在 iPhone 中的 imageView 中设置图像时出错

Question

我已经编写了一段用于设置 imageview 的代码...根据单击的按钮...但我的问题是当我按下按钮时它给了我这样的错误...函数 ImageNamed 的参数太多...执行此操作时line...

UIImage* photo = [UIImage imageNamed:@"robort %d.jpg",x];

整个代码是...

-(void)buttonClicked:(UIButton*)button
{
NSLog(@"%d",button.tag);
int x;
x=button.tag;
// Create the image view.
UIImage* photo = [UIImage imageNamed:@"robort %d.jpg",x];
NSLog(@"%d",x);
CGSize photoSize = [photo size];
UIImageView *imageView = [[UIImageView alloc] initWithFrame:CGRectMake(0.0, 0.0, photoSize.width, photoSize.height)];
[imageView setImage:photo];
[self.view addSubview:imageView]; }

Answer 1

使用 UIImage* photo = [UIImage imageNamed:[NSString stringWithFormat:@"robort de nero%d.jpg",x]];

干杯:)

Answer 2

试试这个

-(void)buttonClicked:(UIButton*)button
{
NSLog(@"%d",button.tag);
int x;
x=button.tag;
// Create the image view.
NSString *imageName = [NSString stringWithFormat:@"robort %d.jpg",x];
UIImage* photo = [UIImage imageNamed:imageName];
NSLog(@"%d",x);
CGSize photoSize = [photo size];
UIImageView *imageView = [[UIImageView alloc] initWithFrame:CGRectMake(0.0, 0.0, photoSize.width, photoSize.height)];
[imageView setImage:photo];
[self.view addSubview:imageView]; 
}

Answer 3

您需要使用：

UIImage* photo = [UIImage imageNamed:[NSString stringWithFormat:@"robort %d.jpg",x]];

Answer 4

用这个

UIImage* photo = [UIImage imageNamed:[NSString stringWithFormat:@"robort %d.jpg",x]];