PHP变量不适用于load函数

Question

This works fine in when using echo/print:

<?php echo ('http://test.com/%%CENTER_NAME_LINK%%/feed/'); ?>

But when I use load, which I need, it does not add in the variable:

<?php
$rss = new DOMDocument();
$rss-> load ('http://test.com/%%CENTER_NAME_LINK%%/feed/');
$feed = array(); ...

Here's the setup:

$center_name_link = $center['center_name'];
$center_name_link = str_replace(' ', '-', $center_name_link);
$center_name_link = strtolower($center_name_link);
$html = str_replace('%%CENTER_NAME_LINK%%', $center_name_link, $html);

How can I get the load to work with the variable?! I've tried using ' . $center_name_link . ', but that does not show up in either echo/print or load. I've tried creating a variable with the link already in it. I've tried a lot of other things, but I just can't get it to work out...

Answer 1

You need to use double quotes when you want to add a variable into a string eg

<?php
$rss = new DOMDocument();
$rss->load ("http://test.com/$CENTER_NAME_LINK/feed/"); // NOTE $ sign
$feed = array(); ..

You can also add variables with the dot operate eg .

$rss->load ("http://test.com/".$CENTER_NAME_LINK."/feed/"); // NOTE . between varialbe and text