这是我的课:
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
最佳答案
当上一页(或类)中的输入连接尚未关闭时，就会出现此类问题。检查你是否关闭了上一个类中的输入连接(通过给出connection.close())。
当您离开 Activity 并保持 HTTP 连接打开时会出现此问题。为了防止这种情况，您可以尝试调用
httpClient.getConnectionManager().closeIdleConnections();
您可以阅读此方法的文档 here .
关于android - W/IInputConnectionWrapper(1066) : showStatusIcon on inactive InputConnection，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/16170449/