Hi panyam and Franklin
i followed the method suggested by panyam its giving useful result.
here it goes as in the order i requested.
Code:
1.
awk '$4=="BSNLSERVICE1"&&$6=="2128" { count++ } END { print "BSNLSERVICE1-->2128-->" count }' 20090604*
2.
awk '$4=="BSNLSERVICE1"&&$6=="2128"{ b[$14]++} END {for(i in b){print i, b[i]} }' 20090604*
3.
awk '$4=="BSNLSERVICE1"&&$6=="2128"{ b[$9]++} END {for(i in b){print i, b[i]} }' 20090604*
one thing i need to know whether its possible to derive filename from date command.
actually i need to schedule it every night at 2:00 am and i need to derive the filename from the date command
lets take the example
tonight 2.00 am the ouput of the
Code:
date +'%Y%m%d'
is 20090606
i need to give the file name as 20090605* in the filename part of the awk ...
and for franklin ....i used this way
Code:
#!/usr/xpg4/bin/awk
awk -v day="20090605" -v serv="BSNLSERVICE1" -v val="2128" '
$1 ~ day && $4==serv && $6==val {
s1++;a[$14]++;b[$9]++
}
END{
print serv,val, "=" s1 "\n"
print serv,val, a["SUCCESS"]
print serv,val, a["SYSTEM"] "\n"
for(i in b){print i, b[i]}
}' 200906051859
and got the response as
Quote:
./reconcil.sh
/usr/xpg4/bin/awk: syntax error Context is:
>>> ./ <<<
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thank u
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