I'm completely stumped. There must be SQL statement created to accomplish the link for the subform. I found error detail on the sql statement. Here is the statement for one of the errors: The SQL command leading to this error is: SELECT * FROM "aquarium"."corrections" WHERE ( "aquarium"."corrections"."testID" = :link_from_ID ) The statement for the other database is: The SQL command leading to this error is: SELECT * FROM "running"."workout" WHERE ( "running"."workout"."ID" = :link_from_workoutID ) For both of these statements the master field is the last portion of the generated name "ID" for the first and "workoutID" for the second. Now if that is how the Now if I run one of those statements directly from the sql view of query design, I'm asked to provide a parameter. If I provide a valid value, I get the exact error message that we got before. Now if I replace the variable name with the parameter I just gave in the dialogue, I now get the result I expect. So the bug must have to do with the link parameter not getting replaced with the appropriate value. Does this help? On Thu, 2019-08-01 at 18:31 +0200, Rene Engelhard wrote: Hi again,On Thu, Aug 01, 2019 at 12:04:58AM +0200, Rene Engelhard wrote:On Wed, Jul 31, 2019 at 04:00:04PM -0500, william l-k wrote:have linked sub-forms from two seperate tables for entering related data.One of the databases started out as a libreoffice base then was convertedto a mariadb connection. The other started out as a connection to mariadb.Maybe it's not related at all but connection to mariadb how? MySQL ODBC?MySQL JDBC? (both of which are removed because being buggy), MariaDBJDBC? Or libreoffice-mysql-connector?If the latter, 5.2 removed the extension and need for libmysqlcppconnbut integrated it properly into the "main" packages |