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| Тема |
 supplied argument is not a valid MySQL result res. |
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| Автор | CaptainFoo (Нерегистриран) | |
| Публикувано | 15.09.03 20:13 |
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privet na vsichki.
opitvam se da razbera zashto scripta (sled tozi class) ne iska da sraboti s MySQL.
iavno ne haresva resource-a, no ne useshtam kyde e problema..
do 'while($row = mysql_fetch_array($dbquery))' niama greshki (mai!;-).
koito ima malko vreme i idei da pishe :-)
blagodaria za otgovorite.
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /www/project/oo/run.php on line 14
// m.class.php
<?
class Common
{
var $m_dbHost = 'localhost';
var $m_dbUser = 'dbuser';
var $m_dbPass = 'dbpass';
var $m_dbName = 'mdb';
function Connect()
{
$this->m_link = mysql_connect($this->m_dbHost, $this->m_dbUser, $this->m_dbPass);
if (!$this->m_link)
{
die($this->m_error = "Не мога да се свържа: " . mysql_error());
return FALSE;
}
echo 'connected';
return TRUE;
}
function Select()
{
$this->tbl = mysql_select_db($this->m_dbName, $this->m_link);
if (!$this->tbl)
{
die($this->m_error = "Не мога да използвам базата: " . mysql_error());
return FALSE;
}
echo 'db selected';
return TRUE;
}
function Query($dbquery)
{
$this->query = mysql_query($dbquery);
if (!$this->query)
{
die($this->m_error = "Не мога да изпълня запитването: " . mysql_error());
return FALSE;
}
echo 'query executed';
return TRUE;
}
function Close()
{
if ($this->m_link) mysql_close($this->m_link);
}
}
?>
// run.php
<?
require 'm.class.php';
$m_db = new Common;
$m_db->Connect();
$m_db->Select();
$m_db->Query('SELECT name, address FROM students WHERE id = 1');
while($row = mysql_fetch_array($dbquery))
{
echo $row['name'].$row['address'];
}
$m_db->Close();
?>
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