On Oct 12, 2006, at 3:44 PM, A. Kretschmer wrote:
Can you show us your SQL? The message is clear: you create a new table
with a foreign key to an other table that doesn't exist. An example:
Yes, I know that part. The error message is bad though, because it
doesn't tell me exactly where the error is.
I got as an error
ERROR: column "id" referenced in foreign key constraint does not exist
I should have gotten something like
ERROR: column "id" referenced in foreign key constraint on column
"xyz" table "abc" does not exist
( the table "abc" is not necessary, i just wanted to be explicit
about the message )
In that create table statement, i had 10 columns each referencing an
'id' in another column. I like very normalized DBs.
I had to go through each column individually to see where my error
was. Postgres should have immediately told me which of the source
table columns that constraint failed on-- not just about the target
column name.
---------------------------(end of broadcast)---------------------------
TIP 9: In versions below 8.0, the planner will ignore your desire to
choose an index scan if your joining column's datatypes do not
match
