SingleClientConnManager

SingleClientConnManager

我有这段代码,当我单击一个按钮时,该函数运行,但是此函数向php发送两个请求,而php在数据库中保存了两个记录...?

public void send()
{
    //get message from message box
    final String  msg = msgTextField.getText().toString();
    InputStream is = null;
    StringBuilder sb = null;
    String result = null;
    //check whether the msg empty or not
    StrictMode.ThreadPolicy policy = new StrictMode.ThreadPolicy.Builder().permitAll().build();
    StrictMode.setThreadPolicy(policy);
    if(msg.length() > 0) {
        try {
            HttpClient httpclient = new DefaultHttpClient();
            HttpPost httppost = new HttpPost("URL");
            httppost.setHeader("Content-Type", "application/x-www-form-urlencoded;charset=UTF-8");
            List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
            nameValuePairs.add(new BasicNameValuePair("Type", "MSG"));
            nameValuePairs.add(new BasicNameValuePair("Message", msg));
            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
            httpclient.execute(httppost);
            HttpResponse response = httpclient.execute(httppost);
            HttpEntity entity = response.getEntity();
            is = entity.getContent();
        } catch (Exception e) {
            Log.e("log_tag", "Error in http connection" + e.toString());
        }
        try {
            BufferedReader reader = new BufferedReader(new InputStreamReader(is, "UTF-8"), 8);
            sb = new StringBuilder();
            sb.append(reader.readLine());
            String line = "0";
            while ((line = reader.readLine()) != null) {
                sb.append(line);
            }
            is.close();
            result = sb.toString();
            LinearLayout Lay = (LinearLayout)findViewById(R.id.MSGes);
            TextView MyView = new TextView(this);
            MyView.setLayoutParams(new ViewGroup.LayoutParams(
                    ViewGroup.LayoutParams.WRAP_CONTENT,
                    ViewGroup.LayoutParams.WRAP_CONTENT));
            MyView.setText(msg);
            MyView.setId(Integer.valueOf(result.toString()));
            Lay.addView(MyView);

            } catch (Exception e) {
                Log.e("log_tag", "Error converting result " + e.toString());
            }

    } else {
        Toast.makeText(getBaseContext(),"Empty",Toast.LENGTH_SHORT).show();
    }


和logcat:

09-15 17:50:32.328:W / SingleClientConnManager(3651):无效使用SingleClientConnManager:连接仍被分配。
09-15 17:50:32.328:W / SingleClientConnManager(3651):确保在分配另一个之前释放连接。
09-15 17:50:32.758:E / log_tag(3651):转换结果java.lang.ArrayIndexOutOfBoundsException时出错:length = 1;索引= 1
09-15 17:50:35.188:W / SingleClientConnManager(3651):无效使用SingleClientConnManager:连接仍被分配。
09-15 17:50:35.188:W / SingleClientConnManager(3651):确保在分配另一个之前释放连接。
09-15 17:50:38.168:W / SingleClientConnManager(3651):无效使用SingleClientConnManager:仍分配了连接。
09-15 17:50:38.168:W / SingleClientConnManager(3651):确保在分配另一个之前释放连接。
09-15 17:50:38.538:E / log_tag(3651):转换结果java.lang.ArrayIndexOutOfBoundsException时出错:length = 1;索引= 1

最佳答案

我认为您在这里输入错误:

httpclient.execute(httppost);
HttpResponse response = httpclient.execute(httppost);


排除第一行,您就可以了。

09-04 23:17