Question

我想通过一个简单的URL连接到Web服务器（页面），该URL已包含我要发送的任何参数，例如：www.web-site.com/action.php/userid/42/secondpara/ 23 /然后获取网站生成的页面内容（不会是简单的OK / NOK）。我该如何设法做到这一点？我找不到任何符合我问题的示例代码或文档。

帮助你。

Answer 1

试试这个：

public static void connect(String url)
{

    HttpClient httpclient = new DefaultHttpClient();

    // Prepare a request object
    HttpGet httpget = new HttpGet(url); 

    // Execute the request
    HttpResponse response;
    try {
        response = httpclient.execute(httpget);
        // Examine the response status
        Log.i("Praeda",response.getStatusLine().toString());

        // Get hold of the response entity
        HttpEntity entity = response.getEntity();
        // If the response does not enclose an entity, there is no need
        // to worry about connection release

        if (entity != null) {

            // A Simple JSON Response Read
            InputStream instream = entity.getContent();
            String result= convertStreamToString(instream);
            // now you have the string representation of the HTML request
            instream.close();
        }


    } catch (Exception e) {}
}

    private static String convertStreamToString(InputStream is) {
    /*
     * To convert the InputStream to String we use the BufferedReader.readLine()
     * method. We iterate until the BufferedReader return null which means
     * there's no more data to read. Each line will appended to a StringBuilder
     * and returned as String.
     */
    BufferedReader reader = new BufferedReader(new InputStreamReader(is));
    StringBuilder sb = new StringBuilder();

    String line = null;
    try {
        while ((line = reader.readLine()) != null) {
            sb.append(line + "\n");
        }
    } catch (IOException e) {
        e.printStackTrace();
    } finally {
        try {
            is.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
    return sb.toString();
}

Android webrequest简单的解决方案

1 个答案: