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

**chriswattsuk** · 06-30-2008, 07:36 PM

Hi all,

I'm a newbie so please forgive me! I've tried finding the answer to this problem for a good couple of hours but not got anywhere...

I'm getting the following error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\wamp\www\chris\petdisplay.php on line 28

Code:

<?php
/* Program: petdisplay.php
* Desc: Displays all pets in selected category.
* /
?>
<html>
<head><title>Pet Catalog</title></head>
<body>
<?php
	$user="catalog";
	$host="localhost";
	$password="p4ssword";
	$database = "PetCatalog";
	$connection = mysql_connect($host,$user,$password)
		or die ("Couldn't connect to server");
	$db = mysql_select_db($database,$connection)
		or die ("Couldn't select database");
	$pettype = "horse"; //horse was typed in a form by user
	$query = "SELECT * FROM Pet WHERE petType='$pettype'";
	$result = mysql_query($query)
		or die ("Couldn't execute query.");

	/* Display results in a table */
	$pettype = ucfirst($pettype)."s";
	echo "<h1>$pettype</h1>";
	echo "<table cellspacing='15'>";
	echo "<tr><td colspan='3'><hr></td></tr>";
	while ($row = mysql_fetch_array($result))
	{
		extract($row);
		$f_price = number_format($price,2);
		 echo "<tr>\n
				<td>$petName</td>\n
				<td>$PetDescription</td>\n
				<td align='right'>\$$f_price</td>\n
				</tr>\n";
			echo "<tr><td colspan='3'><hr></td></tr>\n";
	}
	echo "</table>\n";
?>
</body></html>

Can anybody suggest what the problem might be? I'm working through my PHP + MySQL For Dummies book, but unfortunately the book doesn't offer any troubleshooting at this point! I've retyped the code and checked it many times but can't fathom the problem!

The offending line (28) is:

Code:

while ($row = mysql_fetch_array($result))

Many thanks in anticipation of any help!

**Nile** · 06-30-2008, 07:39 PM

This means that your connection to the database failed. Make sure that thats correct.

**chriswattsuk** · 06-30-2008, 08:01 PM

Thanks for your reply - I thought this piece of code throws up an error if the database connection fails?

Code:

$connection = mysql_connect($host,$user,$password)
		or die ("Couldn't connect to server");

I've checked the db connection details and they're correct - I've also tried with the root localhost user. I've also double checked the case and it all seems to tally.

Any other ideas?

**Nile** · 06-30-2008, 08:04 PM

Try this:

PHP Code:


mysql_connect($host,$user,$password)
or die ("Couldn't connect to server because: ".mysql_error());
mysql_select_db($database)
or die ("Couldn't select database because: ".mysql_error());

And then make sure the database has also the correct name.

**chriswattsuk** · 06-30-2008, 08:18 PM

I've inserted that code as described, still getting exactly the same output though on the page... no more details about the error.

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\wamp\www\chris\petdisplay.php on line 28

Thanks for your help.

**Nile** · 07-01-2008, 12:39 AM

Well, I'm just cleaning up your code here:

Code:

<?php
/* Program: petdisplay.php
* Desc: Displays all pets in selected category.
* /
?>
<html>
<head><title>Pet Catalog</title></head>
<body>
<?php
	$user="catalog";
	$host="localhost";
	$password="p4ssword";
	$database = "PetCatalog";
mysql_connect($host,$user,$password)
or die ("Couldn't connect to server because: ".mysql_error());
mysql_select_db($database)
or die ("Couldn't select database because: ".mysql_error());
	$pettype = "horse"; //horse was typed in a form by user
	$query = "SELECT * FROM `Pet` WHERE `petType`='{$pettype}'";
	$result = mysql_query($query)
		or die ("Couldn't execute query because: ".mysql_error());

	/* Display results in a table */
	$pettype = ucfirst($pettype)."s";
	echo <<< TABLE
        <h1>$pettype</h1>
	<table cellspacing='15'>
	<tr><td colspan='3'><hr></td></tr>
        TABLE;
	while ($row = mysql_fetch_array($result))
	{
		extract($row);
		$f_price = number_format($price,2);
		 echo <<<TEXTA
                               <tr>\n
				<td>$petName</td>\n
				<td>$PetDescription</td>\n
				<td align='right'>\$$f_price</td>\n
				</tr>\n
                                 TEXTA;
			echo "<tr><td colspan='3'><hr></td></tr>\n";
	}
	echo "</table>\n";
?>
</body></html>

**Jas** · 07-01-2008, 03:46 AM

Nile: that post seems pointless, unless you made some corrections I am missing?

chriswattsuk:

Not to intrude, but that error usually has nothing to do with the connection (some times it does, though), it has to do with the query. Specifically, the message is saying that the results you are trying to turn into an array (the var $results) is not a valid MySQL result, probably because the query was wrong. Try echo-ing out the query and check for syntax errors. You could also try:

Code:

echo $query;
$results = MySQL_Query($query) or die(MySQL_Error());

And then let us know what it echo's out.

If that doesn't work, consider the table/database privileges. Do you have access to the table/database?

**allahverdi** · 07-01-2008, 05:11 AM

umm. I think you all missed one thing.

change this:

Code:

$row = mysql_fetch_array($result)

to this:

Code:

$row = mysql_fetch_array($result, MYSQL_ASSOC)

**kuau** · 07-01-2008, 07:11 AM

I'm just a newbie but I thought you weren't allowed to use capitals in table names (?) in MySQL.

Code:

$query = "SELECT * FROM Pet WHERE petType='$pettype'";

**allahverdi** · 07-01-2008, 09:59 AM

kuau,

I'm not agree with you. I think it depends on table and column name.

I have used capitals and they are working fine. For example:

Code:

$result = mysql_query("SELECT * FROM tablename WHERE Id = '$id'");

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