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New to UNIX ... Date parsing

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# 8  
Old 01-22-2013
I typed linedat="12/12/2013"| echo ${linedat:6:4}/${linedat:0:2} on the command line and all I got back was /

How do I tell what version of bash I'm using
Last edited by Scrutinizer; 01-22-2013 at 12:29 PM.. Reason: code tags
# 9  
Old 01-22-2013
Code:
bash --version

# 10  
Old 01-22-2013
okay ... running
Code:
GNU bash, version 3.2.51(1)-release (x86_64-suse-linux-gnu)
Copyright (C) 2007 Free Software Foundation, Inc.

Moderator's Comments:
Mod Comment Please use code tags
# 11  
Old 01-22-2013
Not sure what is wrong! You can use Scrutinizer's solution in post #3 or use an awk program:
Code:
echo $linedate | awk -F"/" '{ print $3,$1,$2 } ' OFS="/"
2013/12/21

# 12  
Old 01-22-2013
Trying to learn this on my own through trial and error, think it may be time to ask the big guns here ... will let you know .. thanks for your help ... This is not like Cobol Smilie
# 13  
Old 01-22-2013
Quote:
Originally Posted by MJKeeble
I typed linedat="12/12/2013"| echo ${linedat:6:4}/${linedat:0:2} on the command line and all I got back was /
[..]
The | should not be there.

bash/ksh93:
Code:
linedat=01/14/2013
echo "${linedat:6:4}/${linedat:0:5}"

# 14  
Old 01-22-2013
I thought that might be needed, I was trying to type it in native unix, so thought it might be needed to type on one line

---------- Post updated at 11:58 AM ---------- Previous update was at 11:41 AM ----------

okay ... so the code works in native unix, not sure why it is not working in my script. Back to the drawing board ... Thank you for your help!
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