logging - Grep a log file with current date -

I need to griddle a log file with today's date but the output is shown for more than today's date. grep date + "20% y-% m-% d" /path/log/General.log | Grep "EmpID # 106496" /path/log/Genral.log

Output:

  2013-06-19 14: 47: 05,996 - Information EmpID # 106496 2013-06-19 14: 47: 05,996 - Information EmpID # 106496 2013-06-21 00: 01: 24,915 - Information EmpID # 106496 2013-06-21 00: 01: 24,915 - Information EmpID # 106496   
 
 
 just type  date  in the form of a pattern in  grep Use in :  
  $ grep "$ (date +"% Y-% m-% d ")" file 2013-06-21 00: 01: 24,915 - Information 2013-06-21 00: 01: 24,915 - info   
 That is, to include the code, you have to include the  date  Will. Also, note that I have used  y  instead of your  20% y .  
 
 
 I have a valid  EmpID          
 -% d ") file | Grep" EmpID # 106496 "2013-06-21 00: 01: 24,915 - Information EmpID # 106496 2013-06-21 00: 01: 24,915 - Information EmpID # 106496