swapping pages...

Anon

Hi,

I am building a site that runs from a MySQL database, via PHP. This is my first PHP project, so please bear with me.

I have several pages, home, profile, studio etc. When I click on a link to home, the link reads "index.php?page=home", or for profile page I link to "index.php?page=profile". Simple enough.

Now I have a function that reads what page we are on, and hopefully returns the data for that page. It works fine for home, but no other pages... I can't figure it out.

function get_Page_Data ($page) {

$page_img_details=get_Page_Img_Details($page_img);

switch ($page) {
		case "home":$page_data[0]=$page_img_details["page_title"];
					$page_data[1]=$page_img_details["page_title_img"];
					$page_data[2]=$page_img_details["page_img_filename01"];
					$page_data[3]=$page_img_details["page_text_title01"];
					$page_data[4]=$page_img_details["page_text01"];
					break;
		case "profile":$page_data[0]=$page_img_details["page_title"];
					$page_data[1]=$page_img_details["page_title_img"];
					$page_data[2]=$page_img_details["page_img_filename01"];
					$page_data[3]=$page_img_details["page_text_title01"];
					$page_data[4]=$page_img_details["page_text01"];
					break;
} 

return $page_data;

} // End function get_Page_Data

So, why does it only display data for Home and not for Profile etc? The PHP I am using to read this info in the index.php file is:

<? echo $page_data[3]; ?>

Can anyone help me please?

Thanks,

herve

What's the content of $page_img variable in :

$page_img_details=get_Page_Img_Details($page_img);

Have you check if it's good in either case ?

Anon

This is the original Function, and I swapped $page to $page_img as I thought it might be interfering with the second, switch $page Function.

function get_Page_Img_Details ($page) {

if (!($link=mysql_connect(DB_SERVER, DB_USER, DB_PASSWORD))) {
	displayErrMsg(sprintf("error connecting to db %d:%s\n", mysql_errno(), mysql_error()));
	exit();
}
$page=addslashes($page);
if (!($result=mysql_db_query(DB, "select pages_img.*, page_id, page_name, page_order, page_grouping, page_grouping_order, page_title, page_title_img, page_img_filename01, page_text_title01, page_text01, page_available from pages_img"))) {
	displayErrMsg(sprintf("error doing select %d:%s\n", mysql_errno(), mysql_error())); 
	exit();
}

$row=mysql_fetch_array($result);

return $row;

} // End function get_Page_Img_Details

herve

You should check the value of $page & $pageimg variable (by printing their
value on screen)

also I don't see
the
mysql_select_db(DB,$link);
call in the Get_Page_Img_Details function
so your mysql_db_query may not work at all

Anon

Hi,

thanks for that, printing the value of $page is helping me somewhat.

As for mysql_select_db(DB,$link);, I didnt include it in the example, but my code is working fine as it displays data from the database, but only for the home page.

Thanks,

Yari