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Command to grep the service with in a timeframe

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Operating Systems Solaris Command to grep the service with in a timeframe
# 8  
Old 10-28-2015
Given your log file has date and time in the first two fields of every entry, try:
Code:
awk '/^2010-06-10.*Loadservice/ {CNT[substr($2, 1, 2)]++} END {for (c in CNT) print c, CNT [c]}' log.06102013.svr1

# 9  
Old 10-28-2015
Corona,

Here is the file info:

Code:
2010-06-10 00:00:35,222|B|stack|sys|||gprs|true|1|{http:///}LoadService|info
2010-06-10 00:00:35,333|B|stack|sys|||gprs|true|9|{http:///}LoadService|info
2010-06-10 01:00:00,333|B|stack|sys|||gprs|true|2|{http:///}LoadService|info

when I execute the script then it gives me '3'
Last edited by Corona688; 10-28-2015 at 05:44 PM..
# 10  
Old 10-28-2015
The first time in the entire thread "LoadService" has an upper case "S". That built in,
Code:
awk '/^2010-06-10.*LoadService/ {CNT[substr($2, 1, 2)]++} END {for (c in CNT) print c, CNT [c]}' file
00 2
01 1

This User Gave Thanks to RudiC For This Post:
kriss.gv
# 11  
Old 10-28-2015
Thanks Rudy.
I am not that good at scripting, so can you explain me the script and out come?

00 2
01 1

I am getting something like this when I execute the awk command. so, does 00 means hours and 2 is the count?
# 12  
Old 10-28-2015
Yes, that's the hour and its count.
The awk looks for lines that contain the target date AND the search word "LoadService". If found, a counter array indexed by the hour (two digits from time string) is incremented. In the end, the result is printed.
# 13  
Old 10-28-2015
Quote:
Originally Posted by kriss.gv
Corona,

Here is the file info:

Code:
2010-06-10 00:00:35,222|B|stack|sys|||gprs|true|1|{http:///}LoadService|info
2010-06-10 00:00:35,333|B|stack|sys|||gprs|true|9|{http:///}LoadService|info
2010-06-10 01:00:00,333|B|stack|sys|||gprs|true|2|{http:///}LoadService|info

when I execute the script then it gives me '3'
The nice thing about dates in YYYY-MM-DD HH:MM:SS order is they compare alphabetically. An exact match is no longer necessary -- you can use >, >=, <, <= to compare the date and get logical answers. awk will take one or more files, too.

Code:
$ awk -F"|" '{ sub(/,.*/, "", $1) } ; ($1 >= S) && ($1 <= E) && /LoadService/ { L++ } END { print L+0 }' S="2010-06-10 00:00:00" E="2010-06-10 01:00:00" inputfile1 inputfile2

3

$

# 14  
Old 10-29-2015
Thanks much Corona
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