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I'm struggling. I'm trying to make a website using the google weather API. Here's where I'm at:

1) I've got the PHP code working, and it gets the current temperature & images showing the forecast for the next 4 days. Simple enough

2) I decided I wanted to have a map which you could hover over to find out the temperature/forecast in seperate divs. I tried this with php then realised you can't do that without refreshing the whole page.

3) I realised I was probably going to have to use AJAX. I've managed to get the AJAX to do a httprequest and return all of the content in the PHP file, but what I really want to do is get the AJAX to call the PHP function, let it execute, then return the data for theat country/city.

I got stuck here. I'll show you what I have:

HTML FILE:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<html>
<head>
<script type="text/javascript" src="ajax.js">
</script>

<title>Weather HTML</title>
</head>
<body>

<p onmouseover='MakeRequest(0)' value='Madrid'>Madrid</p>
<p onmouseover='MakeRequest(1)' value='Brussels'>Brussels</p>
<p onmouseover='MakeRequest(2)' value='London'>London</p>

<div id='responseDiv'> 
This is a div to hold the response. I've not used any elements so this text will disappear.
</div>

</body>
</html>

AJAX FILE:

// JavaScript Document

function getXMLHttp()
{
  var xmlHttp

  try
  {
    //Firefox, Opera 8.0+, Safari
    xmlHttp = new XMLHttpRequest();
  }
  catch(e)
  {
    //Internet Explorer
    try
    {
      xmlHttp = new ActiveXObject("Msxml2.XMLHTTP");
    }
    catch(e)
    {
      	try
      		{
        		xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");
      		}
      	catch(e)
      		{
      		  	alert("Your browser does not support AJAX!")
        		return false;
      		}
    	}
  	}
	return xmlHttp;
}
 function MakeRequest(cityIndex)
{
  var xmlHttp = getXMLHttp();
 
  xmlHttp.onreadystatechange = function()
  {
    if(xmlHttp.readyState == 4)
    {
      HandleResponse(xmlHttp.responseText);
    }
  }

  xmlHttp.open("GET", "weather.php", true);
  xmlHttp.send(null);
}

function HandleResponse(response)
{
  document.getElementById('responseDiv').innerHTML = response;
}

PHP FILE:

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>


<?php

function getWeather($cityIndex) {
	
	$capitals=array("London","Madrid","Brussels","Paris","Rome");
	//Defines the Cities
	
	echo '<div class="cityTitle">';
	echo '<p>' . $capitals[$cityIndex] . '</p>';
	echo '</div>';
	//Capital City Title
	
	
	$requestAddress = "http://www.google.com/ig/api?weather=" . $capitals[$cityIndex];
	// Downloads weather data based on location (e.g. hull).
	
	$xml_str = file_get_contents($requestAddress,0);
	// Parses XML
	
	$xml = new SimplexmlElement($xml_str);
	// Loops XML
	
	$count = 0;
	echo '<div id="weather">';

    foreach($xml->weather as $item) {

        foreach($item->forecast_conditions as $new) {

            echo '<div class="weatherIcon">';
            echo '<img src="http://www.google.com/' .$new->icon['data'] . '"/><br/>';
            echo $new->day_of_week['data'];
            echo '</div>';
            }
			
		foreach($item->current_conditions as $new) {
			
            echo '<div class="temperature">';
			echo '<p>Current Temperature</p>';
            echo '<p>' .$new->temp_c['data'] . '°C<br/></p>';
            echo '</div>';
            }
    }

echo 	'</div>';
/*
echo 	'<p onmouseover="getWeather(1)" onmouseout="getWeather(0)">Madrid</p>'
echo	'<p onmouseover="getWeather(2)" onmouseout="getWeather(0)">Brussels</p>'
*/
}

getWeather(0);
getWeather(1);
getWeather(2);
getWeather(3);
getWeather(4);

?>

</body>
</html>

Hope that kinda makes sense, I would really appreciate any help.

Mike

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Contributors
3
Replies
4
Views
7 Years
Discussion Span
Last Post by electrogear
1

Hey Mike.

To get your current code to only show the city you are hovering over, you need to have AJAX send the index of the city to the PHP file, and have PHP only return the data for that city.

Your code is almost there, actually. There are only two modifications needed to make that work:

  1. Add the index of the city from your MakeRequest JavaScript function to the URI of the AJAX request
    xmlHttp.open("GET", "weather.php?cityindex=" + cityIndex, true);
  2. Have your PHP fetch the ID from the $_GET array and use that to call your getWeather function once.
    getWeather(intval($_GET['cityindex']));

And, as a minor tweak. Your PHP code returns only a part of your HTML markup, so it should not include the doctype declaration and the <html>, <head> and <body> tags. It should only return what you want added to your page.

Votes + Comments
Great help, thankyou
0

Thanks SOOOOOOO much for your help :) it was in the back of my brain somewhere but I don't do regular programming so I forget things. Anyway it works great. I was scratching my head for a while wondering why it didn't quite work then noticed in the PHP 'cityindex' instead of 'cityIndex' - serves me right for copy & pasting :D

Mike

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