issue_comments: 1683118376

The problem here is that the table created by this line: python fresh_db.create_table("breeds", {"name": str}) Has this schema: sql CREATE TABLE [breeds] ( [name] TEXT ); SQLite creates an invisible rowid column for it automatically.

On the main branch with the old implementation that table ends up looking like this after the foreign key has been added: sql (Pdb) print(fresh_db.schema) CREATE TABLE [dogs] ( [name] TEXT , [breed_id] INTEGER, FOREIGN KEY([breed_id]) REFERENCES [breeds]([rowid]) ); CREATE TABLE [breeds] ( [name] TEXT ); But I think this validation check is failing now: https://github.com/simonw/sqlite-utils/blob/842b61321fc6a9f0bdb913ab138e39d71bf42e00/sqlite_utils/db.py#L875-L884

Here's what the debugger reveals about this code: python for fk in foreign_keys: if fk.other_table == name and columns.get(fk.other_column): continue if not any( c for c in self[fk.other_table].columns if c.name == fk.other_column ): raise AlterError( "No such column: {}.{}".format(fk.other_table, fk.other_column) ) (Pdb) fk ForeignKey(table='dogs', column='breed_id', other_table='breeds', other_column='rowid') (Pdb) self[fk.other_table].columns [Column(cid=0, name='name', type='TEXT', notnull=0, default_value=None, is_pk=0)]