How to feed the results of date command into grep to filter results of another command

Question

I need to use the result of a formatted date command (date +"%m/%d") as the grep filter to filter the results of another command that will display alerts on a system so that I only see alerts from the given day.

Lets say alertcommand gives the alert readout in the console. I know I can do alertcommand | grep 8/10 to get the logs from today, but I want to be able to feed date +"%m/%d"into that so that I don't have to specify the date each time I run it, but I can't figure out how to link it into one single command (preferably without having to create temp files or anything as this will be run on customer hardware).

I looked at this question that was similar, but couldn't figure out how to make it work the same way.

Answer 1

In bash, just use something like alertcommand | grep $(date +"%m/%d")

$() executes a command in a subshell and returns the output of the command as string. Alternatively you can enclose the command with backticks to the same effect.

Answer 2

You could simply assign the output of your date command to another variable, and use that as an argument:

myDateVariable=`date +"%m/%d"`
alertcommand | grep $myDateVariable

This way, you could re-use the date value in the future and also debug any intermediate steps

Answer 3

In FreeBSD's shell csh I can't use both of yours methods.

1)

cat /var/log/exim/mainlog | grep $(date +%Y-%m-%d)
Illegal variable name.

2)

d=`date +%Y-%m-%d`;cat /var/log/exim/mainlog|grep $d
d=2011-11-13: Command not found.

But I can use directly backticked code like this:

cat /var/log/exim/mainlog | grep `date +%Y-%m-%d`