0

Hi

I have used this file numerous time and it always worked, and I out of the blue I get this error now i have checked the script but i cant seem to find anything wrong.

<? 
function upload_image($img, $ext, $id) {
	
	$uploadpath = $_SERVER["DOCUMENT_ROOT"].'/css/news-images/';
	
	if ( ($img != 'none') && ($img != '' ) ) {
		
		$imagesize = getimagesize($img); 
		switch ( $imagesize[2] ) {
			
			case 0: //cannot use this image type.
				break;
			case 1: //gif.
				$dest = $uploadpath."News_".$id.'_'.$ext.'.gif';
				$img_fname = "News_".$id.'_'.$ext.'.gif';
				break;
			case 2: //jpg.
				$dest = $uploadpath."News_".$id.'_'.$ext.'.jpg';
				$img_fname = "News_".$id.'_'.$ext.'.jpg';
				break;
			case 3: //png.
				$dest = $uploadpath."News_".$id.'_'.$ext.'.png';
				$img_fname = "News_".$id.'_'.$ext.'.png';
				break;
        }
		if ( isset($dest) ) {
			
			if ( move_uploaded_file( $img, $dest ) ) {
				
				chmod ($dest, 0777);
			}
		}
	}
	
	return ( isset($img_fname) ) ?  $img_fname : "";
}

if ($_POST['Submit']) {
	
	if ($_POST['Title']=="") {
		$title_error = "<span class='admn-error'>You must enter a title for the news article.</span><br />";
	}
	if ($_POST['News']=="") {
		$story_error = "<span class='admn-error'>You must enter the content for the news article.</span><br />";
	}
	
	if ( !isset($title_error) && !isset($story_error) ) {
		
		//first we insert the data from the form.
		$sql = "INSERT INTO latest_news (
				ID, 
				Title, 
				News, 
				Date
			) VALUES (
				'', 
				'".$_POST['Title']."',
				'".$_POST['News']."',
				str_to_date('".$_POST['date']."', '%d-%m-%Y')
			)";
		$result = mysql_query($sql) or die (mysql_error());
		$id = mysql_insert_id();
		
		
		//now we upload the images to the hdd and get the image name(s).
		$image1 = upload_image($_FILES['image1']['tmp_name'], 1, $id);
		$image2 = upload_image($_FILES['image2']['tmp_name'], 2, $id);
		
		//and we put the image name(s) into the database so we can see them later.
		$sql = "UPDATE latest_news SET
				Image_name1 = '".$image1."',
				Image_name2 = '".$image2."'
				WHERE ID = '".$id."'";
		$result = mysql_query($sql) or die (mysql_error());
		
		//if everything was successful we thank the user for the submission.
		if ($id) {
			
			$msg = "<span class='error'>Thank you! The new news article has been created. You can see it <a href='view-news.php?ID=".$id."' target='_blank'>here</a>.</span>";
		} else {
			
			$msg = "<span class='error'>Sorry! Unfortunately something was wrong with your submission, please try again later.</span>";
		}
	}
}
?>
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7 Years
Discussion Span
Last Post by kireol
0

Hi

I have used this file numerous time and it always worked, and I out of the blue I get this error now i have checked the script but i cant seem to find anything wrong.

try <?php instead of <?

maybe someone changed the server

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