Re: [PHP-DB] auto_increment

Hello!

I didn't use any DROP statements. Do exactly as following:
mysql> CREATE TABLE stats2 LIKE stats;
mysql> ALTER TABLE stats2 ADD COLUMN id INT NOT NULL AUTO_INCREMENT, 
ADD PRIMARY KEY(id);
mysql> INSERT INTO stats2 SELECT *, 0 FROM stats ORDER BY

The first line copies the structure of table stats to a new table 
called stats2;
The second one, create a new field called id;
The third inserts all records from stats to stats2, in correct order.

After all, you need drop the table stats and rename the table stats2 
to stats:
mysql> DROP TABLE stats;
mysql> RENAME TABLE stats2 TO stats;

Be careful, I don't know your environment.


[]'s
Adriano!
> > > > > > > > > 
> On Tue, 07 Sep 2010 18:10:34 +0200
> Kapu <[email protected]> wrote:
>  You are missing a semicolon after the first alter statement.
> 
> Kapu
> 
> On 7. 9. 2010 18:05, Ron Piggott wrote:
>> I am receiving the following error Adriano:
>>
>> SQL query:
>>
>> ALTER TABLE `stats` DROP `visits` CREATE TABLE `stats2` LIKE `stats` 
>>;
>>
>> MySQL said: Documentation
>> #1064 - You have an error in your SQL syntax; check the manual that
>> corresponds to your MySQL server version for the right syntax to use 
>>near
>> 'CREATE TABLE `stats2` LIKE `stats`' at line 2
>>
>> The complete commands were:
>>
>> ALTER TABLE `stats` DROP `visits`
>> CREATE TABLE `stats2` LIKE `stats`;
>> ALTER TABLE `stats2` ADD COLUMN `visits` INT( 25 ) NOT NULL
>> AUTO_INCREMENT, ADD PRIMARY KEY(`visits`) FIRST;
>> INSERT INTO `stats2` SELECT *, 0 FROM `stats` ORDER BY 
>>`initial_access`;
>>
>> I don't understand the error, your way of creating a table is new to 
>>me.
>> Did something small get missed?
>>
>> Ron
>>
>>
> 
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