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delete block of lines when pattern does not match

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# 1  
Old 05-06-2010
Question delete block of lines when pattern does not match
I have this input file that I need to remove lines which represents more than 30 days of processing.

Input file:
Code:
On 11/17/2009 at 12:30:00, Program started processing...argc=7
Total number of bytes in file being processed is 390
Message buffer of length=390 was allocated successfully
Number of bytes read from input file is 121756 = 390
Number of bytes put  = 390
Total Number messages  = 1
On 11/17/2009 at 12:30:00, Program has completed processing. RC = 0
****************************************************************************
On 05/05/2010 at 12:30:00, Program started processing...argc=7
Total number of bytes in file being processed is 390
Message buffer of length=390 was allocated successfully
Number of bytes read from input file is 121756 = 390
Number of bytes put  = 390
Total Number messages  = 1
On 05/05/2010 at 12:30:00, Program has completed processing. RC = 0
****************************************************************************

Output file should look like this:
Code:
On 05/05/2010 at 12:30:00, Program started processing...argc=7
 Total number of bytes in file being processed is 390
 Message buffer of length=390 was allocated successfully
 Number of bytes read from input file is 121756 = 390
 Number of bytes put  = 390
 Total Number messages  = 1
 On 05/05/2010 at 12:30:00, Program has completed processing. RC = 0
 ****************************************************************************

On a previous thread, I have used the below lines of code to check on the date on the line if it is greater than or equal to 30 and to print a number of lines when the date condition was met.

Code:
Days=30
Before=$(printf "%4d%02d%02d" $(datecalc -a $(date +'%Y %m %d') - ${Days}))
awk -F '[/ ]*' -v Before=${Before} '
{
   date = $4 $2 $3;
   if (date >= Before) {
      print;
      printSepLine = 7;
   }
} 
/^\*+/ && printSepLine {
   print;
   printSepLine = 0;
}
' log

However I'm getting this result:

Code:
Total number of bytes in file being processed is 390
Message buffer of length=390 was allocated successfully
Number of bytes read from input file is 121756 = 390
Number of bytes put  = 390
Total Number messages  = 1
**************************************************************************** 
On 05/05/2010 at 12:30:00, Program started processing...argc=7
Total number of bytes in file being processed is 390
 Message buffer of length=390 was allocated successfully
Number of bytes read from input file is 121756 = 390
Number of bytes put  = 390
Total Number messages  = 1
On 05/05/2010 at 12:30:00, Program has completed processing. RC = 0
****************************************************************************

The block of lines in between dates that were older than 30 days were retained from the output (blue lines above). How do I delete those block of lines?
# 2  
Old 05-06-2010
Try this:
Code:
Days=30
Before=$(printf "%4d%02d%02d" $(datecalc -a $(date +'%Y %m %d') - ${Days}))

awk -F"[/ ]" -v Before=${Before} '
/^On/{d=$4 $2 $3; if(d < Before){f=1}}
/^\*/ && f {f=0; next}
!f
' log

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