adding a date column in csv file

thepubba1 November 22, 2010 at 21:44:40

If you start out with an input file that looks like this: sl.no, name 1, har 2, kal 3, mal 4, pal and you run this script: #/bin/ksh integer counter=0 exec 3< data.csv while read -u3 line do if [[ $counter -eq 0 ]] then print "$line, date" (( counter +=1 )) else print "$line, $(date +%d/%m/%Y) (coz data \"$line\" has loaded into csv on $(date +%d/%m/%Y))" fi done You will get output that looks like this: sl.no, name, date 1, har, 23/11/2010 (coz data "1, har" has loaded into csv on 23/11/2010) 2, kal, 23/11/2010 (coz data "2, kal" has loaded into csv on 23/11/2010) 3, mal, 23/11/2010 (coz data "3, mal" has loaded into csv on 23/11/2010) 4, pal, 23/11/2010 (coz data "4, pal" has loaded into csv on 23/11/2010) If this is not what you want, then modify what I've written to suit your needs. Jerry Lemieux

Thanks Jerry. But the O/P should not be the way you have shown. I just have a script as mentioned below. This is working, but as many times I run the script those many times the data is loading. the script is: #!/bin/ksh cd /data n=`wc -l < file1` i=1 while [ "$i" -le "$n" ] do line=`cat file1 | head -$i | tail -1` a=`echo $line | nawk 'BEGIN{FS=","};{print NF}'` if [ "$a" -eq "2" ] then echo $line `date +%d%m%y` >> file2 else echo $line fi i=`expr $i + 1` done (where file1 has 1,har 2,mal and file2 is O/P file) So, when I run the script for the first time, file2 is 1,har,221110 2,mal,221110 Again when i load the data into file1 with "3, kar" on 23/11/10 then again when i run the script file2 is 1,har,221110 2,mal,221110 1,har,221110 2,mal,221110 3,kar,231110 I need the unique data. Please try to find a solution for these Thanks, Sunkoju

nails November 23, 2010 at 10:56:27

The biggest problem that I see is that you are not appending a comma for the third field so the file always has two fields. You probably want to do something like: echo "$line,`date +%d%m%y`" I have some comments about your script: 1) Why do you cycle thru file1 like this? Since you want to read file1 a line at a time, either use Jerry's method or this: #!/bin/ksh while read line do echo "$line" done < file1 2) I recommend that when you echo strings that you surround them with double quotes to retain the white space: echo "$line" instead of: echo $line 3) Finally, although it probably doesn't hurt, there is no reason to surround integr variables in an iff comparison: if [ $a -eq 2 ] then

But when I am using Jerrys script, am getting an error as 1) integer not found 2) -u3 is not an identifier What I need to do to remove this error And Please let me get cleared about where can i see the output data from? Is it data.csv itself becomes an output file? Please let me know what to do...

nails November 25, 2010 at 16:07:29

Jerry's script sets file descriptor 3 to data.csv file. Reading file descriptors is a korn shell invention. This doesn't work with the Bourne shell. And if you have not defined your script with the korn shell invocation at line 1, column 1: #!/bin/ksh the script will probably be defined a a bourne shell script and fail.