variables


 
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# 1  
Old 07-25-2007
variables

Here is my code:

DATETIME=1214163003

echo $DATETIME | sed 's/....//' | sed 's/....$//' | read DATEHOUR
echo $DATETIME | sed 's/......//' | sed 's/..$//' | read DATEMIN

echo $DATEHOUR
echo $DATEMIN

The idea is that i need the 16 stored in DATEHOUR and the 30 stored in DATEMIN but when i run the code it just prints 2 blank lines.

What am i doing wrong?
# 2  
Old 07-25-2007
If the DATETIME variable is static then you can make use of cut

Code:
echo $DATETIME | cut -c5-6 | read DATEHOUR
echo $DATETIME | cut -c7-8 | read DATEMIN

Or sed like

Code:
echo $DATETIME | sed 's/[0-9]\{4\}\([0-9]\{2\}\).*/\1/' | read DATEHOUR
echo $DATETIME | sed 's/[0-9]\{6\}\([0-9]\{2\}\).*/\1/' | read DATEMIN

# 3  
Old 07-25-2007
my problem isnt with getting to the numbers (altho your suggestions are nicer!). The code i used with sed got the numbers but it doesnt seem to store them in the new variables. when i run the script in order to print out the new variables it just returns 2 blank lines
# 4  
Old 07-25-2007
Might be you should be using the bash shell to execute the commands. I think your code works perfectly with Ksh

You code

Code:
$ >ksh -x user_c.sh
+ DATETIME=1214163003
+ echo 1214163003
+ sed s/....//
+ sed 's/....$//'
+ read DATEHOUR
+ echo 1214163003
+ sed s/......//
+ sed 's/..$//'
+ read DATEMIN
+ echo 16
16
+ echo 30
30

If you using bash shell then your code should look like


Code:
DATEHOUR=`echo $DATETIME | sed 's/....//' | sed 's/....$//'`
DATEMIN=`echo $DATETIME | sed 's/......//' | sed 's/..$//'`

# 5  
Old 07-25-2007
yeh i was being a bit special there! thanks
 
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