Variable returning empty

$result = mysql_query("SELECT Text
FROM `Pages`
WHERE Page = 'Projects'");
if($result)
{
	$text = $row['Text'];
}
else
{
	die("Error!");
}
?>

PHP:

Alright, so, above is the code I am using for my website. It is meant to set a variable ($text) to the value from my MySQL Database.

The problem I am having, is that when I use the code:

		<?
echo $text;
?>

PHP:

it doesn't show any text.

What am I doing wrong?
The value of Text in the database is "Test."


It is probably something simple to fix, but I just can't wrap my head around it.

 

can you dump your sql database (just sample)? So we can have a look on the content. It may be wrong query or incomplete code.

I know this sounds silly, do you have mysql_connect before that mysql_query right?

 

i think you are better off having a mysql_error() line in your main query line

ex : $result = mysql_query("SELECT Text FROM `Pages` WHERE Page = 'Projects'") or die('cant select '.mysql_error());

if there are errors during query, you can easily debug it

cheers

 

Yes, I do. Otherwise it would return as an error. My problem is that the Echo code doesn't show anything.

Here is the information, as per requested.

When I put

Into PHPMyAdmin it shows the information I am after.

Serialcoder- But I'm not having any errors right now with SQL. I will add that before my next upload, though. Thanks

 

hmm i dont see any mysql_fetch_row() or mysql_fetch_assoc() or even mysql_result(0, $rs)

 

do it like this

$result = mysql_query("SELECT Text FROM `Pages` WHERE Page = 'Projects'");
if($result)
{
$row = mysql_fetch_assoc($result);
$text = $row['Text'];
}
else
{
die("Error!");
}
?>

 

Oh, thanks mate
This is my first project using mySQL that I have created from scrath. I tend to use just files, but this case calls for SQL.