Emmett Bishop <[EMAIL PROTECTED]> wrote:
> 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?
>
Could you provide a test case?
>
> --- 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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