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Can anyone help me to spot my mistake?

 
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Top Forums UNIX for Dummies Questions & Answers Can anyone help me to spot my mistake?
# 1  
Old 08-05-2012
Can anyone help me to spot my mistake?
Hi there can anyone help me to spot my mistake and please explain why it appears
My code :
Code:
#!/usr/bin/gawk -f
BEGIN { bytes =0}
{ temp=$(grep "datafeed\.php" | cut -d" " -f8)
bytes += temp}
END { printf "Number of bytes: %d\n", bytes }

when I am running ./q411 an411
an411:

Code:
152.78.74.111 [13/Feb/2011:06:52:00 +0000] "GET /datafeed.php HTTP/1.1" 200 2826
152.78.74.211 [13/Feb/2011:06:53:00 +0000] "GET /index.html HTTP/1.1" 200 1190
152.78.74.121 [13/Feb/2011:06:55:00 +0000] "GET / HTTP/1.1" 200 3000

It is complaining about cut and closing bracket ")" but I don't understand why.

Any help????
Thanks in advance Smilie

Moderator's Comments:
Mod Comment edit by bakunin: please use CODE-tags not only for code but also for file output. Thank you.
Last edited by bakunin; 08-05-2012 at 07:31 AM..
# 2  
Old 08-05-2012
Code:
temp=$(grep "datafeed\.php" | cut -d" " -f8)

is a shell command; not a gawk command.
This User Gave Thanks to Don Cragun For This Post:
FUTURE_EINSTEIN
# 3  
Old 08-05-2012
Quote:
Originally Posted by Don Cragun
Code:
temp=$(grep "datafeed\.php" | cut -d" " -f8)

is a shell command; not a gawk command.
Thanks a lot!! So is there a way to assing output of the command to a variable in gawk ???
# 4  
Old 08-05-2012
I don't know what the construct $(temp...) is doing in your awk program - first, it is a shell variable that is assigned, second, the grep command does not know on what input it should work. If I guess your intentions right to sum up all byte info in the datafeed line, try the following in your awk program:
Code:
/datafeed\.php/ {bytes += $8} END { printf "Number of bytes: %d\n", bytes }

This User Gave Thanks to RudiC For This Post:
FUTURE_EINSTEIN
# 5  
Old 08-05-2012
If you're trying to get the output RudiC suggested, you don't need awk. You can just do:
Code:
printf "Number of bytes: %d\n" $(grep datafeed.php | cut -d" " -f8) < an411

in the shell. You already had all of the shell code in your awk program.
This User Gave Thanks to Don Cragun For This Post:
FUTURE_EINSTEIN
# 6  
Old 08-05-2012
Quote:
Originally Posted by FUTURE_EINSTEIN
So is there a way to assing output of the command to a variable in gawk ???
In fact there is: use the "system()" function in awk. But to be honest RudiC provided a much better solution if he got your intention right and you should first try to avoid the system()-call.

Every call to an external program is very costly in terms of resources and therefore you should try to solve as much as possible using one command. There is nothing inherently "better" or "worse" in using shell code or awk code, but if you start with awk then stick to it or abandon it altogether. The same goes for shell code.

I hope this helps.

bakunin
This User Gave Thanks to bakunin For This Post:
FUTURE_EINSTEIN
# 7  
Old 08-05-2012
I was mislead by the OP using bytes+=temp into thinking he/she wanted to sum up several lines' byte counts. If there is only one line expected, use Don Cragun's one line shell version.
This User Gave Thanks to RudiC For This Post:
FUTURE_EINSTEIN
 
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