Must be an old or broken version of mysql.  I'm running 3.23.50 and it won't
even allow the `id' field to be a varchar(255).  Change it to int() and
watch the magic.

-----Original Message-----
From: Jas [mailto:[EMAIL PROTECTED]] 
Sent: Monday, June 24, 2002 1:05 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Re: never seen this before..


Here is the db table structure...
CREATE TABLE demo_sessions (
   id varchar(255) NOT NULL auto_increment,
   ipaddy varchar(255) NOT NULL,
   date varchar(255) NOT NULL,
   PRIMARY KEY (id)
);
Let me know if this is correct, I have a feeling that the varchar is my
problem but I could be wrong. TIA Jas

<[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> TIA --
>
>
> column in your table denotes your primary key? After that, how is it 
> incremented? Is it auto or do you have some other form of it.  I think 
> your code is correct so the error seems to be how the data ID is
generated.
>
> gl -- Seth
>
> Jas wrote:
>
> > I am getting an error when inserting data into a mysql database 
> > table.
It
> > is telling me that there is a duplicate entry in the database 
> > everytime
it
> > gets 10 records inserted to the table, if I flush the table and try 
> > it
again
> > it works fine.  Is there some kind of limit on some tables?  Here is 
> > the error php outputs once the table gets 10 records: Duplicate 
> > entry '10'
for
> > key 1
> > Here is the code...
> > require '/home/web/b/bignickel.net/scripts/admin/db.php';
> >      $table = "demo_sessions";
> >      $sql_insert = "INSERT INTO $table (ipaddy,date) VALUES 
> > ('$ipaddy','$date')" or die('Could not insert client info');
> >      $result = @mysql_query($sql_insert,$dbh) or die(MYSQL_ERROR());
> >
> > Any help on this would be great
> > TIA
> > Jas
> >
> >
> >
>



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