Question

这是我的班级：

public class Insert extends Activity 
{

    EditText name,surname,age;
    Button insert;

    @Override
    protected void onCreate(Bundle savedInstanceState) 
     {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.insert);
        StrictMode.ThreadPolicy policy = new StrictMode.ThreadPolicy.Builder().permitAll().build();

        StrictMode.setThreadPolicy(policy);
        name =(EditText)findViewById(R.id.name);
        surname =(EditText)findViewById(R.id.surname);
        age =(EditText)findViewById(R.id.age);
        insert=(Button)findViewById(R.id.click);
        insert.setOnClickListener(new OnClickListener()
        {
              public void onClick(View arg0)
            {
                // TODO Auto-generated method stub
                String nm = name.getText().toString();
                String ct = surname.getText().toString();
                String emailid = age.getText().toString();
                 insertRecords(nm, ct, emailid);
              }
            private void insertRecords(String nm,String ct,String emailid)
            {
                 ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(5);
                    nameValuePairs.add(new BasicNameValuePair("name", nm));
                    nameValuePairs.add(new BasicNameValuePair("surname",ct));
                    nameValuePairs.add(new BasicNameValuePair("age",emailid));
                    sendData(nameValuePairs);
            }
            private void sendData(ArrayList<NameValuePair> data)
            {
                try 
                {
                    HttpClient httpclient = new DefaultHttpClient();
                    HttpPost httppost = new HttpPost("http://10.0.2.2:81/new/insert.php");
                    httppost.setEntity(new UrlEncodedFormEntity(data));
                    HttpResponse response = httpclient.execute(httppost);

                }
                catch (Exception e) {
                    // TODO: handle exception
                    Log.e("log_tag", "Error:  "+e.toString());
                }
            }
        });
    }
    ...
}

在LogCat中

04-23 12:12:10.263: W/IInputConnectionWrapper(1066): showStatusIcon on inactive InputConnection

Answer 1

当前一页（或类）中的输入连接尚未关闭时，会出现此类问题。检查您是否已关闭上一课程中的输入连接（通过提供connection.close（））。

当您离开活动并保持HTTP连接打开时，会出现此问题。为防止这种情况，您可以尝试拨打

httpClient.getConnectionManager().closeIdleConnections();

您可以阅读此方法的文档here。

W / IInputConnectionWrapper（1066）：showStatusIcon在非活动的InputConnection上

1 个答案: