Class Error

[Vbabiy]

Hello i am try to you a db Class and i keep getting this error:

Here is the DB Class:

<?php

class mysql_db
{
	var $dblink;


function mysql_db()
{
	$this->dblink = mysql_connect('localhost','root','') or die(mysql_error());
	mysql_select_db('hotels') or die(mysql_error());
}

function query($query)
{
	return mysql_query($query) or die(mysql_error());

}

function fetchArray($results)
{
	return mysql_fetch_array($results) or die(mysql_error());
}

}
?>

When i call it this way:

<?php
include('includes/db.inc.php');

$db = new mysql_db();
$sql = 'SELECT * FROM travel_checklist';
$a = $db->query($sql);
var_dump($a);
$row =mysql_fetch_array($a);
var_dump($row);
?>

for the first time it does a Var dump it says it a bool not no a resourse call any idea why...

this is what it dispalys:

Content-type: text/html
X-Powered-By: PHP/4.3.11
Set-Cookie: ZDEDebuggerPresent=php,phtml,php3; path=/

bool(true)
<br />
<b>Warning</b>:  mysql_fetch_array(): supplied argument is not a valid MySQL result resource in <b>C:\InetPub\htdocs\Hotel\test.php</b> on line <b>8</b><br />
NULL

[glcarlstrom]

Looks like it's not a problem with your PHP code so much as the MySQL database.

Make sure the mysql_db() constructor is actually connecting to the database (is the login information correct?), and that you do indeed have a table called travel_checklist with results to be retrieved... but the PHP code looks fine.

BTW, so you don't reinvent the wheel: PHP5 includes an OOP MySQL implementation (look at the PHP documentation for "Improved MySQL API").

[Vbabiy]

glcarlstrom wrote:

Looks like it's not a problem with your PHP code so much as the MySQL database.

Make sure the mysql_db() constructor is actually connecting to the database (is the login information correct?), and that you do indeed have a table called travel_checklist with results to be retrieved... but the PHP code looks fine.

BTW, so you don't reinvent the wheel: PHP5 includes an OOP MySQL implementation (look at the PHP documentation for "Improved MySQL API").

Well this is in PHP4 and it does connect but if i rewrite the code like this it works:

<?php
include('includes/db.inc.php');

$db = new mysql_db();
$sql = 'SELECT * FROM travel_checklist';
$a = mysql_query($sql);
var_dump($a);
$row =mysql_fetch_array($a);
var_dump($row);
?>

it displays:

resource(4) of type (mysql result)
array(86) {
  [0]=>
  string(1) "1"
  ["Event_ID"]=>
  string(1) "1"
  [1]=>
  string(4) "Area"
  ["Metropolitan_Area"]=>
  string(4) "Area"
  [2]=>
  string(4) "Area"
  ["City"]=>
  string(4) "Area"
  [3]=>
  string(4) "Area"
  ["State"]=>
  string(4) "Area"
  [4]=>
  string(4) "Area"
  ["Date"]=>
  string(4) "Area"
  [5]=>
  string(4) "Area"
  ["Hotel"]=>
  string(4) "Area"
  [6]=>
  string(4) "Area"
  ["Hotel_Address"]=>
  string(4) "Area"
  [7]=>
  string(4) "Area"
  ["Hotel_Zip"]=>
  string(4) "Area"
  [8]=>
  string(4) "Area"
  ["Contact_Name"]=>
  string(4) "Area"
  [9]=>
  string(5) "12312"
  ["Contact_Num"]=>
  string(5) "12312"
  [10]=>
  string(6) "123213"
  ["Contact_Fax"]=>
  string(6) "123213"
  [11]=>
  NULL
  ["Front_Desk_Num"]=>
  NULL
  [12]=>
  NULL
  ["Front_Desk_Fax_Num"]=>
  NULL
  [13]=>
  NULL
  ["Contact_Email"]=>
  NULL
  [14]=>
  NULL
  ["Contact_other"]=>
  NULL
  [15]=>
  string(5) "12312"
  ["Size_Room_1"]=>
  string(5) "12312"
  [16]=>
  string(6) "123123"
  ["Size_Room_2"]=>
  string(6) "123123"
  [17]=>
  NULL
  ["Size_Room_3"]=>
  NULL
  [18]=>
  NULL
  ["Size_Room_4"]=>
  NULL
  [19]=>
  string(6) "123123"
  ["Cost_per_Weekend"]=>
  string(6) "123123"
  [20]=>
  string(6) "123213"
  ["Team_Leader"]=>
  string(6) "123213"
  [21]=>
  string(8) "12312312"
  ["Num_Sleeping_Rooms"]=>
  string(8) "12312312"
  [22]=>
  string(6) "123123"
  ["Fri_Double"]=>
  string(6) "123123"
  [23]=>
  string(3) "123"
  ["Sat_Double"]=>
  string(3) "123"
  [24]=>
  string(2) "23"
  ["Sun_Double"]=>
  string(2) "23"
  [25]=>
  string(6) "123123"
  ["Fri_Sng"]=>
  string(6) "123123"
  [26]=>
  string(3) "123"
  ["Sat_Sng"]=>
  string(3) "123"
  [27]=>
  string(6) "123123"
  ["Sun_Sng"]=>
  string(6) "123123"
  [28]=>
  string(4) "true"
  ["Do_Not_Book"]=>
  string(4) "true"
  [29]=>
  string(6) "123123"
  ["Num_Mailed"]=>
  string(6) "123123"
  [30]=>
  NULL
  ["Notes"]=>
  NULL
  [31]=>
  string(4) "1234"
  ["Times_Saturday"]=>
  string(4) "1234"
  [32]=>
  string(6) "213123"
  ["Times_Sunday"]=>
  string(6) "213123"
  [33]=>
  string(5) "12312"
  ["Sleeping_RM_Total"]=>
  string(5) "12312"
  [34]=>
  NULL
  ["See_notes"]=>
  NULL
  [35]=>
  string(2) "no"
  ["Parking"]=>
  string(2) "no"
  [36]=>
  string(2) "no"
  ["Shuttle"]=>
  string(2) "no"
  [37]=>
  string(6) "123123"
  ["Rooming_List_Due_Date"]=>
  string(6) "123123"
  [38]=>
  string(5) "12312"
  ["Last_updated"]=>
  string(5) "12312"
  [39]=>
  NULL
  ["Received_Con"]=>
  NULL
  [40]=>
  NULL
  ["Returned_Con"]=>
  NULL
  [41]=>
  NULL
  ["Faxed_RM"]=>
  NULL
  [42]=>
  NULL
  ["Faxed_RM_Setup"]=>
  NULL
}

What is going wrong that when it return the query it converts it to a bool or something

[Vbabiy]

dont mean to double post but any one have any idea

[wilsonodk]

Maybe try passing the connection link id through each method call, like this...

<?php 

class mysql_db 
{ 
    var $dblink; 


function mysql_db() 
{ 
    $this->dblink = mysql_connect('localhost','root','') or die(mysql_error()); 
    mysql_select_db('hotels') or die(mysql_error()); 
} 

function query($query) 
{ 
    return mysql_query($query,$this->dblink) or die(mysql_error()); 

} 

function fetchArray($results) 
{ 
    return mysql_fetch_array($results,$this->dblink) or die(mysql_error()); 
} 

} 
?>

[Vbabiy]

oddly enought this is what fixed it:

 <?php 

class mysql_db 
{ 
    var $dblink; 


function mysql_db() 
{ 
    $this->dblink = mysql_connect('localhost','root','') or die(mysql_error()); 
    mysql_select_db('hotels') or die(mysql_error()); 
} 

function query($query) 
{ 
    $q =  mysql_query($query) or die(mysql_error()); 
    return $q;
} 

function fetchArray($results) 
{ 
    $array = mysql_fetch_array($results) or die(mysql_error()); 
	return $array;
} 

} 
?> 

if i return a var it worked...