How to grep lines from a log file that have the current date?

Question

I need to grep a log file with today\'s date, but the output is showing for more than today\'s date.

grep date +\"20%y-%m-%d\" /path/log/General.log | grep \         


        

Answer 1

If you need to add space delimited fields in date command use double quotes around $() :

$ grep "$(date +"%Y-%m-%d %H:%M")" file
2013-06-21 00:01:24,915 - INFO   
2013-06-21 00:01:24,915 - INFO

Answer 2

Just use the date output as a pattern in grep:

$ grep "$(date +"%Y-%m-%d")" file
2013-06-21 00:01:24,915 - INFO   
2013-06-21 00:01:24,915 - INFO

That is, you need to enclose the date sentence to make it be processed. Also, note I used Y instead of your 20%y.

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I am looking for a sepcific EmpID in the logs with current date.

Then pipe to another grep:

$ grep $(date +"%Y-%m-%d") file | grep "EmpID#106496"
2013-06-21 00:01:24,915 - INFO  EmpID#106496 
2013-06-21 00:01:24,915 - INFO EmpID#106496 

Answer 3

You're not actually executing 'date', because it's not been surrounded by backticks. The command should be

grep `date +"20%y-%m-%d"` /path/log/General.log
#    ^--                ^--

Right now, as written, you are searching for the word date in two different files (+"20%Y-%m-%d", and general.log).