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How do I Substring ??


 
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# 1  
Old 04-25-2007
How do I Substring ??

Hello everyone.

I'm writing a script in UNIX. The purpose is to get the second character from a variable that stores the system year.

This is the code:
Code:
unix_year_yy=`date "+%g"`

This will return "07" in variable unix_year_yy. How can I get the second character (7)??
# 2  
Old 04-25-2007
You could do this

Code:
let unix_year_yy=`date "+%g"`

# 3  
Old 04-25-2007
Quote:
Originally Posted by vino
You could do this

Code:
let unix_year_yy=`date "+%g"`

Thanks vino!! Smilie
# 4  
Old 04-25-2007
Next year that may fail. You will effectively be doing:
let yy=08
the leading zero, according to posix, indicates an octal constant and 08 is not a legal octal constant. Also you should be using %y.
Code:
$ yy=$(date +%y)
$ echo $yy
07
$ yy=${yy#0}
$ echo $yy
7
$

# 5  
Old 04-25-2007
Quote:
Originally Posted by Perderabo
Next year that may fail. You will effectively be doing:
let yy=08
the leading zero, according to posix, indicates an octal constant and 08 is not a legal octal constant. Also you should be using %y.
Code:
$ yy=$(date +%y)
$ echo $yy
07
$ yy=${yy#0}
$ echo $yy
7
$

Thanks Perderabo. I'm not familiar to UNIX, so any help is appreciated.

Thank you both!! Smilie
# 6  
Old 04-25-2007
Quote:
Originally Posted by Rigger
Code:
unix_year_yy=`date "+%g"`

This will return "07" in variable unix_year_yy. How can I get the second character (7)??

If you really want the second character of a 2-character value:

Code:
unix_year_yy=${unix_year_yy#?}

On the other hand, if what you really want is to remove a leading zero, but leave a number greater than 10 as 2 digits, read on.

If you have GNU date (standard on Linux systems):

Code:
unix_year_yy=`date "+%-g"`

Otherwise, remove the leading zero (if there is one) with parameter expansion:

Code:
unix_year_yy=`date "+%g"`
unix_year_yy=${unix_year_yy#0}

# 7  
Old 04-26-2007
Thank u 2 cfajohnson! Smilie

Very helpfull post!!

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