[SOLVED] php server configuration different from xampp ?

j33baS · January 19, 2008

Hi i have just tried to upload code onto a brand new web server, the code is an exact copy and the database is exactly the same too, all I changed was my config.php file where the username, databasename, host etc are stored.

The connection works as i dont get any error messages, but when queries which run perfect on my local xampp server dont work on the webserver ... for instance ....

<table width="600" border="0" cellpadding="0" cellspacing="0">

	  	<?php

	  	$titlequery = "SELECT * FROM entries
	  	ORDER BY dateposted DESC
	  	LIMIT 0, 5;";

	  	$titleresult = mysql_query($titlequery);

	  	while($row = mysql_fetch_array($titleresult)){

	  	echo "<tr><td><div align=\"left\">";


	  	//echo $row['subject'];
	  	echo "<li><a href='viewentry.php?id=" . $row['id'] . "'>" . $row	['subject'] . "</a></li>";

	  	echo "</td></tr>";
	  	      }
	  	    ?>
	  	</table>

will give me my lovely links to news stored in my DB, but on the webserver installation i get ....

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/timmycl/public_html/index.php on line 82

Like i said this is my first time trying to get a site working which was not local to my machine! any help would be greatly appreciated!

Thanks

GingerRobot · January 19, 2008

Whenever you are having mysql issues, add an or die statement to the query:

$titleresult = mysql_query($titlequery) or die(mysql_error());

This will tell you the error mysql is giving.

j33baS · January 19, 2008

ok thanks, i just added that ...

Parse error: syntax error, unexpected T_LOGICAL_OR in /home/timmycl/public_html/index.php on line 81

im really confused as to why it should be different ? Any suggestions?

trq · January 19, 2008

The problem is simply that you fail to check your queries return value prior to using any returned result. This is a common error, but one that is easily avoided using the proper syntx. eg;

<?php

  $sql = "YOUR QUERY";
  if ($result = mysql_query($sql)) {
    if (mysql_num_rows($result)) {
      // now it is safe to use mysql_fetch_assoc
      while ($row = mysql_fetch_assoc($result)) {
        // diplsay data
      }
    } else {
      // no records found
    }
  } else {
    // query failed, handle error.
  }

?>

j33baS · January 19, 2008

yea fair enough, but i'm still failing to see why code works on my local installation and doesnt on my websever, when they r both connecting to identical databases and are both running off php 5.2.5 ???

really confused ...

Fyorl · January 20, 2008

Post the new code.

j33baS · January 20, 2008

<?php
require("header.php");
?>


<table width="800" border="0" cellspacing="0" cellpadding="0">
  <tr>
    <td width="600"><table width="600" border="0" cellspacing="0" cellpadding="0">
        <tr>
          <td height="400" id="mainContent">
	<h2>Welcome</h2>

	<table width="600" border="0" cellspacing="0" cellpadding="0">
	  <tr>
	    <td width="170" height="20">
		<table width="170" border="0" cellspacing="0" cellpadding="0">
	        <tr>
	          <td><div align="center"><img src="images/wasps.gif"></div></td>
	        </tr>
	        <tr>
	          <td><div align="center"><img src="images/munster.gif"></div></td>
	        </tr>
	      </table> </td>
	    <td width="430">
	    
	      <p> </p>
	      </td>
	  </tr>
</table>

	<table width="600" border="0" cellspacing="0" cellpadding="0">
	  <tr>
	    <td><div align="center"><b>x x x</b> </div></td>
	  </tr>
	  <tr>


	  </tr>
</table>
<p>

          </td>
        </tr>
        <tr>
          <td><img src="images/news_banner.jpg"></td>
        </tr>
        <tr>
          <td><div align="center">

          <!-- this is code for the headlines -->
	  	<table width="600" border="0" cellpadding="0" cellspacing="0">

	  	<?php

	  	$titlequery = "SELECT * FROM entries
	  	ORDER BY dateposted DESC
	  	LIMIT 0, 5;";

	  	$titleresult = mysql_query($titlequery);

	  	while($row = mysql_fetch_array($titleresult)){

	  	echo "<tr><td><div align=\"left\">";


	  	//echo $row['subject'];
	  	echo "<li><a href='viewentry.php?id=" . $row['id'] . "'>" . $row	['subject'] . "</a></li>";

	  	echo "</td></tr>";
	  	      }
	  	    ?>
	  	</table>
<!-- end headline code -->

            </div></td>
        </tr>
      </table></td>
    <td width="200" bgcolor="#990000" id="adspace">
    <div align="center">
    <a href="hospitality.php"><img src="images/hospitalityad.jpg" width="185" height="450" border="0"></a>
    <p>
    <a href="ticket.php"><img src="images/ticket_ad.jpg" border="0"></a>
    </td>
  </tr>
</table>

<?php
require("footer.php");
?>

... thats the whole lot, and i know my connection to the DB is being set up fine cos dont get any error messages back (its made in header.php).

Like i said iv got xampp installed and having been using it for developing, and just went to upload all the code to the web server i just bought so i was kinda hoping it would work the same !

Thanks

Fyorl · January 20, 2008

I can't see an 'or die(mysql_error());' at all...

j33baS · January 20, 2008

yea sorry i copy n pasted the old file.

i put it in where ginger robot said, and i got the error:

Parse error: syntax error, unexpected T_LOGICAL_OR in /home/timmycl/public_html/index.php on line 81

regardless of error handling, you cant think of any reason why 2 sets of identical code would perform differently on different servers which to me appear identical in setup!?

Fyorl · January 20, 2008

At a guess I would say the problem is with your database. I would be tempted to say maybe differing MySQL versions but the query's so simple that's probably not the problem. The mysql_error() will reveal all though so could you please just post line 81 so we can fix your syntax error and find out what mysql_error() says.

j33baS · January 20, 2008

<?php

$titlequery = "SELECT * FROM entries
ORDER BY dateposted DESC
LIMIT 0, 5;";

$titleresult = mysql_query($titlequery); or die(mysql_error());   <---- [b]line 81[/b]
while($row = mysql_fetch_array($titleresult)){

echo "<tr><td><div align=\"left\">";


//echo $row['subject'];
echo "<li><a href='viewentry.php?id=" . $row['id'] . "'>" . $row ['subject'] . "</a></li>";

echo "</td></tr>";
}
?>

lol sorry dude thanks for trying to help im gettin so frustrated and getting myself all mixed up !

Fyorl · January 20, 2008

Ah well the syntax error comes from having a semicolon in-between 'mysql_query()' and 'or'. Remove it and you should be good to go with checking the SQL error.

j33baS · January 20, 2008

ahh see! im retarded i could have been looking at that all night ...

ok it says no DB selected ....

so here is my header...

<?php
require("config.php");
$db = mysql_connect($dbhost, $dbuser, $dbpassword);
mysql_select_db($dbdatabase, $db);
?>

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<title>title</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<link rel="stylesheet" href="stylesheet.css" type="text/css" />
<script type="text/javascript" src="countdown.js" defer="defer"></script>
</head>

<body background="images/ed_tile.gif">
<div id="main" align="center">
  <table width="800" height="209" border="3" cellpadding="0" cellspacing="0" bordercolor="#CCCCCC">
    <tr>
      <td height="209" bgcolor="#F4F4F4"><table width="800" height="165" border="0" cellpadding="0" cellspacing="0">
          <tr>
            <td><div align="center"><img src="images/banner.jpg" width="800" height="100"></div></td>
          </tr>
          <tr>
            <td><div align="center"><font color="#FFFFFF" size="4" face="Verdana, Arial, Helvetica, sans-serif"><img src="images/nav.jpg" width="800" height="29" border="0" usemap="#Map2"></font></div>
             <map name="Map2">
		    <area shape="rect" coords="11,6,106,24" href="index.php">
		    <area shape="rect" coords="140,7,190,24" href="news.php">
		    <area shape="rect" coords="217,4,314,24" href="schedule.php">
		    <area shape="rect" coords="346,6,432,25" href="pictures.php">
		    <area shape="rect" coords="459,6,551,24" href="contacts.php">
		    <area shape="rect" coords="581,7,675,24" href="lastyear.php">
		    <area shape="rect" coords="698,6,786,24" href="sponsor.php">
  </map>
            </td>
          </tr>
          <tr>
            <td>

and here is my config.php

<?php

$dbhost = "localhost";  <-- this shud work right ?
$dbuser = "timmycl_eri";    <-- this is how the DB appears in phpMyadmin
$dbpassword = "my password";
$dbdatabase = "my username";

$config_sitename = "site name";
$config_author = "my name";
$config_basedir = "http://www.my domain.com/";

?>

j33baS · January 20, 2008

for the love of god i've just seen my own error... had the username and database variables mixed up...Once again i owe my life to you guys ! and sorry for wasting your time.

Thanks a million, only wish i could return the favor in some way but I doubt that will happen, cos i am stuuuuuuupid

Fyorl · January 20, 2008

It happens to us all at some point in one way or another...

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[SOLVED] php server configuration different from xampp ?

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