Question

滑动时调用此操作：

-(IBAction) sliding:(id)sender{
    UISlider *temp = (UISlider *) sender;
    int value = (int)temp.value;
    val.text= [[NSString alloc]initWithFormat:@"%i",value ]; // val is the name oflabel   
}

但是当我这样做时：

-(IBAction) sliding:(id)sender{

        int value = (int)sender->value;// ERROR(no member named value) WHY ?
        val.text= [[NSString alloc]initWithFormat:@"%i",value ]; // val is the name of label   
    }

如何在不创建UISlider类

的情况下获取滑块的值

Answer 1

您必须创建UISlider的IBOutlet，然后您才能获得该值。

@property(nonatomic,weak) IBOutlet UISlider *slider;
-(IBAction) sliding:(id)sender
{
val.text= [[NSString alloc]initWithFormat:@"%i",self.slider.value];    
}

Answer 2

试试这个：

int value = (int)((UISlider *)sender->value);

Answer 3

我将Slider变量设为一个出口，并从.value获得了值，即myslider的getter。

INTERFACE

{  
   UISlider *slide;
   UILabel *val; 
  }

@property(nonatomic,retain) IBOutlet UISlider *slide;

实现：

-(IBAction) sliding:(id)sender{
   //UISlider *temp = (UISlider *) sender;
    int value = (int)slide.value;
    val.text= [[NSString alloc]initWithFormat:@"%i",value ];    
}

但我仍然希望它知道如何通过发送者???

UISlider通过发件人和类访问

3 个答案: