I am pulling in weather data from a Yahoo weather RSS feed for both London and New York. I'm trying to reduce code duplication by reusing a PHP file which contains functions for pulling in the weather data.
Below is the function I am calling - get_current_weather_data() . This function goes off to various other functions such at the get_city() and get_temperature() functions.
<?php
function get_current_weather_data() {
// Get XML data from source
include_once 'index.php';
$feed = file_get_contents(if (isset($sourceFeed)) { echo $sourceFeed; });
// Check to ensure the feed exists
if (!$feed) {
die('Weather not found! Check feed URL');
}
$xml = new SimpleXmlElement($feed);
$weather = get_city($xml);
$weather = get_temperature($xml);
$weather = get_conditions($xml);
$weather = get_icon($xml);
return $weather;
}
In my index.php I set the URL for the RSS feed as the variable called $sourceFeed .
<?php
$tabTitle = ' | Home';
$pageIntroductionHeading = 'Welcome to our site';
$pageIntroductionContent = 'Twinz is a website which has been created to bring towns together!
Our goal is to bring communities around the world together, by providing
information about your home town and its twin town around the world. Our
site provides weather, news and background information for London and
one of its twin cities New York.';
$column1Heading = 'Current Weather for New York';
$column2Heading = 'Current Weather for London';
$column3Heading = 'Current Weather for Paris';
$sourceFeed = "http://weather.yahooapis.com/forecastrss?p=USNY0996&u=f";
include_once 'header.php';
include_once 'navigationMenu.php';
include_once 'threeColumnContainer.php';
include_once 'footer.php';
?>
I attempt to call the feed in my get_current_weather_data() function using:
(if (isset($sourceFeed)) { echo $sourceFeed; }).
However, I receive the following error
"Warning: file_get_contents() [function.file-get-contents]: Filename cannot be empty in C:\xampp\htdocs\Twinz2\nyWeather.php on line 10
Weather not found! Check feed URL".
If I replace
(if (isset($sourceFeed)) { echo $sourceFeed; })
with the URL for the feed it works but this will stop me from reusing the code. Am I trying to do the impossible or is my syntax just incorrect?
This isset method works fine where used elsewhere like for example the $tabTitle and $pageIntroductionHeading variables just need for the RSS feed.
Thanks in advance.
You're attempting to access a global variable $sourceFeed inside your function. Pass it as a parameter to the function instead:
ReplyDelete// Pass $sourceFeed as a function parameter:
function get_current_weather_data($sourceFeed) {
// Get XML data from source
include_once 'index.php';
$feed = file_get_contents($sourceFeed));
// Check to ensure the feed exists
if (!$feed) {
die('Weather not found! Check feed URL');
}
$xml = new SimpleXmlElement($feed);
$weather = get_city($xml);
$weather = get_temperature($xml);
$weather = get_conditions($xml);
$weather = get_icon($xml);
return $weather;
}
And call the function as:
$sourceFeed = "http://weather.yahooapis.com/forecastrss?p=USNY0996&u=f";
$weather = get_current_weather_data($sourceFeed);
Try to make $sourceFeed global like this:
ReplyDeletefunction get_current_weather_data() {
global $sourceFeed
OR
get_current_weather_data($sourceFeed)
Your issue is a variable scope issue. The variable that goes inside of the function is a new variable that is used only for the function. It will inaccessible once the function is returned/completed. So, you need to let the function know what the value is:
ReplyDeletefunction get_current_weather_data( $sourceFeed ) { // this tells the function to
// read that variable when it starts. You also need to pass it when you call the function.
get_current_weather_data( $sourceFeed );
// OR
get_current_weather_data( 'http://myurl.com' );
The problem is in the following line:
ReplyDelete$feed = file_get_contents(if (isset($sourceFeed)) { echo $sourceFeed; });
it has to be:
if(isset($sourceFeed)){ $feed = file_get_contents($sourceFeed); }
And when you call the function, you also have to pass $sourceFeed as function parameter, like that:
get_current_weather_data($sourceFeed);
I am fairly sure that what you have done there will not even parse. It certainly won't parse in PHP 5.2.
ReplyDelete$feed = file_get_contents(if (isset($sourceFeed)) { echo $sourceFeed; });
This is not valid. You cannot place an if statement inside a call to a function, and even if that did work, it would not affect the way the function is called.
You can use a ternary expression:
$feed = file_get_contents((isset($sourceFeed)) ? $sourceFeed : '');
...but even this will not help you here, since you have to pass a filename to file_get_contents() or you will get the error you see.
Your approach to this is all wrong, rather than including index.php to define your variable, you should pass the variable to the function as an argument. For example:
function get_current_weather_data($sourceFeed) {
// Get XML data from source
$feed = file_get_contents($sourceFeed);
// Check to ensure the feed exists
if (!$feed) {
die('Weather not found! Check feed URL');
}
$xml = new SimpleXmlElement($feed);
$weather = get_city($xml);
$weather = get_temperature($xml);
$weather = get_conditions($xml);
$weather = get_icon($xml);
return $weather;
}