Posted by Daz on 02/25/07 16:07

On Feb 25, 3:55 pm, Jerry Stuckle <jstuck...@attglobal.net> wrote:
> Daz wrote:
> > Hi everyone.
>
> > I am trying to create an extension of the mysqli class within PHP, and
> > I am finding it quite difficult. I am fairly new to PHP classes, and
> > decided to give them a go. Here's what I have to far:
>
> > <?php
> > class sql_db extends mysqli
> > {
> > var $connection = false;
> > function sql_db($username, $password, $database="",
> > $server="localhost")
> > {
> > $this->connection = new mysqli($server, $username,
> > $password, $database);
> > if (mysqli_connect_errno()) {
> > printf("Connect failed: %s\n",
> > mysqli_connect_error());
> > exit();
> > }
> > return = $this->connection;
> > }
> > }
> > ?>
>
> > All I am trying to achieve for now is for a simple error to be printed
> > if the connection to the database fails. Rather than have error
> > checking for each within each PHP file, I would just like to have it
> > all done systematically. I would also like to add a few of my own
> > methods, such as having the object pass back a PHP array, rather than
> > a MySQL array. I know I can do this with a separate function, but
> > really I am doing this for the learning experience more than anything.
>
> > At the present time, this works:
>
> > $db =& new sql_db("myUsername", "myPass", "someDB");
>
> > But as soon as I try to query a table, like so:
>
> > $res = $db->query("SELECT * FROM `some_table`;");
>
> > I get:
>
> > Warning: mysqli::query(): Couldn't fetch sql_db in - on line 21
>
> > Evidently I am doing something wrong, and I would really appreciate
> > any pointers. I suspect that I can't actually do what I want to do,
> > although I can't see why not. In any case, I am sure it's possible,
> > but I am going about it completely wrong.
>
> > Thanks in advance.
>
> > Daz.
>
> Daz,
>
> If you're going to extend a class, you don't use:
>
> $this->connection = new mysqli($server, $username,
> $password, $database);
>
> You already have an instance of mysqli via inheritance; this would be
> creating another one.
>
> Rather, you should be using
>
> parent::mysqli($server, $username, $password, $database);
>
> (Assuming PHP4 from your previous post).
>
> You also don't need to keep the $connection - mysqli will maintain that
> info..
>
> --
> ==================
> Remove the "x" from my email address
> Jerry Stuckle
> JDS Computer Training Corp.
> jstuck...@attglobal.net
> ==================


Thanks for that Jerry. That gives me something to go on. I am using
PHP5, as I believe PHP4 doesn't support the mysqli object (but I am
most likely wrong).

Thanks again for your assistance.

Daz.

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