How to create a static html menu from a database


Posted on 16th Feb 2014 07:03 pm by admin

Hi,

I have built a small cms which allows me to create simple html pages and then upload them to an ftp.

Did you know?Explore Trending and Topic pages for more stories like this.
Everything is working apart from the menu, I cant get my head round how to create the menu system. The outputted file is a static html file that wont be connected to a database, so I need to retrieve the information from the database then write this information in standard html to a page before it is sent to the ftp.

I have looked through the forum and the net and can't find a solution, can anyone help?

CODE//// (which I know is incorrect, but kind of works and should hopefully show what I mean)

$page_name = mysql_query("SELECT * FROM pages WHERE page_display = 'yes' ORDER BY page_order ASC")
or die(mysql_error());


while($info = mysql_fetch_array( $page_name ))
{
$page[] = $info[page_name];
}

$navigation = <<<EOD


<li>$page[0]</li>
<li>$page[1]</li>
<li>$page[2]</li>
<li>$page[3]</li>


EOD;
No comments posted yet

Your Answer:

Login to answer
170 Like 51 Dislike
Previous forums Next forums
Other forums

How to show a complete textarea ??
I enter Client Case Notes notes in a textarea field on a PHP form. The field is 5 rows deep and 70 c

How a counter of users ? such as ---> (231 Viewing)
I want to count how many users are actually viewing the page, How is that possible?
Thank you guy

RSS poster script?
I have found a script that posts RSS's for me on a site that I'm building. However I would like to a

Intrastat Report Config in ECC 6.0 - goods Movement in EU countries
Colleagues, need ur help to understand the changes require in ECC 6 standard Report configuration se

Login Control?!
I have a Web Site that uses the login control also I have set the destinationurl to the page I want

Undefined index: username HELP NEWBIE
I am trying a simple login/logout for my website. It works well with checking if the username exists

Doubles are giving me problems
Ok so, first of all i made a double = 0.05, but when running the debugger it shows up as 0.04999. I

How to display value in drop down list after form has been reloaded
Please bear with me as I am very new to php and html.

I have a form with several drop down me

Call Screen statement Error
Dear Experts,

is there any setting that needs to be done before creating any dialog progr

split values
I have values that are returned to me in this format:

name=>test,age=>49

Sign up to write
Sign up now if you have flare of writing..
Login   |   Register
Follow Us
Indyaspeak @ Facebook Indyaspeak @ Twitter Indyaspeak @ Pinterest RSS



Play Free Quiz and Win Cash