Question

我正在尝试从服务器请求响应，但是当我使用“HttpResponse response = client.execute（request）;”时，程序进入异常情况。

这是我的代码：

从服务器获取响应的功能

public String executeHttpGet（String username，String password）抛出异常{

    BufferedReader in = null;
    try {
        HttpClient client = new DefaultHttpClient();
        HttpGet request = new HttpGet();
        request.setURI(new URI("http://emapzoom.com/setting/device_login"+ "?device_id=" +password+ "&login_name="+ username));



        HttpResponse response = client.execute(request);
        in = new BufferedReader (new InputStreamReader(response.getEntity().getContent()));
        StringBuffer sb = new StringBuffer("");
        String line = "";
        String NL = System.getProperty("line.separator");
        while ((line = in.readLine()) != null) {
            sb.append(line + NL);
        }
        in.close();
        String page = sb.toString();
        System.out.println(page);
        return page;
    } finally {
        if (in != null) {
            try {
                in.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
    }
}

活动中使用的代码

     try{

            test=executeHttpGet(name,pass);

        }catch(Exception e){

        }

执行时，程序进入catch块！

请帮帮我!!! 提前thx！

Answer 1

如果您要针对任何版本的Android＆gt; = Honeycomb构建，则无法在主线程上进行网络调用。尝试将其放入异步任务中，看看它是否有效。

Answer 2

dell116的答案是对的。

我在ICS上遇到了同样的问题，并使用此代码异步解决它：

private void getResponseThread(final String url) {
    new Thread(new Runnable() {
        public void run() {
            String cadHTTP = getResponse(url);
            Message msg = new Message();
            msg.obj = cadHTTP;
            handlerHTTP.sendMessage(msg);
        }
    }).start();
}

private String getResponse(String url) {
    HttpClient httpClient = new DefaultHttpClient();
    HttpGet del = new HttpGet(url);
    del.setHeader("content-type", "application/json");

    String respStr;
    try {
        HttpResponse resp = httpClient.execute(del);
        respStr = EntityUtils.toString(resp.getEntity());
    } catch(Exception ex) {
        Log.e("RestService","Error!", ex);
        respStr = "";
    }

    Log.e("getResponse",respStr);
    return respStr;
}

private Handler handlerHTTP = new Handler() {
    @Override
    public void handleMessage(Message msg) {
        String res = (String) msg.obj;
        //CONTINUE HERE
        nexTask(res);
    }
};

问候！：）

使用HttpResponse response = client.execute（request）时获取异常;

2 个答案: