sqlite php Notice: Undefined offset: 0 in

[mrooks1984]

hello all, i am stuck with a problem and hoping someone can help me.

i have got the following code to sort of work.

the problem i am having is when i put e.g. index.php?page=test and there is no information with test under page in the sqlite db i get this message, Notice: Undefined offset: 0 in C:\xampp\htdocs\test\Frontend\left.php on line 8.

as soon as i type index.php?page=test2 and i have test2 in the database the error is gone and shows the content

heres the code

        if (isset($_GET['page']))
        {
	$page = $_GET['page'];

	// Display Page.
	$query = "SELECT body FROM content WHERE page = '$page' AND location = 'left' AND disabled = 'no' ORDER BY id";
	$results = $base->arrayQuery($query, SQLITE_ASSOC);
	$arr = $results[0];

			echo '<div id="left">';		
                echo '<p>' . $arr['body'] . '</p>';
			echo '</div>';

	}		
        else
        {
                //Show Home Page
	$query = "SELECT body FROM content WHERE page = 'home' AND location = 'left' AND disabled = 'no' ORDER BY id";
	$results = $base->arrayQuery($query, SQLITE_ASSOC);
	$arr = $results[0];

			echo '<div id="left">';
                echo '<p>' . $arr['body'] . '</p>';
			echo '</div>';    
                
}

hope you understand what i am trying to say, thanks very much all.

 

 

 

[bh]

Hi,

 

Heres the problem:

$arr = $results[0];

Your $result variable doesnt have a 0 index (it doesnt exists). Check it before you get.

[mrooks1984]

thanks for you reply i am a newbie learning or tring to learn php, what do you mean by check it?

i have tried removing the 0 as i knew it was something to do with that part but it dident work.

[monkeytooth]

Do a count query.. if the count is 0 then show something else by default. If its 1 (assuming the variable your using would be a unique one) then show the information for that variable. Also since your using _GET/_POST type variables, I would suggest reading up on "Sanitizing your queries" and "mySQL injection prevention"