Hi, I've been trying to do this without awk but it's getting too complicated but I don't know how to use arrays within awk, maybe this is how it could be done best.
I have files something like this that I want to display only lines where the value of the third column is within 2 of each other, in this case I would only want to display lines 3 and 4. Could someone suggest something?
2011-06-25 12:27:59 40 nodea down
2011-06-25 12:28:02 45 nodea up
2011-06-25 12:29:23 70 nodea down
2011-06-25 14:31:14 71 nodea up
2011-06-25 14:31:15 80 nodea down
Init's X (column 3) and Y (entire line) using the first line. Let's go ahead and pretend the first X is 40.
If our next line is 41, it will satisfy this condition. 40=41-1. So this will only match if the next record is +1 or +2... hmm, Did you need it to match -1 and -2 as well?
It appends the entire line to Y. It sets P, which you can think of as a print tag, but the "next" will end processing of this record now and not yet print it right away (because maybe a 3rd line or more follows within the range...)
Lets say line 3 in our example (41,40,...) has a value of 70. So the previous block with the X=$3-1 ... is not satisfied, but this one is since we did set the P flag already. We'll print whats in our buffer Y (which are the lines containing 41 and 40), and clear that flag and start anew.
Still on our 3rd line of text in my example, X=70, will continue through to here, and be set as the next X to be looked for (in case line 4 is 71 or 72)....
If we reach the end of the file and have stuff to print, do it now. Otherwise, we only were actually printing our successful matches at the first non-match and it'd be lost.
Hope this helps.
Last edited by neutronscott; 07-15-2011 at 02:06 PM..
Reason: fixed a code tag
This User Gave Thanks to neutronscott For This Post:
Good evening, Im newbie at unix specially with awk
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