
A       B     C ...... dy    0       xyz
C       R      S .......dx             abc 
E        T      E .......dc             res

The output file sholud look like this..

col1  col2 col3....col20 col21 col22
A       B     C ...... dy    0       xyz
C       R      S .......dx    0         abc 
E        T      E .......dc             res

Without changing any other column data, i need to update only column 20 and print the entire file... please i dont know anything about perl..

Last edited by kvth; 11-29-2010 at 02:25 AM.. Reason: spelling mistake
# 6  
Old 11-29-2010
my @line = split(/\t/, $_);
$line[19] = your_change;
print "$_\t";
print "\n";

If there are other whitespace characters seperating the columns, perl has a match any white space except newline escape character. I can't remember what it is off the top of my head.

Last edited by ilikecows; 11-29-2010 at 03:09 AM.. Reason: added code tags and corrected syntax error
# 7  
Old 11-29-2010
hey thanks Mr.ilikecows... your code is errors free.. but when i run it, i get a blank output file...please help

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