在链接中获取输入

发布于 2024-12-20 09:49:02 字数 1870 浏览 2 评论 0原文

这是一个api：=>

http://www.google.com/ig/api?weather=[城市名称]

当我手动输入 [城市名称] 时，例如 http:// www.google.com/ig/api?weather=dhaka然后，它工作得很好。有没有办法接受用户输入的城市名称。

<?
$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>

提前致谢。

原文

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'] ?>° 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'] ?>° F - <?= $forecast->high['data'] ?>° F,
                <?= $forecast->condition['data'] ?>
            </span>
        </div>  
        <? endforeach ?>
    </body>
</html>

Thanks in advance.

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最冷一天 2024-12-27 09:49:02

您可以创建一个表单

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

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

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']);

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