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How to replace variable inside the variable

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Top Forums Shell Programming and Scripting How to replace variable inside the variable
# 1  
Old 03-21-2007
How to replace variable inside the variable
hi sir,
i need your help for this script

Quote:
USERID=username/passwd
BILLDATE=19-FEB-07
STARTPARTNNUM=101
TOTALPARTN=20
SQLLOG=${BILLDATE}_xMNBDF045_P_CTEL.log
SCRIPT=`cat /rnmucdr/ednms05/ken/xMNBDF045_Script.sql`

sqlplus -s $USERID > $SQLLOG << EOF
WHENEVER SQLERROR EXIT 1
$SCRIPT
EOF
if [ $? -ne 0 ]
then
cat $SQLLOG
else
echo "SUCCESSFULLY FINISHED" > $SQLLOG
fi
inside /rnmucdr/ednms05/ken/xMNBDF045_Script.sql content variable like this

select * from invoice where bill_date=$BILLDATE and startNum=$STARTPARTNNUM and total_partn=$TOTALPARTN

if i just paste this replace with the $SCRIPT it works great,if i put script other place with with the variable,and put in the varaible script,it not works because the variable is not subtitute..
can anyone help me to solve this problem,i need to replace it with the parameter given here
# 2  
Old 03-21-2007
try replacing $script by cat /rnmucdr/ednms05/ken/xMNBDF045_Script.sql

like this

Code:
USERID=username/passwd
BILLDATE=19-FEB-07
STARTPARTNNUM=101
TOTALPARTN=20
SQLLOG=${BILLDATE}_xMNBDF045_P_CTEL.log
SCRIPT=`cat /rnmucdr/ednms05/ken/xMNBDF045_Script.sql`

sqlplus -s $USERID > $SQLLOG << EOF
WHENEVER SQLERROR EXIT 1
cat /rnmucdr/ednms05/ken/xMNBDF045_Script.sql
EOF
if [ $? -ne 0 ]
then
cat $SQLLOG
else
echo "SUCCESSFULLY FINISHED" > $SQLLOG
fi

# 3  
Old 03-21-2007
Add slash
Code:
select \* from invoice where bill_date=$BILLDATE and startNum=$STARTPARTNNUM and total_partn=$TOTALPARTN

Replace
Code:
SCRIPT=`cat /rnmucdr/ednms05/ken/xMNBDF045_Script.sql`

by
Code:
SCRIPT=$( eval echo $(cat /rnmucdr/ednms05/ken/xMNBDF045_Script.sql))

# 4  
Old 03-21-2007
syntax error
0403-057 Syntax error at line 1 : `(' is not expected.

for

Code:
SCRIPT=$( eval echo $(cat /rnmucdr/ednms05/ken/xMNBDF045_Script.sql))

If i do this
Code:
echo $SCRIPT

nothing display..only empty line printed
# 5  
Old 03-21-2007
can you show the code in /rnmucdr/ednms05/ken/xMNBDF045_Script.sql?
# 6  
Old 03-21-2007
here is the code

Code:
USERID=USER/pass
SCRIPTFILE=/rnmucdr/ednms05/ken/xMNBDF045_Script.sql
BILLDATE=19-FEB-07
STARTPARTNNUM=101
TOTALPARTN=20
SQLLOG=${BILLDATE}_xMNBDF045_P_CTEL.log
echo $SQLLOG
echo $SCRIPTFILE
SCRIPT=$( eval echo $(cat $SCRIPTFILE))

sqlplus -s $USERID > $SQLLOG << EOF
WHENEVER SQLERROR EXIT 1
$SCRIPT
EOF

if [ $? -ne 0 ]
then
cat $SQLLOG
else
echo "SUCCESSFULLY FINISHED" > $SQLLOG
fi

HERE IS THE SQL STAMENT /rnmucdr/ednms05/ken/xMNBDF045_Script.sql
Code:
create table acct_to_print_bak as
select BILL_DATE,ACCT_NO,STATUS_CODE,floor((ROWNUM-1)/N)+$STARTPARTNNUM BP_PARTN_NUM,SYS_APPL_ID FROM (
select T1.BILL_DATE,T1.ACCT_NO,T1.STATUS_CODE,T1.BP_PARTN_NUM,T1.SYS_APPL_ID,T3.N from acct_to_print t1,inv_acct_bill_addr T2
,(select ceil(count(*)/$TOTALPARTN) N from inv_acct_bill_addr) T3 where t2.bill_date='$BILLDATE' and t2.bill_date=t1.bill_dat
e and t1.ACCT_NO=t2.ACCT_NO order by T2.postal_code ) T4;
commit;
truncate table acct_to_print;
insert into acct_to_print (select * from acct_to_print_bak);
commit;
drop table acct_to_print_bak;

Any Idea
# 7  
Old 03-21-2007
what's the error you are getting from my code???? And also, remember all the multiple lines will be converted to one line when assignint it to a variable.
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