[SOLVED] banner ad on selected pages

[jarvis]

Hi,

 

We've a cms which has the following pages:

index.php - shows certain article headlines from a db (i.e. last 10 added)

 

article.php shows the actual article by id, i.e. when you click the article title the link is article.php?id=1 this then grabs the info relating to the article with an id of 1

 

contact.php shows the contact info

 

about.php shows the about info

 

What I need to do is assign a banner ad (image and link) which I add via a CMS script. Question is, how can I assign it to a specific page (article, index, about, contact etc)? Is there a way of given a page a hidden value and when that page is loaded it loads the relevant banner?

 

Hope that makes sense?

 

Many thanks

[mikesta707]

if ($_SERVER['PHP_SELF'] == "/index.php"){
//load the correct banner
}

 

Will get kind of clunky with more and more pages, but I would suggest putting the info in a database, or storing the relevant data into an array and getting it that way, like

 

//database way
$get = $_SERVER['PHP_SELF'];

$sql = "SELECT src FROM bannerTable WHERE page='$get'";
$go = mysql_query($sql);
$row = mysql_fetch_array($go);

echo "<img src='".$row['src']"' id='banner' />";

[jarvis]

This is great thanks! I've got a query though, my table allows you to enter the file name or select it from a db (via a drop down of article titles from the cms) if it's a dynamic page. My DB table for this part therefore looks like:

page_id page_name

1 10

2 21

3 index.php

4 3

5 contact_us.php

6 5

7 view_files.php

8 1

 

If I then use my code in the header file to get the image I want to display:

$id = $_GET['aid'];
$query = "SELECT page_id FROM pages WHERE page_name='$aid'";

This works to get the image for any page with an id number (dynamic page), however, it errors on those pages without, such as about_us.php, index.php and view_files.php.

 

How can I get around this?

 

Thanks so much!

[jarvis]

Sorry to bump but any help is much appreciated!

 

Thanks

[dave_sticky]

If I've understood what mean then something like this will allow for differentiation between an ID and a filename:

 

<?PHP
if (!isset($_GET['aid']) {
  $get = $_SERVER['PHP_SELF'];
  // Whatever you want it to do with the file name
} else {
  $aid  = $_GET['aid'];
  // Your SQL
}
?>

 

Is that what you were after?

 

[bundyxc]

if (isset($_GET['aid'])) { //If this there's a $_GET['aid'] in the URL, this page is dynamic. Consult database for the banner. 
    $sql = "SELECT `page_id` FROM `pages` WHERE `page_name`='" . $_GET['aid'] . "'";
    $query = mysql_query($sql);
    $fetch = mysql_fetch_assoc($query);
    $img = $fetch['page_id'];
} else { //Otherwise, this page isn't dynamic.
    if ($_SERVER['PHP_SELF'] == '/about_us.php') {
        $img = '/images/about_us_banner.bmp';
    } elseif ($_SERVER['PHP_SELF'] == '/index.php') {
        $img = '/images/index_banner.jpg';
    } elseif ($_SERVER['PHP_SELF'] == '/view_files.php') {
        $img = '/images/view_files_banner.png';
    }
}

//The $img variable now holds your banner URL.

[jarvis]

@dave_sticky- thanks I think this may work - will have a play & see!

 

@bundyxc - Although I like that, the prob is if the number of pages extends, I need to add to the code & is perhaps a little clumsy? Also, the file name part wont work in that way. As when you add a banner to a page via the cms, it creates a dir with the id number of that entry into the pages table and then the filename remains the same.

For example, if I select from my list of pages index.php and upload istock123.jpg, it then adds this to the pages table and an images table. A dir is created numbered 1 if it's the first entry and the image is uploaded.

 

I wonder if something like:

if (isset($_GET['aid'])) { //If this there's a $_GET['aid'] in the URL, this page is dynamic. Consult database for the banner. 
    $sql = "SELECT `page_id` FROM `pages` WHERE `page_name`='" . $_GET['aid'] . "'";
    $query = mysql_query($sql);
    $fetch = mysql_fetch_assoc($query);
    $img = $fetch['page_id'];
} else { //Otherwise, this page isn't dynamic.
  $get = $_SERVER['PHP_SELF'];

    $sql = "SELECT `page_id` FROM `pages` WHERE `page_name`='" . $get = $_SERVER['PHP_SELF'] . "'";
    $query = mysql_query($sql);
    $fetch = mysql_fetch_assoc($query);
    $img = $fetch['page_id'];
}

Would that work?

[dave_sticky]

$sql = "SELECT `page_id` FROM `pages` WHERE `page_name`='" . $get = $_SERVER['PHP_SELF'] . "'";

 

I wouldn't do that! Could be a copy & paste error though. You've already declared $get, so just use:

 

$sql = "SELECT `page_id` FROM `pages` WHERE `page_name`='" . $get . "'";

[bundyxc]

Well now you've got me all confused. I thought that $_GET['aid'] was the `page_id`. Is it the `page_name`?

[jarvis]

That would be perhaps. I guess what I need to do is determine whether the url is an id or page_name, then from there, pass whichever it is into the sql query!? Would this work

 

 

$sql = "SELECT `page_id` FROM `pages` WHERE `page_name`='";
if (!isset($_GET['aid']) {
  $get = $_SERVER['PHP_SELF'];
  
} else {
  $aid  = $_GET['aid'];
}
"'";

[bundyxc]

Once you have the page name/id, what do you do with it? Is there a seperate column on the database for the banner image, or how are you going to go about finding the banner URL?

[jarvis]

Hi bundyxc, appreciate the time and effort on this one!

 

Basically, all I need to do is grab either the id or page name and pass it to the query, the query will then return the page_id which I tie in with another query (see below), this is used for some code to display the relevant image.

 

When you upload an image to a page, a directory is created and the image uploaded to that dir. So, with the above it would be an image path of something like www.domain.com/banners/1/istock123.jpg

 

Where 1 is the dir created

DROP TABLE IF EXISTS `pages`;
CREATE TABLE IF NOT EXISTS `pages` (
  `page_id` smallint(4) unsigned NOT NULL AUTO_INCREMENT,
  `page_name` varchar(255) NOT NULL,
  PRIMARY KEY (`page_id`)
) ENGINE=InnoDB;

DROP TABLE IF EXISTS `banner_ads`;
CREATE TABLE IF NOT EXISTS `banner_ads` (
  `banner_id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `page_id` int(10) unsigned NOT NULL DEFAULT '0',
  `file_name` varchar(255) NOT NULL,
  `file_size` int(6) unsigned NOT NULL,
  `file_type` varchar(30) NOT NULL,
  `image_description` varchar(255) DEFAULT NULL,
  `web` varchar(150) DEFAULT NULL,
  `date_entered` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
  PRIMARY KEY (`banner_id`),
  KEY `file_name` (`file_name`),
  KEY `date_entered` (`date_entered`)
) ENGINE=InnoDB;

I hope that makes sense?

[bundyxc]

Of course. I'm actually headed to bed after I post this. It's about 5am, and I need to be up by 8am. haha. I swear, I'm almost too passionate about PHP.

 

Anyway, try this:

 

<?
if (isset($_GET['aid'])) {
$id = $_GET['aid'];
$fetch = mysql_fetch_assoc(mysql_query("SELECT `page_name` FROM `pages` WHERE `page_id`='" . $id . "'"));
$name = $fetch['page_name'];
} else {
$name = $_SERVER['PHP_SELF'];
$fetch = mysql_fetch_assoc(mysql_query("SELECT `page_id` FROM `pages` WHERE `page_name`='" . $name . "'"));
$id = $fetch['page_id'];
}
print('Page ID: ' . $id . '<br />Page Name: ' . $name;
?>

[jarvis]

I think that's it! Thank bundyxc.

 

Just one last thing, how do I just get the file name as that returns the whole path. I've the following but it doesn't work, it just says page name is array rather than the page name!

 

$url = $_SERVER["PHP_SELF"];
$name = Explode('/', $url);
$name[count($name) - 1];

#$name = $_SERVER['PHP_SELF'];
$query = "SELECT `page_id` FROM `pages` WHERE `page_name`='" . $name . "'";

If i echo $name it shows the correct file name. Am I being stupid?

[dave_sticky]

$url = $_SERVER["PHP_SELF"];
$name = Explode('/', $url);
$name[count($name) - 1];

#$name = $_SERVER['PHP_SELF'];
$query = "SELECT `page_id` FROM `pages` WHERE `page_name`='" . $name . "'";

 

2 ways of making this work...

 

1. Line 3, change to

$name = $name[count($name) - 1];

 

2. In your SQL Query, change $name to

$name[count($name) -1];

 

Your choice really. Both should work (but not at the same time)!

[jarvis]

Thanks dave_sticky! I think im now getting somewhere. One odd thing is that although the code works in sql if I run the query direct, it wont run in the script.

if (isset($_GET['aid'])) {
$id = $_GET['aid'];
$query = "SELECT `page_name` FROM `pages` WHERE `page_id`='" . $id . "'";
#echo $query;
$result = mysql_query($query);
$name = $row['page_name'];	

while ($row = mysql_fetch_array ($result, MYSQL_ASSOC)) {
	$default_image_sql = 'SELECT image_description, banner_id, page_id, file_name, web FROM banner_ads WHERE page_id = '.$id.'';
	echo $default_image_sql;

	$default_image_rs     =  mysql_query($default_image_sql) or die(mysql_error());
	$default_image_row    =  mysql_fetch_array($default_image_rs);
	$online_images_total_sql = 'SELECT count(file_name) AS total_images FROM banner_ads WHERE page_id = '.$id.'';
	echo $online_images_total_sql;
	$online_images_total_rs  = mysql_query($online_images_total_sql) or die(mysql_error());
	$total_images_uploaded   = mysql_fetch_array($online_images_total_rs);

	$image_available = $total_images_uploaded['total_images'];
	echo $image_available;

	if ($image_available == 0) {
	}else{
			// There is an image so display it
			$imgfile = 'banners/'.$default_image_row['page_id'].'/'.$default_image_row['file_name']; 
			if(file_exists($imgfile) && $image = @getimagesize($imgfile)) { 
				$t_imgfile='banners/'.$default_image_row['page_id'].'/t_'.$default_image_row['file_name']; 

				$web =  $default_image_row['web'];

				if ($web !=""){				
					echo '<a href="http://'.$web.'"><img src="'.$t_imgfile.'" alt="'.$default_image_row['image_description'].'" border="0" /></a>';
				} else {
					echo '<img src="'.$t_imgfile.'" alt="'.$default_image_row['image_description'].'" border="0" />';
				}

			}		
	}		

}


} else {

$url = $_SERVER["PHP_SELF"];
$name = Explode('/', $url);
$name[count($name) - 1];
$name = $name[count($name) - 1];

$query = "SELECT `page_id` FROM `pages` WHERE `page_name`  = '" . $name . "'"; ####THIS LINE###
echo $query.'<hr/>';
$result = mysql_query($query) or die(mysql_error());
$id = $row['page_id'];	

while ($row = mysql_fetch_array ($result, MYSQL_ASSOC)) {
	$default_image_sql = 'SELECT image_description, banner_id, page_id, file_name, web FROM banner_ads WHERE page_id = '.$name.'' ;
	#echo $default_image_sql;
	$default_image_rs     =  mysql_query($default_image_sql) or die(mysql_error());
	$default_image_row    =  mysql_fetch_array($default_image_rs);
	$online_images_total_sql = 'SELECT count(file_name) AS total_images FROM banner_ads WHERE page_id = '.$name.'';
	#echo $online_images_total_sql;
	$online_images_total_rs  = mysql_query($online_images_total_sql) or die(mysql_error());
	$total_images_uploaded   = mysql_fetch_array($online_images_total_rs);

	$image_available = $total_images_uploaded['total_images'];
	echo $image_available;

	if ($image_available == 0) {
	}else{
			// There is an image so display it
			$imgfile = 'banners/'.$default_image_row['page_id'].'/'.$default_image_row['file_name']; 
			if(file_exists($imgfile) && $image = @getimagesize($imgfile)) { 
				$t_imgfile='banners/'.$default_image_row['page_id'].'/t_'.$default_image_row['file_name']; 

				$web =  $default_image_row['web'];

				if ($web !=""){				
					echo '<a href="http://'.$web.'"><img src="'.$t_imgfile.'" alt="'.$default_image_row['image_description'].'" border="0" /></a>';
				} else {
					echo '<img src="'.$t_imgfile.'" alt="'.$default_image_row['image_description'].'" border="0" />';
				}

			}		
	}		

}	

}
print 'Page ID: ' . $id . '<br />Page Name: ' . $name;

I've marked the line which fails. If I browse to the page, it should return the page_id, browse to the dir with that number and then show the image. The query is fine in mysql but wont work above. have tried LIKE %% but that too fails.

 

Why is that? Am at a loss!

 

Thanks

[dave_sticky]

Hmmm, Interesting... See Comments below...

 

$url = $_SERVER["PHP_SELF"];
$name = Explode('/', $url);
$name[count($name) - 1]; // Delete this line... Doesn't need to be there, and don't know if it will be causing problems
$name = $name[count($name) - 1];

Other than that, when you say it doesn't work, what do you mean? Do you get an error message from the

or die(mysql_error());

bit?