Is this efficient?

[toma]

I need a different menu for many pages on my site. I'm using an include file on each page to include the menu. Rather than have a separate menu file to include for each page could I do something like this:

 <div id="menu">
<a href="./" title="home" class="alt">home</a>
<?php
$url = $HTTP_SERVER_VARS["HTTP_HOST"] . $HTTP_SERVER_VARS["REQUEST_URI"];
if ($url = "www.site.com/page1.php") {
?> 
<font class="alt">page1 &raquo; </font>
<a href="link1.php">link1</a> 
<a href="link2.php">link2</a> 
<a href="link3.php">link3</a>     
 
<?php
}
if ($url = "www.site.com/page2.php") {
?> 
<font class="alt">page2 &raquo; </font>
<a href="link4.php">link4</a> 
<a href="link5.php">link5</a> 
<a href="link6.php">link6</a> 
<?php
}
?> 
</div>

If there are eventually many pages on the site, would this be an in
efficient

method? Also, why are both menus being included if I include this file on a page?

Thanks!

[guido2004]

if ($url = "www.site.com/page2.php")

must be

if ($url == "www.site.com/page2.php")


Using a single '=' gives the new value to $url, and it's result is always true

[mark_W]

I need a different menu for many pages on my site. I'm using an include file on each page to include the menu. Rather than have a separate menu file to include for each page could I do something like this:

 <div id="menu">
<a href="./" title="home" class="alt">home</a>
<?php
$url = $HTTP_SERVER_VARS["HTTP_HOST"] . $HTTP_SERVER_VARS["REQUEST_URI"];
if ($url = "www.site.com/page1.php") {
?> 
<font class="alt">page1 &raquo; </font>
<a href="link1.php">link1</a> 
<a href="link2.php">link2</a> 
<a href="link3.php">link3</a>     
 
<?php
}
if ($url = "www.site.com/page2.php") {
?> 
<font class="alt">page2 &raquo; </font>
<a href="link4.php">link4</a> 
<a href="link5.php">link5</a> 
<a href="link6.php">link6</a> 
<?php
}
?> 
</div>

If there are eventually many pages on the site, would this be an in
efficient

method? Also, why are both menus being included if I include this file on a page?

Thanks!

I used a similar method on a site before and it always worked well enough for me

[guido2004]

And yes, i can immagine it becoming an inefficient method if the number of pages, and different menus, grows.
You might put the menu-links in a table (columns 'page' and 'link'). Then you just extract the links for the page you're on and display them.

[toma]

if ($url = "www.site.com/page2.php")

must be

if ($url == "www.site.com/page2.php")


Using a single '=' gives the new value to $url, and it's result is always true

Of course...my oversight. Thanks guido2004, that did the trick!

[toma]

And yes, i can immagine it becoming an inefficient method if the number of pages, and different menus, grows.
You might put the menu-links in a table (columns 'page' and 'link'). Then you just extract the links for the page you're on and display them.

Hmmm...I assume you mean a database table for pages and corresponding links. I'll have to think a bit about how to extract these as needed....

Thanks for the good suggestion.

[AlexBrina]

I would do something like this, because if you need to change the menu format someday, you change only the buildMenu function.

function buildMenu( $page, $links )
{
  $html = "<font class=\"alt\">$title</font>";
  foreach( $links as $href => $text )
  {
    $html .= "<a href=\"$href\">$text</a>";
  }
  
  return $html;
}

switch ( $url )
{
  case "www.site.com/page1.php":
    $page = 'page1';
    $links = array( "link1.php"=>"link1", "link2.php"=>"link2" );
    break;
  case "www.site.com/page2.php":
    $page = 'page2';
    $links = array( "link4.php"=>"link4", "link5.php"=>"link5" );
    break;
}

echo buildMenu( $page, $links ); 

[toma]

I would do something like this, because if you need to change the menu format someday, you change only the buildMenu function.

function buildMenu( $page, $links )
{
$html = "<font class=\"alt\">$title</font>";
foreach( $links as $href => $text )
{
$html .= "<a href=\"$href\">$text</a>";
}
 
return $html;
}
 
switch ( $url )
{
case "www.site.com/page1.php":
$page = 'page1';
$links = array( "link1.php"=>"link1", "link2.php"=>"link2" );
break;
case "www.site.com/page2.php":
$page = 'page2';
$links = array( "link4.php"=>"link4", "link5.php"=>"link5" );
break;
}
 
echo buildMenu( $page, $links ); 

This looks like a good method too. I wonder if it's any more efficient when used with many pages and menus than the original version with "if" statements.

[AlexBrina]

umhh... more efficient... all right

function buildMenu( $menuSource ) {
  $html = "<font class=\"alt\">{$menuSource['page']}</font>";
  foreach( $menuSource['links'] as $href => $text ) {
    $html .= "<a href=\"$href\">$text</a>";
  }
  return $html;
}

function getMenuSource( $url ) {
  $page = substr( $url, strpos( $url, '/' ) + 1 );
  $links = include( 'menu_' . $page );
  return array( 'page'=>$page, 'links'=>$links );
}

echo buildMenu( getMenuSource( $url ) ); 

create one menu file for each page :

//file menu_page1.php
return array( "link1.php"=>"link1", "link2.php"=>"link2" ); 

//file menu_page2.php
return array( "link4.php"=>"link4", "link5.php"=>"link5" ); 

now you can have as many menus as you like, without touching your code, all you need to do is create another file corresponding to the page it belongs to.

[toma]

umhh... more efficient... all right

function buildMenu( $menuSource ) {
$html = "<font class=\"alt\">{$menuSource['page']}</font>";
foreach( $menuSource['links'] as $href => $text ) {
$html .= "<a href=\"$href\">$text</a>";
}
return $html;
}
 
function getMenuSource( $url ) {
$page = substr( $url, strpos( $url, '/' ) + 1 );
$links = include( 'menu_' . $page );
return array( 'page'=>$page, 'links'=>$links );
}
 
echo buildMenu( getMenuSource( $url ) ); 

create one menu file for each page :

//file menu_page1.php
return array( "link1.php"=>"link1", "link2.php"=>"link2" ); 

//file menu_page2.php
return array( "link4.php"=>"link4", "link5.php"=>"link5" ); 

now you can have as many menus as you like, without touching your code, all you need to do is create another file corresponding to the page it belongs to.

This looks great. I just worked vey hard to implement the other version, but I'll certainly have a closer look - AFTER Thanksgiving. Thanks much!

[reminder]

i suggest to insert into database the links with ids..that way u can easly edit the links from database in real time and update or delete it!
so when u go to some page use $_GET variable and switch the id of the links u need for that page!
cheers

[toma]

i suggest to insert into database the links with ids..that way u can easly edit the links from database in real time and update or delete it!
so when u go to some page use $_GET variable and switch the id of the links u need for that page!
cheers

Yes, thank you. I was just realizing the problem with _GET info passed along and I posted a new question about how to strip this out of my $url variable.

$url = $HTTP_SERVER_VARS["HTTP_HOST"] . $HTTP_SERVER_VARS["REQUEST_URI"]; 

How to remove the ? character and anything after?

[reminder]

u havent understand me!
when u pass a variable with get method then make a query...

$a = $_GET["url"];
$ret = mysql_query("SELECT *
                        FROM url_table
                        WHERE sub_id = '$a'
                        ORDER BY id DESC");
while($row = mysql_fetch_array($ret)) {
echo $row["url"];
} 

obviously u have to store all anchor links to db
cheers

ps $_SERVER not http_server_vars couse its deprecated!