I ran into this problem when I installed 4.0.18. All
of the tables in my database are INNODB and the
REPLACE statement was failing on tables that had
foreign key constraints. I just rolled back to 4.0.16
and the problems went away. Not much of a solution,
but it's buying me a little time. Will I have to get
rid of all of the REPLACE INTO statements and replace
them with INSERT/UPDATE statements or is there some
configuration setting that needs to be changed to make
it work?
Cheers,
Tripp
--- Victoria Reznichenko
<[EMAIL PROTECTED]> wrote:
> Kevin Carlson <[EMAIL PROTECTED]> wrote:
> > I have a table with four columns, the first three
> of which are combined
> > into a unique key:
> >
> >
> > create table Test {
> > cid int(9) NOT NULL default '0',
> > sid int(9) NOT NULL default '0',
> > uid int(9) NOT NULL default '0',
> > rating tinyint(1) NOT NULL default '0',
> > UNIQUE KEY csu1 (cid,sid,uid),
> > KEY cid1 (sid),
> > KEY sid1 (sid),
> > KEY uid1 (sid),
> > } TYPE=InnoDB;
> >
> >
> > I am using a REPLACE query to insert a row if it
> doesn't exist and
> > replace an existing row if one does exist:
> >
> > REPLACE into TEST (cid, sid, uid, rating) values
> (580, 0, 205, 1)
> >
> > In the case of this particular row, a row already
> exists with the
> > concatenated key of 580-0-205 and I am getting a
> duplicate key error. I
> > thought REPLACE was supposed to actually replace
> the contents of the row
> > if one exists. Does anyone have any ideas as to
> why this would be
> > causing a duplicate key error?
> >
>
> Works fine for me. The above CREATE TABLE statement
> has some syntax errors. What exactly does CREATE
> TABLE look like?
> What version of MySQL do you use?
>
>
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