Try adding this to your mysql_query() function.
$search = mysql_query()
or die("ERROR: ". mysql_error() . "(" . mysql_errno(). ")");
this will tell you whether or not the query is dieing for one reason or
another.
if it doesn't return the die() statement then you must assume that it was
successfull.
the best way to catch whether or not the result pointer is empty, is to run
a mysql_num_rows() on the pointer.
this will return to you the total number of rows returned by the select
statement.
if you wrap your while clause with an if statement and have the condition be
whether or not the returned value of the mysql_num_rows() is greater then
zero. upon failing, returning zero results, it will never run the while
loop, therefor never give you the error that it's tring to use an invalid
result pointer.
So, try this
<?php
$db = mysql_connect("db30.pair.com", "net3dual", "rosevilleX1");
$db = mysql_select_db("rahs",$db);
$search = mysql_query("SELECT * FROM rahs") or die("ERROR: ". mysql_error()
. "(" . mysql_errno(). ")");
if(mysql_num_rows($search))
{
while($num = mysql_fetch_array($search))
{
echo "<div align='center'>
<table border='0' cellpadding='0'
cellspacing='0' width=\"40%\" align='center'>
<tr>
<td width=\"70%\">Name: ".$num["name"].
" #".$num["number"]."</td>
<td width=\"30%\">Team:
".$num["team"]."</td>
</tr>
<tr>
<td width=\"100%\"
colspan='2'>Penalty".$num["penalty"]."</td>
</tr>
<tr>
<td width=\"50%\">Groundballs:
".$num["groundballs"]."</td>
<td width=\"50%\">Shots:
".$num["shots"]."</td>
</tr>
<tr>
<td width=\"50%\">Faceoffs Won:
".$num["faceoffs_won"]."</td>
<td width=\"50%\">Faceoffs Lost:
".$num["faceoffs_lost"]."</td>
</tr>
<tr>
<td width=\"100%\" colspan='2'><p
align='center'>Date ".$num["date"]."</p></td>
</tr>
</table>
</div>";
}
} else {
die("didn't find any results");
}
?>
----- Original Message -----
From: "Alex Behrens" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Friday, March 01, 2002 5:25 PM
Subject: need help debugging!
> Hey All,
>
> I am new to mysql and I need help debugging this peice of code for my
site:
>
> <?php
> $db = mysql_connect("db30.pair.com", "net3dual", "rosevilleX1");
> mysql_select_db("rahs",$db);
> $search = mysql_query("SELECT * FROM rahs");
> while($num=mysql_fetch_array($search)) {
> printf ("<div align='center'><center><table border='0' cellpadding='0'
> cellspacing='0' width=\"40%\"><tr><td
> width=\"70%\">Name:%s #%s</td><td
> width=\"30%\">Team: %s</td></tr><tr><td width=\"100%\"
colspan='2'>Penalty:
> %s</td></tr><tr><td width=\"50%\">Groundballs: %s</td><td
> width=\"50%\">Shots: %s</td></tr><tr><td width=\"50%\">Faceoffs
> Won:%s</td><td width=\"50%\">Faceoffs Lost: %s</td></tr><tr><td
> width=\"100%\" colspan='2'><p
>
align='center'>Date:%s</td></tr></table></center></div>",$num["name"],$num["
>
number"],$num["team"],$num["penalty"],$num["groundballs"],$num["shots"],$num
> ["faceoffs_won"],$num["faceoffs_lost"],$num["date"]);
> }
> ?>
>
> it says this error:
> Warning: Supplied argument is not a valid MySQL result resource on line 17
> line 17 would be this line:
> while($num=mysql_fetch_array($search)) {
>
> what is wrong with my code?
>
>
> Thanks!
> --------------------------------------------
> -Alex "Big Al" Behrens
> E-mail: [EMAIL PROTECTED]
> Urgent E-mail: [EMAIL PROTECTED] (Please be brief!)
> Phone: 651-482-8779
> Cell: 651-329-4187
> Fax: 651-482-1391
> ICQ: 3969599
> Owner of the 3D-Unlimited Network:
> http://www.3d-unlimited.com
> Send News:
> [EMAIL PROTECTED]
>
>
> ---------------------------------------------------------------------
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> http://www.mysql.com/manual.php (the manual)
> http://lists.mysql.com/ (the list archive)
>
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