recurrence

nodejs 定时任务 node-schedule 库

巧了我就是萌 提交于 2020-08-17 12:01:00
node-schedule 是一个基于时间的调度,而不是基于区间的调度。你可以很容易的让他按照你的意思来干活,比如,你说“每五分钟来运行这个函数",你将发现 setInterval 要更容易使用,也是更适合的。但是如果你想说"运行这个函数在每个月的第三个星期二每个小时的20分和50分",你会发现你更想要Node Schedule组件。此外,Node Schedule 支持windows系统,不像cron并不支持。 注意 Node Schedule 是被设计来进行进程内调度,也就是说调度任务只能在你的脚本运行时才能有效以及调度将在执行成功后消失。如果你需要在你脚步 不 运行的时候调度任务,那就需要考虑使用cron. 任务和调度 每个在Node Schedule的计划任务都会被一个 Job 对象所代表,你可手动创建任务,然后执行 schedule() 方法来应用一个计划,或者使用一个方便的方法 ScheduleJob() 就像下面要说的。 Job 对象是 事件触发器 ,触发一个 run 事件在每次执行之后。 他们也触发一个 scheduled 事件,在每次他们调度运行的时候, canceled 事件可以让一个调用在它执行之前被取消(这两个事件都接受一个JavaScript日期对象作为一个参数). 注意这个任务会第一时间被调度,所以如果你使用 scheduleJob()

Nexus Repository Manager 2.x command injection vulnerability (CVE-2019-5475) bypassed twice

纵然是瞬间 提交于 2020-08-11 01:27:46
Author: Badcode and Longofo@Knownsec 404 Team Date: 2020/02/09 Chinese Version: https://paper.seebug.org/1260/ Foreword At the beginning of September 2019, we responded to the Nexus Repository Manager 2.x command injection vulnerability (CVE-2019-5475). The general reason and steps for recurrence are on Hackerone . It was announced that after emergency response to this vulnerability, we analyzed the patch to fix the vulnerability and found that the repair was incomplete and could still be bypassed. This article records two bypasses of the vulnerability. Although the fix version was released

Is there a formula for upcoming date using formula in Google sheet based on condition?

纵饮孤独 提交于 2020-07-09 06:59:29
问题 I have a google sheet where I need to get the next upcoming date based on the start date set in column A Any pointers are greatly appreciated? I am unable exhibit the efforts as I am completely new to this sort of recurrence using Google sheets https://docs.google.com/spreadsheets/d/1g_UNg4MjDy3gFufpjZtMRkbBz80K3scQdZOaiAnZi7I/edit#gid=0 回答1: This behavior (the next date from today including today) could be implemented manually by this formula: ={ "Next date from today"; ARRAYFORMULA( IFS( A2

Solving the recurrence T(n) = T(n / 2) + O(1) using the Master Theorem? [closed]

大兔子大兔子 提交于 2020-04-11 18:44:25
问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 6 years ago . I'm trying to solve a recurrence relation to find out the complexity of an algorithm using the Master Theorem and its recurrences concepts, how can I prove that: T(n) = T(n/2)+O(1) is T(n) = O(log(n)) ? Any explanation would be apprecciated!! 回答1: Your recurrence is T(n) = T(n / 2) + O(1) Since the Master

Solving the recurrence T(n) = T(n / 2) + O(1) using the Master Theorem? [closed]

╄→гoц情女王★ 提交于 2020-04-11 18:43:10
问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 6 years ago . I'm trying to solve a recurrence relation to find out the complexity of an algorithm using the Master Theorem and its recurrences concepts, how can I prove that: T(n) = T(n/2)+O(1) is T(n) = O(log(n)) ? Any explanation would be apprecciated!! 回答1: Your recurrence is T(n) = T(n / 2) + O(1) Since the Master

Solving the recurrence T(n) = T(n / 2) + O(1) using the Master Theorem? [closed]

不想你离开。 提交于 2020-04-11 18:42:27
问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 6 years ago . I'm trying to solve a recurrence relation to find out the complexity of an algorithm using the Master Theorem and its recurrences concepts, how can I prove that: T(n) = T(n/2)+O(1) is T(n) = O(log(n)) ? Any explanation would be apprecciated!! 回答1: Your recurrence is T(n) = T(n / 2) + O(1) Since the Master

基于arcface 人脸识别demo使用教程

ぐ巨炮叔叔 提交于 2020-02-29 13:28:59
最近在研究虹软家的arcface 人脸识别 demo,现在就给大家分享一下官方的demo ** 工程如何使用? ** 1.下载代码: git clone https://github.com/asdfqwrasdf/ArcFaceDemo.git 或者直接下载压缩包 2.前往官网申请appid和sdkkey。 修改 ArcFaceDemo-master\src\main\java\com\arcsoft\sdk_demo\FaceDB.java 下面的对应的值: public static String appid = "xxxx"; public static String fd_key = "xxxx"; public static String ft_key = "xxxx"; public static String fr_key = "xxxx"; 3.下载sdk包之后,解压各个包里libs中的文件到 ArcFaceDemo-master\libs 下,同名so直接覆盖。 4.Android Studio3.0 中直接打开或者导入Project,编译运行即可。 ** demo如何使用? ** 1.点击第一个按钮 打开图片或者拍一张带人脸的照片,确认后自动执行人脸,弹出注册框,注册第一个人脸。 注册界面底部会展示已注册的信息列表,点击列表项,则可以执行删除操作。 2

python 遍历多级目录下文件的方法

家住魔仙堡 提交于 2020-02-29 06:30:15
最近用Python读取文件夹下所有图片文件时,遇到一点点麻烦:该文件夹包含多级子文件夹。虽然不是什么困难事情,但对新手来说可能是一件抓头的事情。 如图:读取图片及子目录下的图片 废话不多说,直接列出两种方法 递归方法: def recurrence(path,file_list): for file in os.listdir(path): fs = os.path.join(path, file) if os.path.isfile(fs): file_list.append(fs) elif os.path.isdir(fs): recurrence(fs, file_list) if __name__=="__main__": path = 'C:\\Users\\Desktop\\captchaRec\\验证码图片1020' filenames = [] # 带路径的文件名存入列表 recurrence(path, filenames) print(len(filenames)) walk方法: os自带的遍历函数 def walk(path, file_list): ff = os.walk(path) for root, dirs, files in ff: for file in files: file_list.append(os.path.join(root,

Rrule from Datetime collection

南楼画角 提交于 2020-02-25 15:04:01
问题 I'm making a recurrent activity calendar system and i'm facing an issue. My activity has a collection of Period items, which contain a StartedAt Datetime and an EndedAt Datetime. A lot of libs allow me to convert my RRule from a string to Period-like items, but it seems that the reverse (Having multiple "DateTime-couples" converted to a RRule string) isn't common at all. Any option about this ? 回答1: A lot of libs allow me to convert my RRule from a string to Period-like items, but it seems

Solving the recurrence T(n) = 2T(sqrt(n))

别来无恙 提交于 2020-01-24 06:00:23
问题 I would like to solve the following recurrence relation: T(n) = 2T(√n); I'm guessing that T(n) = O(log log n) , but I'm not sure how to prove this. How would I show that this recurrence solves to O(log log n) ? 回答1: One idea would be to simplify the recurrence by introducing a new variable k such that 2 k = n. Then, the recurrence relation works out to T(2 k ) = 2T(2 k/2 ) If you then let S(k) = T(2 k ), you get the recurrence S(k) = 2S(k / 2) Note that this is equivalent to S(k) = 2S(k / 2)