Question

[Joshua F]

I am trying to make it so that it will load if the site if the site is online(EG. $siteonline = 1), and so on. I can't seem to think of how I could do this so it will work from the very top of the page's html, to the very bottom.

 

Like the following code is meant to list all of the stuff in that database.

<?php
$list_q = mysql_query("SELECT * FROM settings") or die (mysql_error());
while($list_f = mysql_fetch_assoc($list_q)) {
}
?>

 

I have my page laid out like this..

<?php 
include "includes/connect.php";
include "includes/config.php"; 
?>
<?php
    $list_q = mysql_query("SELECT * FROM settings") or die (mysql_error());
    while($list_f = mysql_fetch_assoc($list_q)) {
?>
Html content etc..
<?php 
}
include 'includes/footer.php'; ?>

 

But if I do it that way, It only displays what is in footer.php.

[shlumph]

You should redirect at the top, after your necessary includes, if the site is not online. Hope this helps.

[Joshua F]

You should redirect at the top, after your necessary includes, if the site is not online. Hope this helps.

 

This would work, but I'll have the site loading sitename from a database too, and that will be throughout the page(Title, Header, Footer)

[WatsonN]

Would something like this work?

if ($siteonline == 1){
include 'http://www.example.com/file.php?';
}

[fortnox007]

I am not using your code but just a standard one, but i think you get the idea: I am using a database column named active, and a table named settings.

 

$con = mysql_connect("localhost","peter","abc123");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("my_db", $con);

$result = mysql_query("SELECT * FROM settings LIMIT 1"); // i assume you only have 1 row in your settings table

while($row = mysql_fetch_array($result))
  {
  if($row['active']!= 1){ // your condition here, can be anything, you could set it to != 'monkeys' if you like : )
      exit; // or what ever you like die() header location () etc etc
    } else{
   //output your stuff to display here.
   }
  }

mysql_close($con);

 

Hope i could have been of any help.

[Joshua F]

Oh... I found the problem, it was I forgot to add a table into the database. Sorry for wasting your time guys.