Escaping Problem in bash using isql

Posted by latz on Stack Overflow See other posts from Stack Overflow or by latz
Published on 2010-06-02T08:30:20Z Indexed on 2010/06/02 8:33 UTC
Read the original article Hit count: 198

Filed under:
|

Hi there, I am currently working on a little backup script from some firebird databases and I've come up with a weird escaping problem that I don't seem to be able to solve. Here's the thing in my script I create a variable called sqllog in which I would like to put the output of a chain of commands, here it is.

sqllog=echo "SELECT * FROM RDB\$DATABASE;" | isql -u SYSDBA -pass mypasswd localhost:mydatabase | tail -n 2 | head -n 1 | wc -l

if I try to execute this in shell I get the following error

Statement failed, SQLCODE = -204

Dynamic SQL Error -SQL error code = -204 -Table unknown -RDB -At line 1, column 15.

Table unknown RDB means it didn't take my try to escape the $.

thx for any help :)

© Stack Overflow or respective owner

Related posts about bash

Related posts about isql