我試圖通過提供JSON格式數據的Web服務請求天氣。我的PHP請求代碼沒有成功，是：$url="http://www.worldweatheronline.com/feed/weather.ashx?q=schruns,austria&format=json&num_of_days=5&key=8f2d1ea151085304102710";$json = file_get_contents($url);$data = json_decode($json, TRUE);echo $data[0]->weather->weatherIconUrl[0]->value; 這是返回的一些數據。為簡潔起見，部分細節已被截斷，但保留了對象完整性：{ "data": { "current_condition": [ { "cloudcover": "31", ... } ], "request": [ { "query": "Schruns, Austria", "type": "City" } ], "weather": [ { "date": "2010-10-27", "precipMM": "0.0", "tempMaxC": "3", "tempMaxF": "38", "tempMinC": "-13", "tempMinF": "9", "weatherCode": "113", "weatherDesc": [ {"value": "Sunny" } ], "weatherIconUrl": [ {"value": "http:\/\/www.worldweatheronline.com\/images\/wsymbols01_png_64\/wsymbol_0001_sunny.png" } ], "winddir16Point": "N", "winddirDegree": "356", "winddirection": "N", "windspeedKmph": "5", "windspeedMiles": "3" }, { "date": "2010-10-28", ... }, ... ] } }}
3 回答
紅顏莎娜
TA貢獻1842條經驗 獲得超13個贊
如果您使用以下代碼：
$json = file_get_contents($url);
$data = json_decode($json, TRUE);
TRUE返回一個數組而不是一個對象。
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