MySql warning i cant figure

kyla

hey peeps

ive been struggling with this for some time now and i just cant anymore. please can someone explain this and what to do:

Warning: Supplied argument is not a valid MySQL result resource in c:\apache\htdocs\includes\sql_layer.php on line 286

thanks
Kyz

Lars_Berg

Show the code which causes the error, e.g. rows 280-290

kyla

🙁
its a heck load of code to paste, this error almost fills the entire thing. varying with "on line ***" numbers.

<table cellpadding="0" cellspacing="0" width="100%" border="0" align="center" bgcolor="#ffffff">
<tr>
<td bgcolor="#ffffff">
<img height="16" alt="" hspace="0" src="themes/NukeNews/images/corner-top-left.gif" width="17" align="left">
<a href="index.php"><img src="themes/NukeNews/images/logo.gif" align="left" alt=""._WELCOMETO." hashbofh" border="0" hspace="10"></a></td>
<td bgcolor="#999999"><IMG src="themes/NukeNews/images/pixel.gif" width="1" height="1" alt="" border="0" hspace="0"></td>
<td bgcolor="#cfcfbb" align="center">
<center><form action="modules.php?name=Search" method="post"><font class="content" color="#000000"><b>Search </b>
<input type="text" name="query" size="14"></font></form></center></td>
<td bgcolor="#cfcfbb" align="center">
<center><form action="modules.php?name=News&new_topic" method="post"><font class="content"><b>Topics </b>
<select name="new_topic" onChange='submit()'>
<option value="">All Topics</option>

im not sure if i've pasted the right thing here. dont want to waste anyone's time 🙁

Kyz

piersk

Originally posted by Lars Berg
Show the code which causes the error, e.g. rows 280-290

I think he means the PHP code that makes up lines 280-290, not the HTML.

kyla

🙂 thought so, but not wanting to seem like a complete tit i did the opposite and well...

here's row 280 - 290:

function sql_fetch_row(&$res, $nr)
{
global $dbtype;
switch ($dbtype) {

case "MySQL":
    $row = mysql_fetch_row($res);
    return $row;
break;;

case "mSQL":

Then sadly enough there's now:

Warning: Supplied argument is not a valid MySQL result resource in c:\apache\htdocs\includes\sql_layer.php on line 238

and

Warning: Supplied argument is not a valid MySQL result resource in c:\apache\htdocs\includes\sql_layer.php on line 494

and again

Warning: Supplied argument is not a valid MySQL result resource in c:\apache\htdocs\includes\sql_layer.php on line 342

Do I give up while I'm ahead?

Kyz

Lars_Berg

The error you get indicates that the variable $res is 0. This happens normally if you pass an invalid sql statement to mysql_query.

After each call to mysql_query you should check for errors

$res = mysql_query($sql_string) or die(mysql_error()) .  " when  executing " . $sql_string;

kyla

Thank-you Lars

Kyz

merlinti

Hey All,

I'm a newbie and I'm getting the same Warning message:

Warning: Supplied argument is not a valid MySQL result resource in /home/sites/site190/web/test01.php on line 45

Warning: Supplied argument is not a valid MySQL result resource in /home/sites/site190/web/test01.php on line 53

Warning: Supplied argument is not a valid MySQL result resource in /home/sites/site190/web/test01.php on line 69

<?php

// Set the variables for the database access:
$Host = "localhost";
$User = "username";
$Password = "password";
$DBName = "databasename";
$TableName = "Stratosphere";

$CheckIn = "09/01/2002";
$CheckOut = "09/05/2002";

$Link = mysql_connect ($Host, $User, $Password);

$Query = "SELECT from $TableName WHERE (Dates = '$CheckIn')";
$Query01 = "SELECT from $TableName WHERE (Dates = '$CheckOut')";
$Query02 = "SELECT Price FROM $TableName WHERE (Dates BETWEEN '$CheckIn' AND '$CheckOut')";

$Result = mysql_db_query ($DBName, $Query, $Link);
$Result01 = mysql_db_query ($DBName, $Query01, $Link);
$Result02 = mysql_db_query ($DBName, $Query02, $Link);

// Create a table.
print ("<TABLE BORDER=1 WIDTH=\"20%\" CELLSPACING=1 CELLPADDING=1 ALIGN=LEFT>\n");
print ("<TR ALIGN=CENTER VALIGN=TOP>\n");
print ("<TD ALIGN=CENTER VALIGN=TOP>Price</TD>\n");
print ("</TR>\n");

// Fetch the results from the database.
while ($Row = mysql_fetch_array ($Result)) {
print ("<TR ALIGN=CENTER VALIGN=TOP>\n");
print ("<TD ALIGN=CENTER VALIGN=TOP>$Row[Price]</TD>\n");
print ("</TR>\n");
print "<p>Check In Day: $$Row[Price].00<br></p>\n";
}

// Fetch the results from the database01.
while ($Row = mysql_fetch_array ($Result01)) {
print ("<TR ALIGN=CENTER VALIGN=TOP>\n");
print ("<TD ALIGN=CENTER VALIGN=TOP>$Row[Price]</TD>\n");
print ("</TR>\n");
print "<p>Check Out Day: $$Row[Price].00<br></p>\n";
}

print ("</TABLE>\n");

while ($Row = mysql_fetch_array ($Result02)) {
print ("<TR ALIGN=CENTER VALIGN=TOP>\n");
print ("<TD ALIGN=CENTER VALIGN=TOP>$Row[Price]</TD>\n");
print ("</TR>\n");
}

print ("</TABLE>\n");

$a = array($Query02);
echo "sum(a) = ".array_sum($a)."\n";

mysql_close ($Link);

I have this working fine on my computer testing environment, IIS 5.0 , PHP 4.21.

But when I upload it to my web host I get the Warnings.
I can see that they are using PHP 4.12, I'm not sure if that makes a difference.

thanks for any help.