穿甲兵技术社区 asyncins

docker tag 882db3a24154 registry.cn-hangzhou.aliyuncs.com/wqzcir/circles:v1.0.0

在 docker 里安装 gitlab runner，并将 runner 注册到 gitlab 中

gitlab-runner register \
  --non-interactive \
  --executor "docker" \
  --docker-image alpine:latest \
  --url "http://121.36.22.204/" \
 --registration-token "zHkqsD2ffFHNioqkG5r4" \

backup:

$ sudo su

$ cd /

$ tar cvpzf backup.tgz –exclude=/proc –exclude=/lost+found –exclude=/backup.tgz –exclude=/mnt –exclude=/sys /

根目录生成的 backup.tgz 就是备份结果

sudo nano /usr/share/applications/Pycharm.desktop

[Desktop Entry]
Name=Pycharm
Comment=Pycharm:The Python IDE
Exec=/usr/lib/pycharm-community/bin/pycharm.sh
Icon=/usr/lib/pycharm-community/bin/pycharm.png
Terminal=false
Type=Application

	(function() {
	'use strict';

	var t = window.xxxxx // 指定 window 对象
	Object.defineProperty(window, 'xxxxx', {
	get: function(){
	console.log("Get window property");
	return t;
	},
	set: function(val){

	# Definition for singly-linked list.
	# class ListNode:
	# def __init__(self, x):
	# self.val = x
	# self.next = None

	class Solution:
	def reverseList(self, head: ListNode) -> ListNode:
	prev = None
	current = head

	class TreeNode:
	def __init__(self, val):
	self.val = val
	self.left, self.right = None, None

	# 二叉树前中后序遍历
	class Solution:
	def __init__(self):
	self.traverse = []
	# 前序遍历组合

	# 将零移动到末尾，要求在原数组操作且尽可能减少操作次数
	source = [0, 3, 5, 0, 2, 1, 0, 9]

	def solution(source):
	j = 0
	for k, i in enumerate(source):
	if i != 0:
	source[j], source[k] = i, 0
	j += 1
	return source

	package main

	import (
	"errors"
	"fmt"
	"sync"
	"time"
	)

	const (

 /* 判断两个数组是否相同 */
 func ArrayEqual(a, b []string) bool {
 	if (a == nil) != (b == nil) {
 		return false
 	}
 	if len(a) != len(b) {
 		return false
 	}