awk help - Simple question

Shell Programming and Scripting
1

I have what a think is a simple question but I'm just a beginner in scripting. I'm my unix command line I run a date command that returns the following:

Wed Apr  3 10:39:30 EDT 2013

How do I awk out the "10" only in awk? Or is awk the way to do it or is there a better way?

2

You can simply run date with format control to get hour:

date +%H
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4 
echo 'Wed Apr  3 10:39:30 EDT 2013' | awk -F'[ :]' '{print $4}'
date +%H
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5

Thanks. How can I get it to recognize the hour variable?

hour=date +%H -d "1 hour ago"

i=test.`date +%Y%m%d`-$hour3101.txt
6

Use command substitution $(..)

hour=$(date +%H -d "1 hour ago")
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7 
hour=$(date +%H -d "1 hour ago")

i=$(grep Multiple test.`date +%Y%m%d`-$hour3101.txt)

I'm still missing something. I'm getting the following error:

grep: test.20130403-.txt: No such file or directory..

So it's not picking up the hour variable.

8 
i=$( grep Multiple "test.`date +%Y%m%d`-${hour}3101.txt" )

The curly brace form is unambiguous as the variable name is clearly delimited.

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9

Would it not it be better to remove the back ticks complete?

i=$( grep Multiple "test.$( date +%Y%m%d )-${hour}3101.txt" )