Taking input in link

Question

This is an api:=>

http://www.google.com/ig/api?weather=[city name]

When i manually put [city name] like http://www.google.com/ig/api?weather=dhakaThen,It's works perfectly.Is there any way to take user input for city name.

<?
$xml = simplexml_load_file('http://www.google.com/ig/api?weather=dhaka'); //Manually,I put 'dhaka' here
$information = $xml->xpath("/xml_api_reply/weather/forecast_information");
$current = $xml->xpath("/xml_api_reply/weather/current_conditions");
$forecast_list = $xml->xpath("/xml_api_reply/weather/forecast_conditions");
?>
<html>
    <head>
        <title>Google Weather API</title>
    </head>
    <body>
        <h1><?= print $information[0]->city['data']; ?></h1>
        <h2>Today's weather</h2>
        <div class="weather">       
            <img src="<?= 'http://www.google.com' . $current[0]->icon['data']?>" alt="weather"?>
            <span class="condition">
            <?= $current[0]->temp_f['data'] ?>&deg; F,
            <?= $current[0]->condition['data'] ?>
            </span>
        </div>
        <h2>Forecast</h2>
        <? foreach ($forecast_list as $forecast) : ?>
        <div class="weather">
            <img src="<?= 'http://www.google.com' . $forecast->icon['data']?>" alt="weather"?>
            <div><?= $forecast->day_of_week['data']; ?></div>
            <span class="condition">
                <?= $forecast->low['data'] ?>&deg; F - <?= $forecast->high['data'] ?>&deg; F,
                <?= $forecast->condition['data'] ?>
            </span>
        </div>  
        <? endforeach ?>
    </body>
</html>

Thanks in advance.

score 2 · Answer 1 · answered Dec 09 '11 at 12:59

2

Use html forms, get input from $_GET / $_POST and rawurlencode() it to api url.

answered Dec 09 '11 at 12:59

piotrekkr

2,785
2
21
35

score 2 · Accepted Answer · answered Dec 09 '11 at 13:00

You could create a form

<form method="post">
City <input type="text" name="cityName">
<br>
<input type="submit">
</form>

and change $xml = simplexml_load_file('http://www.google.com/ig/api?weather=dhaka'); to $xml = simplexml_load_file('http://www.google.com/ig/api?weather='.$_POST['cityName']);

Taking input in link

2 Answers2