Question

我正在使用此代码为UIImageView生成随机图像。单击按钮时，它会显示另一个图像。我想问你如何创建四个图像视图 - 单击按钮 - ＆gt; apears四个不同的随机图像。

ViewController.h

@interface ViewController : UIViewController {
   IBOutlet UIImageView *imageView;
   // IBOutlet UIImageView *imageView2;
}
-(IBAction) randomImage;

ViewController.m

- (IBAction) randomImage {
NSArray *images = [[NSArray alloc] initWithObjects:@"image1.jpg", @"image2.jpg", @"image3.jpg", @"image4.jpg", nil];
int count = [images count]
int index = arc4random() % count; 

 imageView.image = [UIImage imageNamed:[images objectAtIndex:index]];
// imageView2.image = ....
}

我在故事板中创建了四个imageViews，我尝试了类似上面的内容// imageView2.image ...但它不是正确的解决方案:-) 谢谢你的帮助

Answer 1

列出4个随机图像的代码：

NSMutableArray *images = [[NSMutableArray alloc] initWithArray: @[@"image1.jpg", @"image2.jpg", @"image3.jpg", @"image4.jpg"]];
int countImg = images.count;

for (int i = 0; i < countImg; i++) {
    NSInteger index = arc4random() % images.count;
    NSLog(@"Image%d name = %@",i , [images objectAtIndex:index]);
    [images removeObjectAtIndex:index];
}

Answer 2

使用IBOutlet UIImageView *imageView;而不是使用IBOutletCollection(UIImageView) NSArray *imageViews;，并将所有图片视图连接到此。

然后，在您的代码中：

- (IBAction) randomImage {
    NSArray *images = [[NSArray alloc] initWithObjects:@"image1.jpg", @"image2.jpg", @"image3.jpg", @"image4.jpg", nil];
    NSInteger count = [images count];

    for (UIImageView *imageView in self.imageViews) {
        NSInteger index = arc4random() % count; 

        imageView.image = [UIImage imageNamed:[images objectAtIndex:index]];
    }
}

显示四个随机图像

2 个答案: