iOS UIScrollView与滑动返回冲突

在网上搜罗了一些方法，因为我目前项目中的界面比较复杂，所以一些方法已被pass掉，目前还有以下方法可用。

方案一：

//当前为侧滑手势的时候设置scrollview手势失效才即可

    NSArray *gestureArray = self.navigationController.view.gestureRecognizers;
    for (UIGestureRecognizer *gesture in gestureArray) {
        if ([gesture isKindOfClass:[UIScreenEdgePanGestureRecognizer class]]) {
            [self.bottomView.panGestureRecognizer requireGestureRecognizerToFail:gesture];
        }
    }

+1s

    [_smallScrollView.panGestureRecognizer requireGestureRecognizerToFail:self.navigationController.interactivePopGestureRecognizer];

    [_bigScrollView.panGestureRecognizer requireGestureRecognizerToFail:self.navigationController.interactivePopGestureRecognizer];

手势返回冲突.jpg

方案二：新建继承自UIScrollView的类，然后将当前控制器中的ScrollView替换，在.m中重写一下方法

#import <UIKit/UIKit.h>

@interface BottomScrollView : UIScrollView
{
    
    BOOL isBack;
    
}
@end


#import "BottomScrollView.h"

@implementation BottomScrollView

-(BOOL) gestureRecognizerShouldBegin:(UIGestureRecognizer *)gestureRecognizer{
    if([self panBack:gestureRecognizer]) {
        return NO;
    }
    return YES;
}

- (BOOL)panBack:(UIGestureRecognizer*)gestureRecognizer{
    
    if(gestureRecognizer == self.panGestureRecognizer) {
        
        UIPanGestureRecognizer *pan = (UIPanGestureRecognizer*)gestureRecognizer;
        
        CGPoint point = [pan translationInView:self];
        
        UIGestureRecognizerState state = gestureRecognizer.state;
        
        if(UIGestureRecognizerStateBegan== state ||UIGestureRecognizerStatePossible== state) {
            
            CGPoint location = [gestureRecognizer locationInView:self];

            if(point.x>0&& location.x<150&&self.contentOffset.x<=0) {
                
                isBack=YES;
                return YES;
            }
        }
    }
    return NO;
}


-(BOOL)gestureRecognizer:(UIGestureRecognizer*)gestureRecognizer shouldRecognizeSimultaneouslyWithGestureRecognizer:(nonnull UIGestureRecognizer*)otherGestureRecognizer{
    
    if(isBack) {
        isBack=NO;
        return YES;
    }else{
        return NO;
    }
}

@end

非常感谢各位大佬的分享与交流，也希望我的发布能够帮助到跟多的人。