作者elveily (...)
看板java
标题[问题] airtime问题
时间Tue May 9 20:31:55 2006
我用书中的范例在WTK里run..确出现以下讯息
ConnectorDemo wants to send and receive data
using the network. this will use airtime and may result in charges.
is it ok to use airtime?
这要怎麽解决阿 ??
范例如下：
public ConnectorDemo() // 属性、元件、初值设定(初始化)
{
display = Display.getDisplay(this);
exitCommand = new Command("离开", Command.EXIT, 1);
connCommand = new Command("连线", Command.OK, 1);
}
public void startApp() // 程式进入点
{
showForm =new Form("HTTP 连线");
showForm.append(new String("请按「连线」键") );
showForm.addCommand(exitCommand);
showForm.addCommand(connCommand);
showForm.setCommandListener(this);
display.setCurrent(showForm);
}
public void pauseApp() { }
public void destroyApp(boolean unconditional) { }
public void commandAction(Command c, Displayable s) // 各事件处理函数
{
if(c==connCommand)
{
String showStr="";
HttpConnection httpConn = null;
try // 开启连线
{
httpConn =
(HttpConnection)Connector.open("
http://127.0.0.1:8080/index.htm");
if( httpConn.getResponseCode() == HttpConnection.HTTP_OK)
showStr = "连线成功！";
else
showStr = "连线失败！";
}
catch(Exception O)
{
showStr = O.getMessage();
}
try // 关闭连线
{
if(httpConn != null)
httpConn.close();
}
catch(Exception O)
{
showStr = O.getMessage();
}
// 将连线状态显示在 Alert 物件上
Alert alert = new Alert("连线状态", showStr, null, AlertType.INFO);
alert.setTimeout( Alert.FOREVER );
display.setCurrent( alert );
}
if(c==exitCommand)
{
destroyApp(false);
notifyDestroyed();
}
}
}
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