如果用户点击按钮,使用故事板,它将根据随机数导航到不同的类,我无法准备3个segues所以请帮助我。
我的代码是
- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender {
int randomNumber= arc4random() % 3;
NSIndexPath *indexPath = [tableview indexPathForSelectedRow];
if ([segue.identifier isEqualToString:@"randomsegue"]){
if (randomNumber==0){
//navigate to first view controller
}
else if(randomNumber==1){
//navigate to second view controller
}
else{
//navigate to third view controller
}
}
}
答案 0 :(得分:1)
检查一下,我认为你想要的是这个。
-(IBAction)buttonAction:(id)sender{
int randomNumber= arc4random() % 3;
switch (randomNumber) {
case 0:
[self performSegueWithIdentifier:@"segue1" sender:sender];
break;
case 1:
[self performSegueWithIdentifier:@"segue2" sender:sender];
break;
case 2:
[self performSegueWithIdentifier:@"segue3" sender:sender];
break;
default:
break;
}
}
- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender {
if ([segue.identifier isEqualToString:@"segue1"])
{
//navigate to view controller 1
}
else if ([segue.identifier isEqualToString:@"segue2"])
{
//navigate to view controllr 2
}
else{
//navigate to view controller 3
}
}
答案 1 :(得分:0)
使用标识符设置所有可能的seques,然后在switch语句中设置带有相应标识符的IBAction。