Question

我正在尝试从openweatherapi获取一些xml，但是当url包含æ，ø或å时我遇到了问题。

失败的是：

URL website = new URL(params[0]);
InputStream inputStream = website.openStream(); // here it throws an FileNotFoundException
InputSource input = new InputSource(inputStream);

params [0]可以是：http://api.openweathermap.org/data/2.5/weather?q =århus，＆amp; mode = xml＆amp; units = metric

整个代码是：

public class GetWeatherInfo extends AsyncTask<String, Integer, String> {

    @Override
    protected String doInBackground(String... params) {

        String weatherInfo = null;

        try {
            URL website = new URL(params[0]);
            InputStream inputStream = website.openStream();
            InputSource input = new InputSource(inputStream);

            SAXParserFactory saxp = SAXParserFactory.newInstance();
            SAXParser sp = saxp.newSAXParser();
            XMLReader xmlReader = sp.getXMLReader();

            HandlingXMLStuff handler = new HandlingXMLStuff();
            xmlReader.setContentHandler(handler);               

            xmlReader.parse(input);             

            weatherInfo = handler.info.dataToString();

        } catch (Exception e) {
            e.printStackTrace();
        }
        return weatherInfo;
    }

Answer 1

您需要转义特殊字符。构造一个URI对象，然后使用toASCIIString()方法转义特殊字符。这可以按如下方式完成

    try {
        String url = "http://api.openweathermap.org/data/2.5/weather?q=århus,&mode=xml&units=metric";
        URI uri = new URI(url);
        URL escapedUrl = new URL(uri.toASCIIString());
    } catch (URISyntaxException e) {
        // handle exception
    } catch (MalformedURLException e) {
       // handle exception
    }

现在，这意味着，在您的示例中，您可以执行以下操作：

URL website = new URL(new URI(params[0]).toASCIIString());

使用URL.openStream（Java）打开æøå的URL

1 个答案: