我是新的iPhone應用程序。我想在按下按鈕時每2秒顯示一個消息 爲此,我正在使用此代碼。有關performSelector
此代碼僅工作一次。這意味着只打一次電話。你能幫我解決這個問題嗎?
-(IBAction)fortunecookieAction:(id)sender
{
[self performSelector:@selector(showfortune) withObject:nil afterDelay:2.0];
}
-(void)showfortune
{
int number=arc4random()%5;
switch (number) {
case 0:
[email protected]"A holiday takes you back to the summer of '69";
break;
case 1:
[email protected]"A meal turns erotic muffin";
break;
case 2:
[email protected]"A massage brings";
break;
case 3:
[email protected]"A letter in the pa special delivery";
break;
case 4:
[email protected]"A spillage tuoo";
break;
default:
break;
}
}
我不明白這個問題。輕按按鈕後,是否要重複更改文字? –