Question

我试图按照本教程进行操作： http://www.androidhive.info/2012/01/android-json-parsing-tutorial/

我在我的网址字符串中添加param时遇到问题，我需要将网址设为：

http://api.androidhive.info/contacts?email=params

和params是TextView的内容，我试图采用这样的参数：

String email = emailget.getText().toString();
RequestParams params = new RequestParams();
params.put("email", email);

但是我应该如何以及在哪里使用此参数来确保URL为：

http://api.androidhive.info/contacts?email=params

我需要它，因为在我的Web服务中我的SQL查询是：

SELECT * 
FROM Contact 
WHERE email = ?

我接受这样的查询参数：

@QueryParam("email") String email)

有人可以帮我解决这个问题吗？

提前致谢

Answer 1

使用StringBuilder来表示params ....

您能否显示更多代码？

Answer 2

        // If Your are using a Post Method than add a Params in List Name  value pair.

        List<NameValuePair> params = new ArrayList<NameValuePair>();
        params.add(new BasicNameValuePair("email",xyx@gmail.com));
        params.add(new BasicNameValuePair("password",12345678));


       // Post a request from server

public static  String post(String url,List<NameValuePair> params)
{
    InputStream is = null;
    try {
        DefaultHttpClient httpClient = new DefaultHttpClient();
        HttpPost httpPost = new HttpPost(url);
        httpPost.setEntity(new UrlEncodedFormEntity(params,"UTF-8"));   
        HttpResponse httpResponse = httpClient.execute(httpPost);
        HttpEntity httpEntity = httpResponse.getEntity();
        is = httpEntity.getContent();
    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    } catch (ClientProtocolException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }

    try {
        BufferedReader reader = new BufferedReader(new InputStreamReader(is, "UTF-8"), 8);
        String line = null;
        while ((line = reader.readLine()) != null) {
            respone=line.trim();
        }
        is.close();
    } catch (Exception e) {
        Log.e("Buffer Error", "Error converting result " + e.toString());
    }
    return respone;
}

Android AsyncTask从TextView获取param

2 个答案: