On 14 Sep 2011, at 9:46pm, Fabio Spadaro wrote: > Hello. I use sqlite3 python. > I followed exactly the steps to create a foreign key from the site > http://www.sqlite.org/foreignkeys.html but I still can insert enter values > not allowed for constraint.
So you get no error from the INSERT command, and you can also find the new row using SELECT ? > Before I do the insert properly executed the instruction "PRAGMA > FOREIGN_KEYS =ON" but did not return any data, the site shows that if the > result is not 0 or 1 > then either the version of sqlite is more old of 3.6..19 (MAKES IT IS MY > VERSION 3.6.22) or it has been compiled with SQLITE_OMIT_FOREIGN_KEY > SQLITE_OMIT_TRIGGER ago. > Someone can give me light? Make a connection to the database. Issue "PRAGMA foreign_keys" and look at the value returned. Issue "PRAGMA foreign_keys = OFF". Issue "PRAGMA foreign_keys" and look at the value returned. Issue "PRAGMA foreign_keys = ON". Issue "PRAGMA foreign_keys" and look at the value returned. Please post a follow-up with the three values. Simon. _______________________________________________ sqlite-users mailing list sqlite-users@sqlite.org http://sqlite.org:8080/cgi-bin/mailman/listinfo/sqlite-users
