Shell Programming and Scripting

Hi,

I have a text file as shown below:-

$cat hpdbdv01.out
-rw-r--r-- 1 oracle 35517583 May 4 2005 /tmp/eai/eai_obj1.dmp.Z
-rw-r--r-- 1 oracle 8733 May 4 2005 /tmp/eai/eai_obj2.dmp.Z
-rw-rw-rw- 1 oracle 18869248 Jun 24 2005 /tmp/edw.dmp
-rw-r--r-- 1 oracle 5775360 Jun 24 2005 /tmp/jhr.dmp
-rw-r--r-- 1 oracle 2048 Jun 24 2005 /tmp/expdat.dmp
-rw-r--r-- 1 oracle 132623360 Jun 24 2005 /tmp/omstage.dmp
-rw-r----- 1 svarughe 491520 Jul 19 2005 /tmp/exp_dwprd_of_rapidmart.dmp
-rw-r--r-- 1 oracle 0 Aug 12 14:25 /home/oracle/dba/michael/test.dmp
-rw-r--r-- 1 oracle 16384 Jul 16 2004 /home/oracle/outln.dmp
-rw-r--r-- 1 115 47104 Feb 9 2005 /home/oracle/gl_per.dmp

I need to find the sum of 4th column, so that I can tell so many bytes has been occupied by these files.

I use the below script and am getting error, please help in resolving this. If there's any alternate way also pls let me know.

AWK script:-
cat rd1.awk
+++++++
BEGIN{print "Report Starts" }
{ print " The file size :" $4
sm+=$4
}
END{printf "%ld" $sm}
+++++++

I execute the above script as shown below:-

$awk -f rd1.awk hpdbdv01.out

The error which I get is:-

...
...
...
...
The file size :1943552
The file size :40371200
awk: The field 2147483647 must be in the range 0 to 199.

All I want is, just to find the sum of the fourth column. Please advise.


TIA,
-Thiagu

try

Code:

cat hpdbdv01.out | awk -v s=0 '{ s += $4 } END { print "The file size: " s }'

Remove the "$" from "$sm". You are confusing the shell construct with awk's field reference syntax. If you use $ then the value of the variable tries to reference a field number.

HI fidodido,

Thanks for your reply.

I tried the code, which you've provided. I get the below output, how can I format this number column, I gave %ld it didnt work..!

The file size: 1.33231e+11

TIA,
Thiagu

To use printf with a format string requires two arguments, separated by a comma

Code:

END{printf "%d\n", sm }