Description of problem: Parameters do not substitute correctly when using the LIKE keyword in a WHERE clause of a query. Version-Release number of selected component (if applicable): libreoffice-base-4.4.4.3-6.fc22.x86_64 How reproducible: always Steps to Reproduce: 1. Create a query using LIKE without a parameter and execute it. It returns records. 2. Substitute a parameter for part of the LIKE. 3. Execute it. Receive error. Actual results: Get error. Using More from error dialog shows "SQL Status: 22023 The column index is out of range: 1, number of columns: 0." Expected results: Data should return. Additional info: Here are two queries. One without parameter that works and one with that does not work. The parameter is prompted for by LibreOffice, the query is just not executed properly. SELECT "first_name", "last_name", "spouse", "address", "ml_num", "ml_address" FROM "floridasun"."contacts" "contacts" WHERE lower("address") like '%' || 'venice' || '%' SELECT "first_name", "last_name", "spouse", "address", "ml_num", "ml_address" FROM "floridasun"."contacts" "contacts" WHERE lower("address") like lower('%' || :address || '%')