variable passed to sql from shell is not working - sql

Question

My code is:
#!/bin/sh
cat tmp_ts.log | awk ' {print $8}'
lookup=$8
sqlplus -s "sys/Orcl1234 as sysdba" << EOF
SELECT tablespace_name FROM dba_tablespaces WHERE tablespace_name='$lookup';
exit;
EOF
and my output is:
IAM_OIM
no rows selected
In this variable lookup I have passed to select statement but it's not working.
My end result should be with select statement. See below the output of select query:
See below:
My end result should be this but that variable is not working in select statement.

Answer 1

#!/bin/sh
lookup="$(awk '/tablespace/{print $8;exit}' tmp_ts.log)"
echo "Querying database with lookup = $lookup"
sqlplus -s "sys/Orcl1234 as sysdba" <<EOF
SELECT tablespace_name FROM dba_tablespaces WHERE tablespace_name='$lookup';
exit;
EOF
You have to use awk's output to set lookup. The shell knows nothing about the $8 which was set in awk. Also, I have ensured that awk exits after the first matching line, so that there is no risk of returning multiple values, or simply empty lines as it did in your version.

Answer 2

You can fill lookup with a command like awk, sed or cut.
lookup=$(cut -d" " -f8 tmp_ts.log)
You should add some checks, like #Dario did (with an exit after the first match and only converting lines with tablespace, but what to do when no lines match?).
When you don't add the checks you can skip setting the $lookup:
sqlplus -s "sys/Orcl1234 as sysdba" << EOF
SELECT tablespace_name FROM dba_tablespaces
WHERE tablespace_name='$(sed 's/.*tablespace- //' tmp_ts.log)';
exit;
EOF