在我的一个应用程序中，如果用户在后台点击，我经常希望第一响应器辞职。为了这个目的，我在UIView上写了一个category，我在UIWindow上调用它。

下面的代码基于此，应该返回第一个响应器。

@implementation UIView (FindFirstResponder)
- (id)findFirstResponder
{
    if (self.isFirstResponder) {
        return self;        
    }
    for (UIView *subView in self.subviews) {
        id responder = [subView findFirstResponder];
        if (responder) return responder;
    }
    return nil;
}
@end

iOS 7 +

- (id)findFirstResponder
{
    if (self.isFirstResponder) {
        return self;
    }
    for (UIView *subView in self.view.subviews) {
        if ([subView isFirstResponder]) {
            return subView;
        }
    }
    return nil;
}

迅速:

extension UIView {
    var firstResponder: UIView? {
        guard !isFirstResponder else { return self }

        for subview in subviews {
            if let firstResponder = subview.firstResponder {
                return firstResponder
            }
        }

        return nil
    }
}

在Swift中的使用示例:

if let firstResponder = view.window?.firstResponder {
    // do something with `firstResponder`
}

@thomas-müller的快速回复

extension UIView {

    func firstResponder() -> UIView? {
        if self.isFirstResponder() {
            return self
        }

        for subview in self.subviews {
            if let firstResponder = subview.firstResponder() {
                return firstResponder
            }
        }

        return nil
    }

}

你也可以这样尝试:

- (void) touchesBegan: (NSSet *) touches withEvent: (UIEvent *) event { 

    for (id textField in self.view.subviews) {

        if ([textField isKindOfClass:[UITextField class]] && [textField isFirstResponder]) {
            [textField resignFirstResponder];
        }
    }
} 

我没有试过，但这似乎是个不错的解决办法

这是一个类别，允许您通过调用[UIResponder currentFirstResponder]快速找到第一个响应器。只需添加以下两个文件到您的项目:

UIResponder + FirstResponder.h:

#import <Cocoa/Cocoa.h>
@interface UIResponder (FirstResponder)
    +(id)currentFirstResponder;
@end

UIResponder + FirstResponder.m:

#import "UIResponder+FirstResponder.h"
static __weak id currentFirstResponder;
@implementation UIResponder (FirstResponder)
    +(id)currentFirstResponder {
         currentFirstResponder = nil;
         [[UIApplication sharedApplication] sendAction:@selector(findFirstResponder:) to:nil from:nil forEvent:nil];
         return currentFirstResponder;
    }
    -(void)findFirstResponder:(id)sender {
        currentFirstResponder = self;
    }
@end

这里的技巧是将一个动作发送给nil将它发送给第一个响应者。

(我最初在这里发表了这个答案:https://stackoverflow.com/a/14135456/322427)

下面是一个基于Jakob Egger最优秀的答案在Swift中实现的扩展:

import UIKit

extension UIResponder {
    // Swift 1.2 finally supports static vars!. If you use 1.1 see: 
    // http://stackoverflow.com/a/24924535/385979
    private weak static var _currentFirstResponder: UIResponder? = nil
    
    public class func currentFirstResponder() -> UIResponder? {
        UIResponder._currentFirstResponder = nil
        UIApplication.sharedApplication().sendAction("findFirstResponder:", to: nil, from: nil, forEvent: nil)
        return UIResponder._currentFirstResponder
    }
    
    internal func findFirstResponder(sender: AnyObject) {
        UIResponder._currentFirstResponder = self
    }
}

斯威夫特4

import UIKit    

extension UIResponder {
    private weak static var _currentFirstResponder: UIResponder? = nil
    
    public static var current: UIResponder? {
        UIResponder._currentFirstResponder = nil
        UIApplication.shared.sendAction(#selector(findFirstResponder(sender:)), to: nil, from: nil, for: nil)
        return UIResponder._currentFirstResponder
    }
    
    @objc internal func findFirstResponder(sender: AnyObject) {
        UIResponder._currentFirstResponder = self
    }
}

这是递归的一个很好的候选!不需要添加一个类别到UIView。

用法(来自你的视图控制器):

UIView *firstResponder = [self findFirstResponder:[self view]];

代码:

// This is a recursive function
- (UIView *)findFirstResponder:(UIView *)view {

    if ([view isFirstResponder]) return view; // Base case

    for (UIView *subView in [view subviews]) {
        if ([self findFirstResponder:subView]) return subView; // Recursion
    }
    return nil;
}