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New to UNIX ... Date parsing

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# 1  
Old 01-22-2013
New to UNIX ... Date parsing
Hi ... extremely new to Unix scripting ... I have to get a date field from mm/dd/yyyy to yyyy/mm/dd format ... I have a variable that I want parsed. I don't seem to be doing it right?? Thanks.
# 2  
Old 01-22-2013
Using bash sub-string:
Code:
DT="10/01/2012"
echo ${DT:6:4}/${DT:0:2}/${DT:3:2}
2012/10/01

# 3  
Old 01-22-2013
Code:
echo "${var##*/}/${var%${var##*/}}"

Code:
IFS=/ read m d y << EOF
$var
EOF
echo "$y/$m/$d"

# 4  
Old 01-22-2013
Okay

Here is what I entered

Code:
 
linedate="12/21/2013"
lineyyyy=${linedate:6:4}
linemm=${linedate:0:2}
linedd=${linedate:3:2}
echo $linedate $lineyyyy/$linemm/$linedd

nothing is being echoed ??? Am I doing something wrong?
Last edited by Scott; 01-22-2013 at 12:10 PM.. Reason: Code tags
# 5  
Old 01-22-2013
What OS & shell are you using? Can you run below commands and post results?
Code:
uname
echo $SHELL

# 6  
Old 01-22-2013
Linux
/bin/bash

Told you I was new to this Smilie
# 7  
Old 01-22-2013
Works for me!
Code:
$ linedate="12/21/2013"
$ lineyyyy=${linedate:6:4}
$ linemm=${linedate:0:2}
$ linedd=${linedate:3:2}
$ echo $linedate $lineyyyy/$linemm/$linedd
12/21/2013 2013/12/21

I am using bash version 3.2.25
Code:
$ bash --version
GNU bash, version 3.2.25(1)-release (x86_64-redhat-linux-gnu)
Copyright (C) 2005 Free Software Foundation, Inc.
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