Write Forums

Quick fix: Conditional statement with an array

Posted on 16th Feb 2014 07:03 pm by admin

Hi, I'm getting the temperature value off of the Environment Canada website along with the icon filename, so I can display my own icons and temperature on my site. All this works and I can display the temperature but my conditional statements aren't working to display my icons. I'm new to PHP and I don't see why this is a problem:
Code: <?php
$url = "http://www.weatheroffice.gc.ca/city/pages/on-69_metric_e.html";
$ch = curl_init();//Initialise CURL
curl_setopt($ch, CURLOPT_URL, $url);//Set the url
curl_setopt ($ch, CURLOPT_RETURNTRANSFER, 1);//We want it to return data
curl_setopt ($ch, CURLOPT_CONNECTTIMEOUT, 100);//Of course, we don't want your script to run forever, so set a timeout
$text = curl_exec($ch);//Execute and get the page

$exploded = explode('<p class="temperature">',$text);//Make an array of the part before and after the <title> tag
$result = explode('<sup>',$exploded[1]);//Grab the part after the <title> tag, and split that too

$exploded2 = explode('<img id="currentimg" src="/weathericons/',$text);//Make an array of the part before and after the <title> tag
$result2 = explode('.gif"',$exploded2[1]);//Grab the part after the <title> tag, and split that too

curl_close($ch);//Close cURL

if($result2[0] < "02"){
echo "<img src="images/weather/00.gif" alt="Sunny">";
}
if(($result2[0] > "01") && ($result2[0] < "09")) {
echo "<img src="images/weather/01.gif" alt="Cloudy with chance of showers">";
}
if($result2[0] == "09" || $result2[0] == "19" || $result2[0] == "39") {
echo "<img src="images/weather/03.gif" alt="Storm">";
}
if(($result2[0] > "9" && $result2[0] < "9") || ($result2[0] > "19" && $result2[0] < "25") || ($result2[0] > "26" && $result2[0] < "30") || ($result2[0] > "31" && $result2[0] < "39")) {
echo "<img src="images/weather/02.gif" alt="Cloudy with chance of showers">";
}
if(($result2[0] > "24" && $result2[0] < "27") || ($result2[0] > "29" && $result2[0] < "32") || ($result2[0] > "39") {
echo "";
}
else{
echo "";
}
echo $result[0];//Echo the result
?>

No comments posted yet

212 7 Tweet

 formating when pulling data from a mysql database 
Ok so Im not to sure if this is the right thread to post in but here is my catch 22 issue.

I 
Not sure why this script is not working? 
Hi I am new to php programing and I was trying to make up a simple script like a captcha but I canno
Line break? 
Hi, I'm new to the forum and new to php. I'm not sure if I'm using the correct terminology so here i
Simpler method of getting variables from mysql 
Hi Guys,

I'm trying to streamline my CMS's code and as I was writing a new page it occured to
Backflush 
I am looking for a report in SAP that would give me a list of all parts that are populated with a ba