RE: time in trace file

From: Aponte, Tony <AponteT_at_hsn.net>
Date: Wed, 28 Apr 2004 17:34:14 -0400
Message-ID: <5D48DB5EE592094FBE5373DFF0A4062905D1841A@exchsrv02.stpete.hsn.net>

Try this gawk script to see if works for you. I modified mktime.awk to process my trace files by putting the date text along with the tim value. I ran your sample and got the result show below. I'm be interested in knowing if it works for anybody else.=20

Tony Aponte
Home Shopping Network, Inc.

• sample run -------------------- $ echo "PARSE #30:c=3D2,e=3D15,p=3D0,cr=3D0,cu=3D0,mis=3D1,r=3D0,dep=3D0,og=3D4,tim=3D9= 33354557" | gawk -f mktime.awk PARSE = #30:c=3D2,e=3D15,p=3D0,cr=3D0,cu=3D0,mis=3D1,r=3D0,dep=3D0,og=3D4,tim=3D9= 33354557 04/27/2004 10:16:08 $
• the gawk script --------------- # mktime.awk --- convert a canonical date representation # into a timestamp # Arnold Robbins, arnold_at_gnu.org, Public Domain # May 1993 BEGIN \
  {
  # Initialize table of month lengths _tm_months[0,1] =3D _tm_months[1,1] =3D 31 _tm_months[0,2] =3D 28; _tm_months[1,2] =3D 29 _tm_months[0,3] =3D _tm_months[1,3] =3D 31 _tm_months[0,4] =3D _tm_months[1,4] =3D 30 _tm_months[0,5] =3D _tm_months[1,5] =3D 31 _tm_months[0,6] =3D _tm_months[1,6] =3D 30 _tm_months[0,7] =3D _tm_months[1,7] =3D 31 _tm_months[0,8] =3D _tm_months[1,8] =3D 31 _tm_months[0,9] =3D _tm_months[1,9] =3D 30 _tm_months[0,10] =3D _tm_months[1,10] =3D 31 _tm_months[0,11] =3D _tm_months[1,11] =3D 30 _tm_months[0,12] =3D _tm_months[1,12] =3D 31 } # decide if a year is a leap year function _tm_isleap(year, ret)
  {
  ret =3D (year % 4 =3D=3D 0 && year % 100 !=3D 0) || (year % 400 =3D=3D 0) =20 return ret } # convert a date into seconds function _tm_addup(a, total, yearsecs, daysecs, hoursecs, i, j)
  {
  hoursecs =3D 60 * 60 daysecs =3D 24 * hoursecs yearsecs =3D 365 * daysecs =20 total =3D (a[1] - 1970) * yearsecs =20 # extra day for leap years for (i =3D 1970; i < a[1]; i++) if (_tm_isleap(i)) total +=3D daysecs =20 j =3D _tm_isleap(a[1]) for (i =3D 1; i < a[2]; i++) total +=3D _tm_months[j, i] * daysecs =20 total +=3D (a[3] - 1) * daysecs total +=3D a[4] * hoursecs total +=3D a[5] * 60 total +=3D a[6] =20 return total } # mktime --- convert a date into seconds, # compensate for time zone =20 function mktime(str, res1, res2, a, b, i, j, t, diff)
  {
  i =3D split(str, a, " ") # don't rely on FS =20 if (i !=3D 6) return -1 =20 # force numeric for (j in a) a[j] +=3D 0 =20 # validate if (a[1] < 1970 || a[2] < 1 || a[2] > 12 || a[3] < 1 || a[3] > 31 || a[4] < 0 || a[4] > 23 || a[5] < 0 || a[5] > 59 || a[6] < 0 || a[6] > 60 ) return -1 =20 res1 =3D _tm_addup(a) t =3D strftime("%Y %m %d %H %M %S", res1) =20 if (_tm_debug) printf("(%s) -> (%s)\n", str, t) > "/dev/stderr" =20 split(t, b, " ") res2 =3D _tm_addup(b) =20 diff =3D res1 - res2 =20 if (_tm_debug) printf("diff =3D %d seconds\n", diff) > "/dev/stderr" =20 res1 +=3D diff =20 return res1 } BEGIN { if (_tm_test) { printf "Enter date as yyyy mm dd hh mm ss: " getline _tm_test_date t =3D mktime(_tm_test_date) r =3D strftime("%Y %m %d %H %M %S", t) printf "Got back (%s)\n", r } } /tim=3D/ { timv=3Dsubstr($0,index($0,"tim=3D")+4) timv=3Dsubstr(timv,1,match(timv,/ |$/)-1) stim=3D(6*17895697051) #Epoch in HSecs less the v$timer timv=3Dstim + timv #Add v$timer=20 timv=3Dint(timv/100) #Convert from HSecs to Seconds ##sub(/tim=3D[0-9]*/,"& " strftime("%Y %m %d %H %M %S", timv) " ("timv ")") sub(/tim=3D[0-9]*/,"& " strftime("%m/%d/%Y %H:%M:%S", timv)) } {print $0}

-----Original Message-----
From: DEEDSD_at_Nationwide.com [mailto:DEEDSD_at_Nationwide.com] Sent: Tuesday, April 27, 2004 11:59 AM
To: oracle-l_at_freelists.org
Subject: Re: time in trace file

There's a Perl script in Cary's book that converts from tim to real date/time and back. It's example 6-10 on page 134.

=20

                          "Sultan Syed"

                          <ssyed_at_fine.ae>          T

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                                                   Subject:
time in trace file         =20

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=20

                          04/27/2004 02:48 AM

                          Please respond to

                          oracle-l

=20

=20

Hi guys,
How can I interpret with this time format,I know this is the time of parse call is over,but how can I convert this into date format.Is it possible?
PARSE =
#30:c=3D2,e=3D15,p=3D0,cr=3D0,cu=3D0,mis=3D1,r=3D0,dep=3D0,og=3D4,tim=3D9= 33354557
Thanks in advance
Syed

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Received on Wed Apr 28 2004 - 16:31:46 CDT