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Shell Programming and Scripting

how to assign to each variable after selected?

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Top Forums Shell Programming and Scripting how to assign to each variable after selected?
# 8  
Old 04-13-2009
So,

data_type_SQL="|G |NRT |NRTW |" put double quotas
so you'll got
echo $data_type_SQL
|G |NRT |NRTW |

localty=$(echo $data_type_SQL | cut -d\| -f2| sed -e 's/|//g')
# echo $localty
G

whareH=$(echo $data_type_SQL | cut -d\| -f3| sed -e 's/|//g')
# echo $whareH
NRT

localid=$(echo $data_type_SQL | cut -d\| -f4| sed -e 's/|//g')
# echo $localid
NRTW

this is working in my shell, if i understood

Hope it helps

Bye
# 9  
Old 04-13-2009
Yes.... it is working..
Thank you so... sosososososo much..!!!
i will work hard!!!
terima kasih...
# 10  
Old 04-13-2009
Try:

Code:
### Remove hyphen ###

$CONNECT cat << __EOF__ |

SET NOCOUNT ON
select LOC_TYPE, WAREHOUSE, LOCATION_ID from JEALOCATION where
LOCATION_ID="$data_LocID_SQL"

__EOF__

read LOC_TYPE WAREHOUSE LOCATION_ID

Quote:
got value inside in each LOC_TYPE, WAREHOUSE, LOCATION_ID from data_type_SQL variable..

but when i try to assign each variable... those 3 is empty..

localty=$(echo "$LOC_TYPE")
wareH=$(echo "$WAREHOUSE")
localid=$(echo "$LOCATION_ID")
First of all, you have not declared any variables name LOC_TYPE, WAREHOUSE.. etc...
Those are column names from your SQL and are not automatically exported with
their values in tact to the shell.

Secondly, if they were, you could make the assignments like this, but it would
be unnecessary and better practice just to leave them as their column names.

Code:
localty=$LOC_TYPE
wareH=$WAREHOUSE
localid=$LOCATION_ID

Last edited by quirkasaurus; 04-13-2009 at 11:13 AM.. Reason: slight error in script.
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