ios - 使用一个与 UISwitch 相同的 UIButton

Question

我想将 UIButton 用作开关，第一次点击表示打开，第二次点击表示关闭 - 并继续。

我读了一个stackoverflow article并按照描述创建了一个新类 ToggleButton。通过堆栈溢出的建议，编码级别不会出现错误消息。然而，在运行程序时，按下切换按钮并因错误“[ViewController ToggleOn:]:无法识别的选择器发送到实例”而终止。

是否需要进一步的编码改进？

我的示例代码如下。

//ToggleButton.h:

@interface ToggleButton : UIButton
@property (nonatomic, getter=isOn) BOOL on;
-(void)toggle;
@end


//ToggleButton.m:

#import "ToggleButton.h"
@interface ToggleButton ()
@property (nonatomic, strong) UIImage *offStateImage;
@property (nonatomic, strong) UIImage *onStateImage;
-(void)touchedUpInside:(UIButton*) sender;
@end

@implementation ToggleButton
@synthesize on = _on;
@synthesize offStateImage = _offStateImage;
@synthesize onStateImage = _onStateImage;

-(void)awakeFromNib
{
[super awakeFromNib];

self.offStateImage = [self imageForState:UIControlStateNormal];
self.onStateImage = [self imageForState:UIControlStateHighlighted];

[self addTarget:self
         action:@selector(touchedUpInside:)
forControlEvents:UIControlEventTouchUpInside];
}


-(void)touchedUpInside:(UIButton*) sender
{ [self toggle]; }

-(void)toggle
{self.on = (!_on);}

-(void)setOn:(BOOL) on
{
_on = on;

if (on)
    [self setImage:self.onStateImage forState:(UIControlStateNormal)];
else
    [self setImage:self.offStateImage forState:(UIControlStateNormal)];
}

@end


//  ViewController.h`
#import <UIKit/UIKit.h>
#import "ToggleButton.h"

@interface ViewController : UIViewController
@property (nonatomic, retain) IBOutlet UIActivityIndicatorView *ai;
@property (nonatomic, retain) IBOutlet UISwitch *sw;
@property (nonatomic, retain) IBOutlet ToggleButton *tb;
@end


//ViewController.m

#import "ViewController.h"
@interface ViewController ()
@end

@implementation ViewController
@synthesize ai;
@synthesize sw;
@synthesize tb;

- (void)viewDidLoad
{
[super viewDidLoad];
}

- (void)didReceiveMemoryWarning
{
[super didReceiveMemoryWarning];
}

//IBSwitch implementation

- (IBAction)SwitchChanged:(id)sender {
if(sw.on){    
[ai startAnimating];

}else{

[ai stopAnimating];}

}


//ToggleButton implementation

- (IBAction)ToggleButton:(id)sender {

if(tb.on){

    [ai startAnimating];

}else{

    [ai stopAnimating];}
}
@end

Answer 1

在界面构建器上添加一个按钮并将自定义类设置为ToggleButton(左起第三个选项卡)。

在ViewController.h中导入ToggleButton.h并添加这一行

@property (nonatomic, retain) IBOutlet ToggleButton *tb;

将 tb 和 - (IBAction)ToggleButton:(id)sender 连接到界面构建器上的按钮。

在 ToggleButton.m 类中替换此方法

-(void)toggle
{self.on = (_on);}

至

-(void)toggle
{self.on = (!_on);}

现在它可以正常工作了。