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我在尝试从字符串中获取 HTTP 响应的正文时遇到了困难。所以根据我的代码，我应该在名为“responseString”的字符串中获得来自 PHP 页面的响应，但是因为这是在一个异步任务中，我无法在其他任何地方访问该字符串，那么我该如何接收该字符串？

例如，我想将此字符串拍摄到文本视图中以进行测试，我该怎么做？我是否错误地阅读了代码？我得到的响应是否没有被放入该字符串中？

这是我的代码：

class RequestTask extends AsyncTask<String, String, String>{

    @Override
    protected String doInBackground(String... uri) {
        HttpClient httpclient = new DefaultHttpClient();
        HttpResponse response;
        String responseString = null;
        try {
            response = httpclient.execute(new HttpGet(uri[0]));
            StatusLine statusLine = response.getStatusLine();
            if(statusLine.getStatusCode() == HttpStatus.SC_OK){
                ByteArrayOutputStream out = new ByteArrayOutputStream();
                response.getEntity().writeTo(out);
                out.close();
                responseString = out.toString();

            } else{
                //Closes the connection.
                response.getEntity().getContent().close();
                throw new IOException(statusLine.getReasonPhrase());
            }
        } catch (ClientProtocolException e) {
            //TODO Handle problems..
        } catch (IOException e) {
            //TODO Handle problems..
        }
        return responseString;
    }

    @Override
    protected void onPostExecute(String result) {
        super.onPostExecute(result);



    }
}

我正在使用以下内容执行：

new RequestTask().execute("http://www.mywebsite.com/android/registercheck.php?first=" + first2 + "&last=" + last2 + "&dispname=" + display2 + "&email=" + email2 + "&password=" + password2 );
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2 回答 2

1

执行后将直接启动，您可以doInBackground在onPostExecute此处更新 UI，因为您返回的是 responseString，然后onPostExecute将其放入 TextView

的返回doInBackgound是参数，onPostExecute所以 Stringresult是你的responseString.

@Override
protected void onPostExecute(String result) {
    super.onPostExecute(result);

    tv.setText(result);

}
于 2013-12-29T11:56:08.727 回答
0

我认为你的问题还没有解决。所以，

对于 Log 或 Toast，使用以下代码：

class RequestTask extends AsyncTask<String, String, String>{
String responseString = "";
@Override
protected String doInBackground(String... uri) {
    HttpClient httpclient = new DefaultHttpClient();
    HttpResponse response;

    try {
        response = httpclient.execute(new HttpGet(uri[0]));
        StatusLine statusLine = response.getStatusLine();
        if(statusLine.getStatusCode() == HttpStatus.SC_OK){
            ByteArrayOutputStream out = new ByteArrayOutputStream();
            response.getEntity().writeTo(out);
            out.close();
            responseString = out.toString();

        } else{
            //Closes the connection.
            response.getEntity().getContent().close();
            throw new IOException(statusLine.getReasonPhrase());
        }
    } catch (ClientProtocolException e) {
        //TODO Handle problems..
    } catch (IOException e) {
        //TODO Handle problems..
    }
    return responseString;
}

@Override
protected void onPostExecute(String result) {
    super.onPostExecute(result);
    Log.d("Result is", responseString );
    Toast.makeText(YourClass.this , responseString , Toast.LENGTH_SHORT).show();
 }
}
于 2013-12-29T13:02:49.607 回答