iOS 简单易用的跳转方式

1、跳转方法：

//ViewController为要跳转的界面
id viewController = [[NSClassFromString(@"ViewController")alloc]init]; 
//判断urlString命名的字符串是否存在，如果不存在还进行传值会导致崩溃所以需要进行判断
if ([self checkIsExistValue:viewController andStringName:@"urlString"]) {
    //urlString是NSString为ViewController创建的字符串，此为传值方法
    [viewController setValue:@"https://www.baidu.com" forKey:@"urlString"];
}
[self.navigationController pushViewController:viewController animated:YES];

2、判断是否存在此字符串

- (BOOL)checkIsExistValue:(id)viewController andStringName:(NSString *)string
{
    unsigned int count = 0;
    // 获取对象里的属性列表
    objc_property_t * properties = class_copyPropertyList([viewController class], &count);
    for (int i = 0; i < count; i ++) {
        objc_property_t property = properties[i];
        // 属性名转成字符串
        NSString * propertyName = [[NSString alloc]initWithCString:property_getName(property) encoding:NSUTF8StringEncoding];
        // 判断该属性是否存在
        if ([propertyName isEqualToString:string]) {
            return YES;
        }
    }
    free(properties);
    return NO;
}