im having trouble tou remove some specific lines from a file, but its not that simple. Tthe problem asks to search and then save the lines from a specific date and hour for example 2008-06-08 and 12:40:53.90 AM until the date that the user wants which is a read variable input $dayy. The real issue is that i want the lines that is greater after 12:40:53 until that if the $dayy is for example 2008-06-12 i need the lines that exist before the specific time 12:40:53.90, exactly  after 4 days. i used the bash code below and with the last awk command i wanted to remove the lines that are in that day ($dayy) and their time are after the 12:40:53.90 but it removes all the lines in that date. if you could help i will really appreciate that.
for example if the initial day is 2008-06-08 (spec_date) and time 12:40:53.90 (timee) then i want all the lines from that day, time until the SAME time in two days 2008-06-10 (dayy)
read dayy ;  read spec_date ; read timee ;
awk -v dayy $dayy -v spec_date=$spec_date -v timee=$timee '{ if($1 >=spec_date && $2>=timee  && $1 <= dayy) print $0}' ex1.dat > ex2.dat
awk -v dayy=$dayy '! ( $1==dayy  &&   $2 ="[12-23]:[0-59]:[0-59].[0-50]")' ex2.dat
example of my data
2008-06-08  12:40:53.90  
2008-06-08  12:43:39.80 
.......................
2008-06-11  **12:39:34.22**
2008-06-11    13:21:23:43
 
    