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Using $variables in grep

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# 1  
Old 10-18-2014
Using $variables in grep
Hi, I'm currently trying to use variables in grep on my script. Printing the variable via echo works fine. Also, if I hard coded the date of the appointment it works just fine. But, if I try to use the $DATE as an argument in grep it doesn't do anything.

Code:
#!/bin/bash

DATE=${2}/${3}/${1}     #take in arguments such as ./appoint.sh 2015 01 25

   echo "Looking for appointments on $DATE ...."  # The $DATE prints in format 01/25/2015
   grep -r "$DATE" *


Can anyone help me?
Last edited by nuclearpenguin; 10-18-2014 at 09:08 PM..
# 2  
Old 10-18-2014
Try putting quotes around the values for DATE.

Looks fine to mine.
Are you sure the files contains this specific format?

AFAIK: 'Official' file extension for bash scripts is .sh.
This User Gave Thanks to sea For This Post:
nuclearpenguin
# 3  
Old 10-18-2014
I tried putting quotes around the values of $DATE, but still doesn't seem to do anything.

Also, yes, I believe the format is correct. The string I'm trying to look for that is contained within the files is formatted as "01/25/2015"

If I try
Code:
grep -r "01/25/2015" *

it works just fine, but with $DATE it doesn't seem to work.
# 4  
Old 10-18-2014
Let's debug it. Could you add the following to it and afterward post the result as it shows in your screen?
Code:
#!/bin/bash
set -x

DATE=${2}/${3}/${1}     #take in arguments such as ./appoint.bash 2015 01 25

   echo "Looking for appointments on $DATE ...."  # The $DATE prints in format 01/25/2015

printf "Where am I before executing grep?: %s\n" $PWD
   grep -r "$DATE" *

# 5  
Old 10-18-2014
Ok I ran the script with
Code:
 $ ./appoint.sh 2015 25 01

and got
Quote:
+ DATE=01/25/2015
+ echo 'Looking for appointments on 01/25/2015 ....'
Looking for appointments on 01/25/2015 ....
+ printf 'Where am I before executing grep?: %s\n' /home/nukep/practice/Files/MailArchive
Where am I before executing grep?:
/home/nukep/practice/Files/MailArchive
+ grep -r 01/25/2015 001 002 003 004 005
# 6  
Old 10-18-2014
Code:
+ grep -r 01/25/2015 001 002 003 004 005

The code shows that it is executing as expected.
grep is searching for the string 01/25/2015 in files: 001, 002, 003, 004 and 005. If that's correct, it is not finding any lines with that 01/25/2015 pattern.

You can verify and see if you find any 01/25/2015 in 001, 002, 003, 004 or 005, manually without grep.
Last edited by Aia; 10-18-2014 at 09:38 PM.. Reason: more
This User Gave Thanks to Aia For This Post:
nuclearpenguin
# 7  
Old 10-18-2014
Oh, I see.
I was inputting the arguments wrong. Instead of writing the year month day, I was writing the year day month. It's working now.
Thanks for the help everyone!
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