Question

如何从网站获取XML响应？网站不是我的，所以我要去解析它。我使用WebView并通过JSONP解析代码。但是有一些需要的按钮，当我点击它们时，我得到了XML响应。请帮帮我，我不知道怎么做

Answer 1

要使用AsyncTask或Android-Async-Http从您的java代码发送请求，请将accept标头设置为application / xml并将其传递给您的网络视图。

以下是使用异步Http客户端的通用请求的示例：

import com.loopj.android.http.*;
import cz.msebera.android.httpclient.Header;


public class AppConnector {

    private String url;

    private void setUrl(String url){
        this.url = url;
    }

    public ListenableFuture<String> getSomething(){
        Log.d("App", "get something..");
        return launchHttpCall();
    }


    public ListenableFuture<String> launchHttpCall(){

        final SettableFuture<String> future = SettableFuture.create();
        AsyncHttpClient asyncHttpClient = new AsyncHttpClient();

        Log.d("App", "GET request towards: "+url);

        asyncHttpClient
            .get( url, new TextHttpResponseHandler()
            {
                @Override
                public void onSuccess(int statusCode, Header[] headers, String responseBody)
                {
                    Log.d("App", "HTTP CLIENT SUCCESS..");
                    future.set("Response Success. statusCode:"+statusCode+", Response body: "+responseBody);
                }

                @Override
                public void onFailure(int statusCode, Header[] headers, String responseBody, Throwable error){

                    Log.d("MeeRBA", "HTTP CLIENT FAILURE.. ");
                    future.setException(error);
                }

            });
        return future;
    }

这是 AsyncTask ：

的示例

class WebAsyncTask extends AsyncTask<String, Void, String> {

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

        // params comes from the execute() call: params[0] is the url.
        try {
            return goGoConnect(urls[0]);
        } catch (IOException e) {
            return "Unable to retrieve web page. URL may be invalid.";
        }
    }

    // onPostExecute displays the results of the AsyncTask.
    @Override
    protected void onPostExecute(String result) {
        Log.d("App", "Post execute..");

        // textView.setText(result);
        // or callback
   }

   private String goGoConnect(String myurl) throws IOException {

        InputStream is = null;
        // Only display the first 500 characters of the retrieved web page content.
        int len = 5000;

        try {
            URL url = new URL(myurl);
            HttpURLConnection conn = (HttpURLConnection) url.openConnection();
            conn.setReadTimeout(3000);
            conn.setConnectTimeout(5000); // millis
            conn.setRequestMethod("GET");
            conn.setDoInput(true);

            conn.setRequestProperty("User-Agent", USER_AGENT);
            conn.setRequestProperty("Accept-Language", "en-US,en;q=0.5");
            conn.setRequestProperty("Content-Type", "application/xml");

            // Starts the query
            conn.connect();

            int response = conn.getResponseCode();
            is = conn.getInputStream();

            // Convert the InputStream into a string
            InputStreamReader reader = null;
            reader = new InputStreamReader(is, "UTF-8");        
            char[] buffer = new char[len];
            reader.read(buffer);
            String contentAsString = new String(buffer);

            Log.d("App", "The response content is: "+contentAsString);

            return contentAsString;

        }catch(Exception e){

            Log.d("App", "Exception:"+e);
            return "Eception: >>"+e;

        // Makes sure that the InputStream is closed after the app is finished using it.
        } finally {
            if (is != null) {
                is.close();
            } 
        }
    }
}

如果您想在网页浏览中使用 javascript 发送GET请求，则可以使用以下内容：

function httpGetAsync(theUrl, callback)
{
    var xmlHttp = new XMLHttpRequest();
    xmlHttp.onreadystatechange = function() { 
        if (xmlHttp.readyState == 4 && xmlHttp.status == 200)
            callback(xmlHttp.responseText);
    }
    xmlHttp.open("GET", theUrl, true); // true for asynchronous 
    xmlHttp.send(null);
}

或使用jquery（或其他框架，如angularJS，用于更复杂的webview）。使用 jquery ：

的示例

$.ajax({
    type: "Get",
    url: "http://domain.com/function?Data=1234567890",
    xhrFields: {withCredentials: true},
    dataType: "JSONP text xml",
    contentType: "application/xml",
    cache: false,
    success: function(xml)
    {
    alert($(this).find('ResponseStatus').text());
    }
});

无论如何，来自javascript的跨域xml是禁止的，除非您可以控制正在吐出XML的应用程序并且可以使用格式化技巧来“欺骗”脚本将其解析为JSON。

如何获得XML响应

1 个答案: