iphone - iPhone 上的键盘和起床状态

Question

如何确定键盘是否已启动？

我有一个成为第一响应者的 UISearchbar 实例。

当键盘出现时，通知会作为 API 的一部分发出，但我不想立即对此做出回应。我可以在 bool 状态中记录它，但这看起来很笨重。我想知道是否有一些“ setter/getter ”，我可以打电话了解一下。

Answer 1

我是这样做的:

键盘状态监听器.h

@interface KeyboardStateListener : NSObject {
    BOOL _isVisible;
}
+ (KeyboardStateListener *) sharedInstance;
@property (nonatomic, readonly, getter=isVisible) BOOL visible;
@end

KeyboardStateListener.m

#import "KeyboardStateListener.h"

static KeyboardStateListener *sharedObj;

@implementation KeyboardStateListener

+ (KeyboardStateListener *)sharedInstance
{
    return sharedObj;
}
+ (void)load
{
    NSAutoreleasePool *pool = [[NSAutoreleasePool alloc] init];
    sharedObj = [[self alloc] init];
    [pool release];
}
- (BOOL)isVisible
{
    return _isVisible;
}
- (void)didShow
{
    _isVisible = YES;
}
- (void)didHide
{
    _isVisible = NO;
}
- (id)init
{
    if ((self = [super init])) {
        NSNotificationCenter *center = [NSNotificationCenter defaultCenter];
        [center addObserver:self selector:@selector(didShow) name:UIKeyboardDidShowNotification object:nil];
        [center addObserver:self selector:@selector(didHide) name:UIKeyboardWillHideNotification object:nil];
    }
    return self;
}

-(void) dealloc {
    [[NSNotificationCenter defaultCenter] removeObserver:self];
    [super dealloc];
}

@end

然后用这个算出剩下的:

KeyboardStateListener *obj = [KeyboardStateListener sharedInstance];
if ([obj isVisible]) {
    //Keyboard is up
}