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Article #9828

Re: <? Tabel genereren, hoe? ?>

#9828
From: webmasterATarea32PUNTbe Area32
Date: Tue, 02 Apr 2002 17:38
158 lines
4118 bytes
Ik heb jou class eens geprobeerd, ik kreeg 
2 fouten:
Warning: Supplied argument is not a valid MySQL result resource in
C:\apache\htdocs\mysql.php on line 25


Warning: Supplied argument is not a valid MySQL result resource in
C:\apache\htdocs\mysql.php on line 34

Dit is mijn code met jou aanpassingen:

<?
include("config.inc.php");
include("functies.inc.php");


jumpbox_cat();


 if($categorie){
?>



 <table width="100%">


        <?php
                include ('mysql.php');


                $dbconn = new mySQL;


                $dbstmt = "SELECT kaarten.col1, kaarten.col2,
kaarten.col3 FROM kaarten ";


                $dbres  = $dbconn->RecordSet($dbstmt);
                $count  = $dbconn->RecordCount($dbres);
                $dbdata = $dbconn->GetRows($dbres);


                for ($i = 0; $i < $count; $i++):
        ?>


        <tr>
                <td><?php echo $dbdata[$i]['col1']; ?></td>
                <td><?php echo $dbdata[$i]['col2']; ?></td>
                <td><?php echo $dbdata[$i]['col3']; ?></td>
        </tr>


        <?php endfor; ?>


</table>


 <?


}
  else {
        echo "<br><strong>Er werd geen categorie
geselecteerd.</strong>";
                }
?>
Hopelijk raak je er wijs uit....



Op Tue, 02 Apr 2002 13:42:45 GMT, Philip Paeps <philip@paeps.cx>
schreef:

> Area32 <webmasterATarea32PUNTbe> wrote:
> > Ik probeerde het met een functie maar hoe meer ik probeerde hoe meer
> > fouten ik krijg :-(
> 
> Probeer eens met mijn class:
> 
> 	class mySQL
> 	{
> 		var $dbname = "database";	// database name
> 		var $dbhost = "hostname";	// mySQL server
> 		var $dbuser = "user";		// user for conn
> 		var $dbpass = "password";	// users' pass
> 
> 		// Executes a query and returns the recordset
> 		function RecordSet($SQL) 
> 		{
> 			$link = mysql_connect($this->dbhost, $this->dbuser, $this->dbpass); 
> 			mysql_select_db($this->dbname);
> 			
> 			$rs = mysql_query($SQL);
> 			
> 			return $rs;
> 			
> 			$this->DBClose($link); 
> 		}
> 		
> 		// Counts the number of rows in a recordset
> 		function RecordCount($rs) 
> 		{
> 			return mysql_num_rows($rs);
> 		}
> 
> 		// Returns a 2D associative array of rows
> 		function GetRows($rs) 
> 		{
> 			$arr = array();
> 			$counter = 0;
> 			
> 			while ($row = mysql_fetch_array($rs)) 
> 			{
> 				$arr[$counter] = $row;
> 				$counter++;
> 			}
> 			
> 			return $arr;
> 		}
> 
> 		// Close the database connection
> 		function DBClose($link) 
> 		{
> 			mysql_close($link);
> 		}
> 	
> 		// Print an error
> 		function Status() 
> 		{
> 			$errno = mysql_errno();
> 			$error = mysql_error();
> 
> 			return "($errno)<br/>$error";
> 		}
> 	}
> 
> Hier nooit last mee,
> 
> Wat voor errors krijg je zoal?
> 
>  - Philip
> 
> -- 
> Philip Paeps                                                   philip@paeps.cx
> 
>   You sure have to borrow a lot of money these days to
>   be an average consumer.





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From: webmasterATarea32PUNTbe (Area32)
Newsgroups: be.comp.programming
Subject: Re: <? Tabel genereren, hoe?  ?>
Date: Tue, 02 Apr 2002 17:38:20 GMT
Organization: Area32.be
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