Actually I figured out the problem my db name was wrong, however I am
getting a new code, here is my code now tho and the error (with the pass
this time!):
<?php
$db = mysql_connect("db", "***", "***");
mysql_select_db("net3dual_lax",$db);
$search = mysql_query("SELECT * FROM rahs");
echo "<div align='center'><center><table border='0' cellpadding='0'
cellspacing='0' width=\"40%\">\n";
while($num=mysql_fetch_array($search)) {
printf ("<tr><td width=\"70%\">Name:%s
Number:%s</td><td width=\"30%\">Team: %s</td></tr><tr><td width=\"100%\"
colspan='2'>Penalty: %s</td></tr><tr><td width=\"50%\">Groundballs:
%s</td><td width=\"50%\">Shots: %s</td></tr><tr><td width=\"50%\">Faceoffs
Won:%s</td><td width=\"50%\">Faceoffs Lost: %s</td></tr><tr><td
width=\"100%\" colspan='2'><p
align='center'>Date:%s</td></tr></table></center></div>",$num["name"],$num["
number"],$num["team"],$num["penalty"],$num["groundballs"],$num["shots"],$num
["faceoffs_won"],$num["faceoffs_lost"],$num["date"]);
}
?>
it gives me this error:
Warning: printf(): too few arguments
how do I fix this? I checked the number of "%s" and the number of $num
arguments and they shuld match. what am I doing wrong?
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]
----- Original Message -----
From: "Jim Lucas [jimmysql]" <[EMAIL PROTECTED]>
To: "Alex Behrens" <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Friday, March 01, 2002 10:19 PM
Subject: Re: need help debugging!
> 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]
> >
> >
> > ---------------------------------------------------------------------
> > Before posting, please check:
> > http://www.mysql.com/manual.php (the manual)
> > http://lists.mysql.com/ (the list archive)
> >
> > To request this thread, e-mail <[EMAIL PROTECTED]>
> > To unsubscribe, e-mail
> <[EMAIL PROTECTED]>
> > Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php
> >
> >
>
>
> ---------------------------------------------------------------------
> Before posting, please check:
> http://www.mysql.com/manual.php (the manual)
> http://lists.mysql.com/ (the list archive)
>
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