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[SOLVED] basic question about PHP include in head


suede1976

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Hi all, i am just just learning PHP and wrote my first if statement and it worked and i thought i was the bees knees, but it turns out it does not in fact work:

 

I have a basic index page structure like this:

 

<?php

$page = $_GET['page'];

if (!$page) {
$page = "home";
}

include_once("header.php");
include_once("content/".$page.".php");
include_once("menu.php");
include_once("footer.php");

?>

 

there are several PHP docs in the contents folder. in the header i want certain scripts only to load with a certain page so i did this in the header.php file:

 

<?php

if ($page = "portfolio") {
include_once("content/portScripts.php");
}

?>

 

I was hoping that, the file "portScripts.php" would only load when the page selected had the variable "portfolio" in the URL.

 

the thing is, it loads "portscripts.php" just fine, but every page has the "portfolio" content, no matter what the $page is in the URL it always comes up portfolio now.

 

before I added this IF statement to the header.php everything worked just fine. it seems somehow i made the $page variable always be "portfolio"

 

let me know if i need to include more info.

 

Thanks!

Instead of this:

 

$page = $_GET['page'];

if (!$page) {
$page = "home";
}

 

Just do this:

 

$page = (isset($_GET['page'])) ? $_GET['page'] : "home";

 

It's called a ternary  operator.

 

And this line should use double '=='.  Single is for assigning, double and triple is for comparing. (triple compares datatypes as well as value)

 

if ($page == "portfolio") {

<?php

if ($page = "portfolio") {
include_once("content/portScripts.php");
}

?>

 

In that if statement you're assigning the string 'portfolio' to the variable 'page', so all your if statementes will come true. If you want to test equality you need two equals signs;

 

<?php

if ($page == "portfolio") {
include_once("content/portScripts.php");
}

?>

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