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Output no. of rows and total value of a file to a new file

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Top Forums Shell Programming and Scripting Output no. of rows and total value of a file to a new file
# 1  
Old 03-22-2010
Output no. of rows and total value of a file to a new file
I am relatively new to Unix and I would be grateful if someone could help me with the syntax of a script.

I am trying to write a Unix script against the file ‘ajtest.dat’ (see below) so that it outputs the number of rows and total value of column 3 in a file called ‘aj1’ as follows:

Code:
No. of rows : 15
Total value : 16208.49

File: ajtest.dat:

Code:
3033*94917,PL123000100049,+000097.37,20-Mar-2008,24-Aug-2008
3033*96345,PL091708600108,+001724.64,17-Sep-2008,02-Aug-2009
3033*96479,PL054808900070,+000201.18,22-Oct-2008,24-Feb-2009
3033*98774,PL056901201003,+001313.52,31-Mar-2009,07-Jun-2009
3033*96420,PL090007900061,+003802.54,26-Aug-2008,29-Jun-2009
3033*94089,PL090402200131,+005008.13,26-Feb-2008,25-Mar-2009
3033*100366,PL090708300112,+000141.94,21-Sep-2009,08-Nov-2009
3033*97563,PL090714000249,+000000.84,12-Jan-2009,25-Jan-2009
3033*94845,PL090814400090,+000081.12,20-Feb-2008,31-Jul-2008
3033*90236,PL090912800080,+000116.69,29-Sep-2008,18-Jan-2009
3033*95124,PL015104300203,+000260.41,14-Apr-2008,01-Dec-2008
3033*95474,PL015301000256,+000048.96,04-Jun-2008,30-Oct-2008
3033*78033,PL015304400202,+000181.44,18-Aug-2008,12-Feb-2009
3033*95068,PL015308200270,+001801.71,02-May-2008,24-Aug-2009
3033*98498,PL052402800063,+001428.00,26-Mar-2009,25-Nov-2009

Last edited by radoulov; 03-23-2010 at 07:42 AM.. Reason: Please use code tags!
# 2  
Old 03-22-2010
Try this:

Code:
awk -F ',' '{ sum = sum + $3} { sum1 = sum1 + 1} END { print "No. of rows : "sum1" Total value : "sum}' ajtest.dat >> aj1



---------- Post updated at 11:58 AM ---------- Previous update was at 11:53 AM ----------

Try this (print next line):

Code:
awk -F ',' '{ sum = sum + $3} { sum1 = sum1 + 1} END { print "No. of rows : "sum} NR  { print "Total value : "sum1}' ajtest.dat | tail -2 > aj1



---------- Post updated at 12:02 PM ---------- Previous update was at 11:58 AM ----------

Correct one is :

Code:
awk -F ',' '{ sum = sum + $3} { sum1 = sum1 + 1} END { print "Total Value : "sum} NR  { print "No. of rows: : "sum1}' ajtest.dat   | tail -2 > aj1

Last edited by radoulov; 03-23-2010 at 07:42 AM.. Reason: Please use code tags!
# 3  
Old 03-22-2010
Hello Aj,

You can use my below code to achive this. This script accepts file name from command line.
Code:
#!/bin/ksh

file_name=$1 ;
no_rows=0 ;
total=0 ;

while read line
do
temp_total=`echo $line | cut -d "," -f3 | tr -d '+,-' ''` ;
no_rows=`expr ${no_rows} + 1` ;
total=`echo ${temp_total}+${total} | bc` ;
done < $file_name

echo "Number of rows = $no_rows" ;
echo "Third column total = $total" ;

-Nithin.
# 4  
Old 03-23-2010
Hi Jairaj,
Thank you for your reply.

I have tried the script entitled 'Correct one' but when I execute it, I get the message:
awk: syntax error near line 1
awk: bailing out near line 1

What have I done wrong?
# 5  
Old 03-23-2010
Code:
nawk -F, '{sum+=$3}END{print "No. of rows: " NR"\nTotal value: " sum}' infile

# 6  
Old 03-23-2010
Hi Malcomex999,

Thank you for your reply.

I have run your script and it works except for it rounds the pence up to one decimal place. How do I get it to display to two decimal places?
# 7  
Old 03-23-2010
All are in the single line.

Code:
awk -F ',' '{ sum = sum + $3} { sum1 = sum1 + 1} END { print "Total Value : "sum} NR { print "No. of rows: : "sum1}' ajtest.dat | tail -2 > aj1

Last edited by radoulov; 03-23-2010 at 07:43 AM.. Reason: Code tags, please!
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