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使用json_decode在PHP中解析JSON对象

更新时间:2023-12-04 14:49:46

这似乎有效:

$url = 'http://www.worldweatheronline.com/feed/weather.ashx?q=schruns,austria&format=json&num_of_days=5&key=8f2d1ea151085304102710%22';
$content = file_get_contents($url);
$json = json_decode($content, true);

foreach($json['data']['weather'] as $item) {
    print $item['date'];
    print ' - ';
    print $item['weatherDesc'][0]['value'];
    print ' - ';
    print '<img src="' . $item['weatherIconUrl'][0]['value'] . '" border="0" alt="" />';
    print '<br>';
}

如果将json_decode的第二个参数设置为true,则会得到一个数组,因此无法使用->语法.我还建议您安装 JSONview Firefox扩展,因此您可以以类似于Firefox显示XML结构的漂亮格式的树状视图查看生成的json文档.这使事情变得容易得多.

If you set the second parameter of json_decode to true, you get an array, so you cant use the -> syntax. I would also suggest you install the JSONview Firefox extension, so you can view generated json documents in a nice formatted tree view similiar to how Firefox displays XML structures. This makes things a lot easier.