发出简单的 HTTP 请求并接收内容

Question

我从How can I make a simple HTTP request in MainActivity.java? (Android Studio) 在这里找到了一个简单的HTTP 请求的工作代码，我将在下面发布它（有一些更改，如果我没记错的话，现在必须使用try{} catch{}）。但是我想问我如何才能收到内容？我通过以下方式处理代码：

GetUrlContentTask req = new GetUrlContentTask();
req.execute("http://192.168.1.10/?pin=OFF1");
textView3.setText(req.doInBackground("http://192.168.1.10/?pin=OFF1")); 

GetUrlContentTask

private class GetUrlContentTask extends AsyncTask<String, Integer, String> {
    protected String doInBackground(String... urls) {
        // String content1 = "";
        try {
            URL url = new URL(urls[0]);
            HttpURLConnection connection = (HttpURLConnection) url.openConnection();
            connection.setRequestMethod("GET");
            connection.setDoOutput(true);
            connection.connect();


            BufferedReader rd = new BufferedReader(new InputStreamReader(connection.getInputStream()));
            String content = "", line;
            while ((line = rd.readLine()) != null) {
                content += line + "\n";
            }
            // content1 = content;
        }
        catch (Exception e) {
            e.printStackTrace();
        }

        // return content1; - returns "", wrong
        return "aaa";
        //does not work    return content;
    }

    protected void onProgressUpdate(Integer... progress) {
    }

    protected void onPostExecute(String result) {
        // this is executed on the main thread after the process is over
        // update your UI here    
    }
}

Answer 1

将此添加到您的onPostExecute(String result) 通话中

    protected void onPostExecute(String result) {
        // this is executed on the main thread after the process is over
        // update your UI here    
        setMyData(result);
    }

然后填写您的文本字段或您需要更新的任何其他数据。

    private void setMyData(String result){
        textView3.setText(result); 
    }

Answer 2

你到底卡在哪里了？您的代码大部分是正确的，尽管您可能需要稍微重新排列它。您的范围有点偏离，您的注释代码几乎可以到达那里。见下文

protected String doInBackground(String... urls) {
    String content = "", line = "";
    HttpURLConnection httpURLConnection;
    BufferedReader bufferedReader;
    URL url;

    try {
        for(String uri : urls) {
            url = new URL(uri);
            url = new URI(url.toURI().getScheme(), url.getAuthority(), url.getPath(), "pin=" + URLEncoder.encode("&OFF1", "UTF-8"), null).toURL();

            httpURLConnection = (HttpURLConnection) url.openConnection();
            httpURLConnection.setRequestMethod("GET");
            httpURLConnection.setDoOutput(true);
            httpURLConnection.connect();


            bufferedReader = new BufferedReader(new InputStreamReader(httpURLConnection.getInputStream()));
            while ((line = bufferedReader.readLine()) != null) {
                content += line + System.lineSeparator();
            }
        }
     } catch (Exception e) {
         e.printStackTrace();
     } finally {
         try {
            bufferedReader.close();
         } catch(Exception e) {
             e.printStackTrace();
         }
     }

     return content;

}

这将返回您页面的字符串表示形式。如果页面包含 HTML，它将返回 text/html，如果它包含文本，它将仅返回文本。

然后正如之前的用户所说，您可以在 GUI 上设置响应

protected void onPostExecute(String result) {
    textView3.setText(result);
}

上面的例子在技术上是可行的。但是，我强烈建议在 HttpURLConnection 上使用 http://loopj.com/android-async-http/。您的代码会更好，这将很重要。

我发送给您的代码假定参数始终为 pin=OFF1，因为它更像是一个概念证明