mysql_fetch_array(): supplied argument is not a valid MySQL result

[SeanKann]

<?php
$DBhost = "localhost";
$DBuser = "root";
$DBName = "ooguildwww";
$table = "banner";
mysql_connect($DBhost,$DBuser,$DBpass);
@mysql_select_db($DBName) or die( "Unable to select database");
$result="SELECT url image FROM $table ORDER BY id RAND";
while ($row = mysql_fetch_array($result)) {


$url=mysql_result("url");
$image=mysql_result("image");
echo "URL: " . $row["url"];
echo "<br />";
echo "Image: " . $row["image"];
}
#echo "<a href=\"" . $row["url"] . "\"><img src=\"" . $row["image"] . "\" border=0 alt=\"" . $row["url"] . "\"></a>";
#}
mysql_close();
?>

On this line

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

Apache is reporting an error of mysql_fetch_array(): supplied argument is not a valid MySQL result

I've tried everything i can think of to get this to work but nothing .. Any ideas would be great

[planetsim]

That would be because you havent run the query try this

<?php
$DBhost = "localhost";
$DBuser = "root";
$DBName = "ooguildwww";
$table = "banner";
mysql_connect($DBhost,$DBuser,$DBpass);
@mysql_select_db($DBName) or die( "Unable to select database");
$result=mysql_query("SELECT url image FROM $table ORDER BY id RAND");
while ($row = mysql_fetch_array($result)) {


$url=mysql_result("url");
$image=mysql_result("image");
echo "URL: " . $row["url"];
echo "<br />";
echo "Image: " . $row["image"];
}
#echo "<a href=\"" . $row["url"] . "\"><img src=\"" . $row["image"] . "\" border=0 alt=\"" . $row["url"] . "\"></a>";
#}
mysql_close();
?> 

Also why are you using mysql_result("image"); etc they are not required

[johnnyv]

you need to perform a query before you can fetch the array. Try:

$connection = mysql_connect($DBhost,$DBuser,$DBpass); 
$sql="SELECT url image FROM $table ORDER BY id RAND";
$result = mysql_query($sql,$connection);

<edit>
Too late.
Also if you are only fetching an associative array then use mysql_fetch_assoc($result) instead of mysql_fetch_array($result)
</edit>

[Tracer]

You also need a comma here

$sql="SELECT url image FROM $table ORDER BY id RAND"; 

Like

$sql="SELECT url, image FROM $table ORDER BY id RAND"; 

[SeanKann]

Tried everything you guys said.. Still no go for me..

I don't know if this might have an effect on it.. I'm running Mysql 4.0.18 with php 4.3.8

Thanks for the help so far btw

[bodzan]

like johnyv said change your query to:

$connection = mysql_connect($DBhost,$DBuser,$DBpass);//connects to mysql db
$sql="SELECT url image FROM $table ORDER BY id RAND";//this is a query you're about to ask
$result = mysql_query($sql,$connection);//ask a query to the db from $connection

and just then use the while loop, like:

while ($row = mysql_fetch_assoc($result)) {

And then when you want to echo the result of your query, inside the loop, do:

echo $row["url"] and echo $row["image"]...

Like the guys said this is the way that it should be done... Have you tried everything that I summed up here?

[Tracer]

Additionally.. Try Echoing your SQL to the screen.... Maybe something wierd is going in there... Cut and Past it from your Screen into PhpMyAdmin's SQL and execute and make sure your getting a good return from MySQL..

$connection = mysql_connect($DBhost,$DBuser,$DBpass);//connects to mysql db
$sql="SELECT url image FROM $table ORDER BY id RAND";//this is a query you're about to ask
$result = mysql_query($sql,$connection);//ask a query to the db from $connection

echo "<br>My SQL: $sql<br>";