Extending the mysqli class

[Daz]

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 $connection = false;
function sql_db($usernam e, $password, $database="",
$server="localh ost")
{
$this->connection = new mysqli($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("myUsern ame", "myPass", "someDB");

But as soon as I try to query a table, like so:

$res = $db->query("SELEC T * 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.

[Jerry Stuckle]

Re: Extending the mysqli class

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 $connection = false;
function sql_db($usernam e, $password, $database="",
$server="localh ost")
{
$this->connection = new mysqli($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("myUsern ame", "myPass", "someDB");
>
But as soon as I try to query a table, like so:
>
$res = $db->query("SELEC T * 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.
jstucklex@attgl obal.net
=============== ===

[Daz]

Re: Extending the mysqli class

On Feb 25, 3:41 pm, "Daz" <cutenfu...@gma il.comwrote:

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 $connection = false;
function sql_db($usernam e, $password, $database="",
$server="localh ost")
{
$this->connection = new mysqli($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("myUsern ame", "myPass", "someDB");
>
But as soon as I try to query a table, like so:
>
$res = $db->query("SELEC T * 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.

OK, I've done some more thinking, and I think that what I am trying to
do, would involve overriding the constructor of the mysqli class,
which I guess wouldn't be wise, and might not even be possible.

Is there any way to override the contructor and find out what the
constructor code is? I guess that it wouldn't be a PHP class as such,
but more of a hybrid, but I am uncertain.

I can see that what I am doing is correct for adding extra functions,
but I would appreciate it if anyone can help me figure out the best
way to do some error checking when the connection is opened.

I would also appreciate any input on overriding the default methods,
i.e whether it's possible or not, or whether it's ill advised, or if I
can actually use the original constructor and carefully modify it.

I think the initial error checking might require an external function
which instantiates the class, and does the checking, then returns the
mysqli object if all went well, or returns false if it didn't.

[Daz]

Re: Extending the mysqli class

On Feb 25, 3:55 pm, Jerry Stuckle <jstuck...@attg lobal.netwrote:

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 $connection = false;
function sql_db($usernam e, $password, $database="",
$server="localh ost")
{
$this->connection = new mysqli($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("myUsern ame", "myPass", "someDB");

>

But as soon as I try to query a table, like so:

>

$res = $db->query("SELEC T * 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...@attgl obal.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.

[Jerry Stuckle]

Re: Extending the mysqli class

Daz wrote:

On Feb 25, 3:55 pm, Jerry Stuckle <jstuck...@attg lobal.netwrote:

>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 $connection = false;
>> function sql_db($usernam e, $password, $database="",
>>$server="loca lhost")
>> {
>> $this->connection = new mysqli($server, $username,
>>$password, $database);
>> if (mysqli_connect _errno()) {
>> printf("Connect failed: %s\n",
>>mysqli_connec t_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("myUsern ame", "myPass", "someDB");
>>But as soon as I try to query a table, like so:
>> $res = $db->query("SELEC T * 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...@attg lobal.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.
>

Daz,

Actually, PHP4 does support the mysqli object, also.

But in PHP5 your constructors are named __construct, not the class name.
So you need to change the constructor name in your class, and call

parent::constru ct(...)

instead.


--
=============== ===
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attgl obal.net
=============== ===

[Jerry Stuckle]

Re: Extending the mysqli class

Daz wrote:

On Feb 25, 3:41 pm, "Daz" <cutenfu...@gma il.comwrote:

>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 $connection = false;
> function sql_db($usernam e, $password, $database="",
>$server="local host")
> {
> $this->connection = new mysqli($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("myUsern ame", "myPass", "someDB");
>>
>But as soon as I try to query a table, like so:
>>
> $res = $db->query("SELEC T * 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.

>
>
OK, I've done some more thinking, and I think that what I am trying to
do, would involve overriding the constructor of the mysqli class,
which I guess wouldn't be wise, and might not even be possible.
>
Is there any way to override the contructor and find out what the
constructor code is? I guess that it wouldn't be a PHP class as such,
but more of a hybrid, but I am uncertain.
>
I can see that what I am doing is correct for adding extra functions,
but I would appreciate it if anyone can help me figure out the best
way to do some error checking when the connection is opened.
>
I would also appreciate any input on overriding the default methods,
i.e whether it's possible or not, or whether it's ill advised, or if I
can actually use the original constructor and carefully modify it.
>
I think the initial error checking might require an external function
which instantiates the class, and does the checking, then returns the
mysqli object if all went well, or returns false if it didn't.
>

Daz,

Overriding the constructor is quite common. But you need to ensure you
also call the parent class constructor.

As to seeing the code - it's like anything else in PHP. If it's a PHP
file, you can see the code. If, like with mysqli, it's a compiled
extension, you need to look at the source code for the extension.

If you want some error checking when you open the connection, see if the
connection is open after calling the mysqli constructor (see if
mysqli_connect_ error() is non-zero).

You don't need an external function - you just need to check for errors.
And if you don't want to keep checking the responses, check you
exception handling.

--
=============== ===
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attgl obal.net
=============== ===

[Daz]

Re: Extending the mysqli class

On Feb 25, 4:17 pm, Jerry Stuckle <jstuck...@attg lobal.netwrote:

Daz wrote:

On Feb 25, 3:41 pm, "Daz" <cutenfu...@gma il.comwrote:

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 $connection = false;
function sql_db($usernam e, $password, $database="",
$server="localh ost")
{
$this->connection = new mysqli($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("myUsern ame", "myPass", "someDB");

>

But as soon as I try to query a table, like so:

>

$res = $db->query("SELEC T * 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.

>

OK, I've done some more thinking, and I think that what I am trying to
do, would involve overriding the constructor of the mysqli class,
which I guess wouldn't be wise, and might not even be possible.

>

Is there any way to override the contructor and find out what the
constructor code is? I guess that it wouldn't be a PHP class as such,
but more of a hybrid, but I am uncertain.

>

I can see that what I am doing is correct for adding extra functions,
but I would appreciate it if anyone can help me figure out the best
way to do some error checking when the connection is opened.

>

I would also appreciate any input on overriding the default methods,
i.e whether it's possible or not, or whether it's ill advised, or if I
can actually use the original constructor and carefully modify it.

>

I think the initial error checking might require an external function
which instantiates the class, and does the checking, then returns the
mysqli object if all went well, or returns false if it didn't.

>
Daz,
>
Overriding the constructor is quite common. But you need to ensure you
also call the parent class constructor.
>
As to seeing the code - it's like anything else in PHP. If it's a PHP
file, you can see the code. If, like with mysqli, it's a compiled
extension, you need to look at the source code for the extension.
>
If you want some error checking when you open the connection, see if the
connection is open after calling the mysqli constructor (see if
mysqli_connect_ error() is non-zero).
>
You don't need an external function - you just need to check for errors.
And if you don't want to keep checking the responses, check you
exception handling.
>
--
=============== ===
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstuck...@attgl obal.net
=============== ===

Thanks for that Jerry. You're a star! :)

[peter]

Re: Extending the mysqli class

Actually, PHP4 does support the mysqli object, also.
>
But in PHP5 your constructors are named __construct, not the class name.
So you need to change the constructor name in your class, and call
>
parent::constru ct(...)

Just a correction of Jerry's post. As of php 5 the __constructor method was
made available BUT the old constructor name is still useable.

The class you are actually using (mysqli) does in fact use a contructor
called mysqli as the following code snippet will demonstrate if you run it:-

<?php
function output_methods( $obj)
{
$methods = get_class_metho ds($obj);
foreach ($methods as $method)
{
echo "function $method()\n";
}
}

output_methods( "mysqli");
?>

[peter]

Re: Extending the mysqli class

Actually, PHP4 does support the mysqli object, also.

>
But in PHP5 your constructors are named __construct, not the class name.
So you need to change the constructor name in your class, and call
>
parent::constru ct(...)

Also just thought of another clarification. If you do not declare a
constructor for a class that extends another class, the parent constructor
will be called. for example the following code snippet will output "parent
constructor and the input was THIS TEXT":-

<?php
class c1
{
function __construct($in put)
{
echo "parent constructor and the input was ".$input;
}
}
class c2 extends c1
{
}
$c = new c2('THIS TEXT');
?>

[Daz]

Re: Extending the mysqli class

On Feb 25, 5:15 pm, "peter" <sub...@flexiwe bhost.comwrote:

Actually, PHP4 does support the mysqli object, also.

>

But in PHP5 your constructors are named __construct, not the class name.
So you need to change the constructor name in your class, and call

>

parent::constru ct(...)

>
Just a correction of Jerry's post. As of php 5 the __constructor method was
made available BUT the old constructor name is still useable.
>
The class you are actually using (mysqli) does in fact use a contructor
called mysqli as the following code snippet will demonstrate if you run it:-
>
<?php
function output_methods( $obj)
{
$methods = get_class_metho ds($obj);
foreach ($methods as $method)
{
echo "function $method()\n";
}
>
}
>
output_methods( "mysqli");
?>

Excellent! Just what I needed :)

I am wondering why constructors are different in PHP5. Surely anyone
coding portable scripts, would just use the PHP4 constructor
technique, as there's no guarantee that the server it needs to run on
is running PHP5. Some people have no control over what version of PHP
is installed if they are simply paying for a basic hosting account.
Fortunately for me, I have the choice between 4 and 5, but others
won't necessarilly be quite so fortunate.

Thanks for your help Peter. That's sure going to save me digging
through source code and scratching my head to to point of baldness.

Best wishes.

Daz.

[Daz]

Re: Extending the mysqli class

On Feb 25, 5:15 pm, "peter" <sub...@flexiwe bhost.comwrote:

Actually, PHP4 does support the mysqli object, also.

>

But in PHP5 your constructors are named __construct, not the class name.
So you need to change the constructor name in your class, and call

>

parent::constru ct(...)

>
Just a correction of Jerry's post. As of php 5 the __constructor method was
made available BUT the old constructor name is still useable.
>
The class you are actually using (mysqli) does in fact use a contructor
called mysqli as the following code snippet will demonstrate if you run it:-
>
<?php
function output_methods( $obj)
{
$methods = get_class_metho ds($obj);
foreach ($methods as $method)
{
echo "function $method()\n";
}
>
}
>
output_methods( "mysqli");
?>

I have just one more question. Seeing that you can print methods, how
would I print the code from a function?

Many thanks.

Daz.

[Jerry Stuckle]

Re: Extending the mysqli class

peter wrote:

>Actually, PHP4 does support the mysqli object, also.
>>
>But in PHP5 your constructors are named __construct, not the class name.
>So you need to change the constructor name in your class, and call
>>
> parent::constru ct(...)

>
Just a correction of Jerry's post. As of php 5 the __constructor method was
made available BUT the old constructor name is still useable.
>

That's true if there is no __constructor method. However, from what I
understand, the old classname constructor is going away. If you're on
PHP5, you should be using the new way.

The class you are actually using (mysqli) does in fact use a contructor
called mysqli as the following code snippet will demonstrate if you run it:-
>
<?php
function output_methods( $obj)
{
$methods = get_class_metho ds($obj);
foreach ($methods as $method)
{
echo "function $method()\n";
}
}
>
output_methods( "mysqli");
?>
>
>
>

--
=============== ===
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attgl obal.net
=============== ===

[Jerry Stuckle]

Re: Extending the mysqli class

Daz wrote:

On Feb 25, 5:15 pm, "peter" <sub...@flexiwe bhost.comwrote:

>>Actually, PHP4 does support the mysqli object, also.
>>But in PHP5 your constructors are named __construct, not the class name.
>>So you need to change the constructor name in your class, and call
>> parent::constru ct(...)

>Just a correction of Jerry's post. As of php 5 the __constructor method was
>made available BUT the old constructor name is still useable.
>>
>The class you are actually using (mysqli) does in fact use a contructor
>called mysqli as the following code snippet will demonstrate if you run it:-
>>
><?php
>function output_methods( $obj)
>{
> $methods = get_class_metho ds($obj);
> foreach ($methods as $method)
> {
> echo "function $method()\n";
> }
>>
>}
>>
>output_methods ("mysqli");
>?>

>
>
Excellent! Just what I needed :)
>
I am wondering why constructors are different in PHP5. Surely anyone
coding portable scripts, would just use the PHP4 constructor
technique, as there's no guarantee that the server it needs to run on
is running PHP5. Some people have no control over what version of PHP
is installed if they are simply paying for a basic hosting account.
Fortunately for me, I have the choice between 4 and 5, but others
won't necessarilly be quite so fortunate.
>
Thanks for your help Peter. That's sure going to save me digging
through source code and scratching my head to to point of baldness.
>
Best wishes.
>
Daz.
>

Daz,

It was changed because they're moving away from the old classname
constructor to just __constructor. New code on php5 should be using the
new method.

For compatibility with PHP4, just create your constructor as
__constructor. Then create a classname constructor and have it call
__constructor.

--
=============== ===
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attgl obal.net
=============== ===

[Michael Fesser]

Re: Extending the mysqli class

..oO(Daz)

>I am wondering why constructors are different in PHP5. Surely anyone
>coding portable scripts, would just use the PHP4 constructor
>technique

I don't care about PHP 4 anymore. Personally I consider it dead.

>as there's no guarantee that the server it needs to run on
>is running PHP5.

My scripts make use of many of the new OOP features in PHP 5, so they
won't run at all on PHP 4.

Micha

[admin@fahost.com]

Re: Extending the mysqli class

I havejustone more question. Seeing that you can print methods, how
would I print the code from a function?

As far as I know you cannot print the code for a class method only
retrieve it's name.

(had to use google groups to post a reply as for some some reason my
news provider is not working properly)