Question

我想在选定的延迟时间延迟祝酒，如（15,30,60秒，没有延迟），但它不起作用。这是代码：

public void onItemSelected(AdapterView<?> parent,
            View view, int pos, long id) {
        if(FirstLoad){
            FirstLoad = false;
            return;                         
        }
          Toast.makeText(parent.getContext(), "You chose " + 
              parent.getItemAtPosition(pos).toString()+ " to delay", Toast.LENGTH_LONG).show();

          Message message = new Message();
          Bundle bun = new Bundle();
          bun.putString("delay", parent.getItemAtPosition(pos).toString());
          message.obj = bun;
          if (pos == 0) {
              handler.sendMessageDelayed(message, 0);
          }
          else if (pos == 1) {
              handler.sendMessageDelayed(message, 15000);
          }
          else if (pos == 2) {
              handler.sendMessageDelayed(message, 30000);
          }
          else if (pos == 3) {
              handler.sendMessageDelayed(message, 60000);
          }
          //handler.sendMessageDelayed(message, 15000);
        }

        public void onNothingSelected(AdapterView<?> parent) {
          return;
        }

请帮助。

Answer 1

试试这个：

    final Toast toast = Toast.makeText(parent.getContext(), "You chose "
            + parent.getItemAtPosition(pos).toString() + " to delay",
            Toast.LENGTH_LONG);

    Runnable showToastRunnable = new Runnable() {
        public void run() {
            toast.show();
        }
    };

    if (pos == 0) {
        handler.postDelayed(showToastRunnable, 0);
    } else if (pos == 1) {
        handler.postDelayed(showToastRunnable, 15000);
    } else if (pos == 2) {
        handler.postDelayed(showToastRunnable, 30000);
    } else if (pos == 3) {
        handler.postDelayed(showToastRunnable, 60000);
    }

编辑：

顺便说一句，我想把它转移到发送按钮，我想根据用户选择的延迟延迟“发送消息”的祝酒词。我该如何实施呢？

你是如何得到延迟的？是用户在EditText中输入的内容吗？在这种情况下，您可以像这样得到延迟：

int delay = Integer.parseInt(delayEditText.getText().toString());

然后使用该延迟量将runnable发布到处理程序，如下所示：

handler.postDelayed(showToastRunnable, delay);

在这种情况下，您可以删除整个if-else块。

Answer 2

以这种方式声明你的处理程序：

Hanlder handlder=new Handler() {

    public  void handleMessage (Message msg) {
         Toast.makeText(YOUR_ACTIVITY_CLASS_NAME.this,"You chose"+(Bundle(msg.obj)).getString("delay","DEFAULT_VALUE")+"to delay",Toast.LENGTH_LONG).show();
      }
    };

简单地说，您不必使用捆绑包，但可以调用msg.what = THE DELEY TIME。此外，您可以调用handler.obtainMessage来获取消息。见http://developer.android.com/reference/android/os/Handler.html#obtainMessage%28%29 因此，每次发送消息时，都会在此处理，因此您可以调用show toast。对不起，我没有在这台笔记本电脑上安装Eclipse，所以我无法测试代码。但是，我相信它有效。

Answer 3

for this you can use custom dialog and hide it after a particular time.

    class CustomDialog extends Dialog
    {
             setContentView(R.layout.dialogxml);    
         txtview=(TextView)findViewById(R.id.txtmsg);
    }
    Customdialog dialog= CustomDialog.show();

    dialog.hide();

Answer 4

Handler hl_DelayedToast = new Handler(); // scope global

public void onItemSelected(AdapterView<?> parent,View view, int pos, long id) 
{
    if(FirstLoad)
    {
        FirstLoad = false;
        return;                         
    }

    //if else logic to check the time
    // if 0
    hl_DelayedToast.postDelayed(mytoastshower,0);
    // if 1
    hl_DelayedToast.postDelayed(mytoastshower,1000);
}

public Runnable mytoastshower = new Runnable
{ 
    public void run()
    {
        Toast.show();// show the toast
    }
}

希望它有所帮助。

Android：Toast不会延迟微调器

4 个答案: