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Variable substitution error

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# 1  
Old 12-12-2007
Variable substitution error
Hi
The following command gives me error :The specified substitution is not valid for this command

export DB_ID=${DB_${MART}_USER}

What is the correct syntax for the above ?
# 2  
Old 12-12-2007
something like

Code:
DB_ID=`sh -c "echo DB_${MART}_USER"`
DB_ID=`sh -c "echo \$$DB_ID"`
export DB_ID

or you could try eval
# 3  
Old 12-12-2007
I think you want this:

Quote:
$ MART=wall
$ DB_wall_USER=wallmart
$ eval eval export DB_ID=\${DB_${MART}_USER}
$ echo $DB_ID
wallmart
HTH.
# 4  
Old 12-12-2007
Hi

Thanks for the suggestion but It doesnt seem to retrieve the value of the variable correctly in 2

(1)
DB_ID=`sh -c "echo DB_${MART}_USER"`

Result = DB_XMM_USER
(2)
DB_ID=`sh -c "echo \$$DB_ID"`
Result = 1233DB_ID

I need to be able to get the value of DB_XMM_USER ie ${DB_XMM_USER}
# 5  
Old 12-12-2007
Quote:
Originally Posted by cosec
DB_ID=`sh -c "echo \$$DB_ID"`
Result = 1233DB_ID
Seems you did

DB_ID=`sh -c "echo $$DB_ID"`

where $$ expanded to the current process id.

The slash prefixing the first $ is to have it taken literally.
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