I was trying out this piece of code below
Because i wanted make that script very generic such that in which ever place the pattern "successful" is found it should get the count. Since I want to use the same script for two more diffrent log in which the successful message can be in any place not neccesarily in the last ($NF == "successful").
But the problem i was facing here was i was not able to increment the timestamp values i.e "st" and "et" with 30 minutes and get a logic for that to repeat the piece of code for every 30 minutes.
So can you please suggest me a solution for that.
Assuming you still want the same output format, I would just change the:
in the script I gave you to:
Adding the boilerplate around the piece of code you suggested to produce the same output (calling awk 48 times calling grep 48 times and reading your log file 49 times) instead of calling awk once to read your log file once seems extremely wasteful.
If you really want to process your log file 48 times instead of once, I can come up with a way to do that, but I'll need some explanation as to why doing that would be better than the trivial change to my script that is shown above.
This User Gave Thanks to Don Cragun For This Post:
need to get the sum of the successful batches at the last
Quote:
Originally Posted by Don Cragun
Assuming you still want the same output format, I would just change the:
in the script I gave you to:
Adding the boilerplate around the piece of code you suggested to produce the same output (calling awk 48 times calling grep 48 times and reading your log file 49 times) instead of calling awk once to read your log file once seems extremely wasteful.
If you really want to process your log file 48 times instead of once, I can come up with a way to do that, but I'll need some explanation as to why doing that would be better than the trivial change to my script that is shown above.
Thanks again...
I do not have any specific reason to go with that solution (awk -v rd="2013/03/07" -v st="00:00:00" -v et="00:29:59" '$1==rd && $2>=st && $2<=et' log | grep -c "successful"), I just wanted to let you know that i was trying out in that method. Thanks for the feedback on the same as well.
Its good for me to go with your solution.
But I need one more functionality to be added with this i.e to get the sum of all the successful batches at the last as shown in the below sample output
With the scripts I have given you and the explanation of what one of the scripts does, can you tell us where changes need to be made to the script to add the additional three lines of output at the end?
How close can you come to getting the script to produce the output you want?
This User Gave Thanks to Don Cragun For This Post:
With the scripts I have given you and the explanation of what one of the scripts does, can you tell us where changes need to be made to the script to add the additional three lines of output at the end?
How close can you come to getting the script to produce the output you want?
Please get into the habit of using indentation to show the structure of your code. It makes it a lot easier for humans to see what you're trying to do even if the awk interpreter (and the Korn shell) don't much care about spacing in the scripts they process (as long as reserved words are separated as required by their lexical analyzers).
I guess I confused you by separating the format string used by printf() to format its output from the printf() statement itself. Please look at your system's man page for the awk utility and look closely at its description of the awk printf() function.
As had been mentioned before, $dd in awk is not a reference to an awk variable named dd, it is a reference to the contents of the field specified by the numeric value of the contents of the awk variable dd. References to variables in the awk programming language and references to variables in the shell programming language are different. So, $dd (when dd contains 2013/03/07) is not valid; but, even if it was, $dd inside quotes in the format string will print the literal string $dd not the contents of the dd'th field in the current input line (which has no defined meaning in the END clause).
Try the following:
I would normally split that format string into two parts (so my script would be easily readable on an 80 column output device:
but I didn't want to confuse you by splitting the format string into two parts.
This User Gave Thanks to Don Cragun For This Post:
Please get into the habit of using indentation to show the structure of your code. It makes it a lot easier for humans to see what you're trying to do even if the awk interpreter (and the Korn shell) don't much care about spacing in the scripts they process (as long as reserved words are separated as required by their lexical analyzers).
I guess I confused you by separating the format string used by printf() to format its output from the printf() statement itself. Please look at your system's man page for the awk utility and look closely at its description of the awk printf() function.
As had been mentioned before, $dd in awk is not a reference to an awk variable named dd, it is a reference to the contents of the field specified by the numeric value of the contents of the awk variable dd. References to variables in the awk programming language and references to variables in the shell programming language are different. So, $dd (when dd contains 2013/03/07) is not valid; but, even if it was, $dd inside quotes in the format string will print the literal string $dd not the contents of the dd'th field in the current input line (which has no defined meaning in the END clause).
Try the following:
I would normally split that format string into two parts (so my script would be easily readable on an 80 column output device:
but I didn't want to confuse you by splitting the format string into two parts.
Great Thanks for your time, comments and solution...!!!
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