我想要创建一个脚本来安装我的raspberry pi上的所有东西(在重新安装之后或在另一个pi上,但这通常是linux问题)。我将命令逐个写入脚本中,但当我安装samba sudo apt-get install samba -y时,将打开以下问题并等待答复:
Samba server and utilities
If your computer gets IP address information from a DHCP server on the network,
the DHCP server may also provide information about WINS serve
我正在做一个学校项目,我必须在linux上设置一个基本的samba服务器,以便在两台计算机之间共享文件(我只需要显示它是否工作),一台计算机是一台在虚拟控制台上运行linux的膝上型计算机,我在虚拟控制台上安装了服务器,另一台是我的windows桌面计算机,在那里我试图访问samba服务器上的文件,但我无法让它工作!我有服务器运行,但我只是不能连接我的桌面计算机到它。我的smb.conf文件看起来像这样。
[global]
; General server settings
netbios name = YOUR_HOSTNAME
server string =
workgroup = YOU
我最近开始学习Python,我想测试一下自己。那么,你能告诉我我会从10中得到多少吗?我如何改进我未来的编码?
import random
import time
print('Welcome to Rock, Paper, Scissors')
print(' '*25)
wins = 0
loses = 0
draws = 0
point= int(input('How many rounds do you want to play?'))
list = ['r','p','s']
for x
def move(list, wins_1, wins_2):
global turn
if turn % 2 == 0:
sign = "| x "
else:
sign = "| o "
y_1 = int(input("Type the value of y: "))
x_1 = int(input("Type the value of x: "))
if list[y_1 - 1][x_1 - 1] == "| x " or list[y_1 - 1
我的职能是:
wins = {}
players={}
function Wins(name)
for i,wins in ipairs(wins) do
if name==wins then
return true
end
end
return false
end
function eventNewGame()
for name in pairs(tfm.get.room.playerList) do
counter[name]
对于以下子查询:
;WITH results as (
SELECT 'DAL' as team, 2010 as season, 7 as wins union
SELECT 'DAL' as team, 2011 as season, 11 as wins union
SELECT 'DAL' as team, 2012 as season, 11 as wins union
SELECT 'NE' as team, 2012 as season, 15 as wins union
SE
在下面的岩布剪刀程序中,我的循环出了点问题。使用设置长度的输入,每组游戏将玩N次。如果结果是玩家2,计算机0;玩家0,计算机2;玩家2,计算机5;玩家0,计算机4;一个额外的游戏被添加到集合中。我已经多次更改了函数,但找不到哪里出了问题。
def rpsls_play():
print("Welcome to the Rock-Scissors-Paper-Lizard-Spock game!")
player_sets=0
N=int(input("Select set length: "))
times=0
playe
嗨,我试着从4个玩家中计算出胜利者(这个游戏是经典的纸牌大战),但我只能让它做两个。代码的第一位是我的卡片类,第二位是主程序,其中有更多的if语句来计算胜利者。我遇到的问题是使用"getWinner“方法。如果需要的话,我还包括了甲板类(在底部)。包装卡;
public class Card
{
private int value, suit;
private String result, suitStr;
/**
* @param num represents value of card
* @param type represents the s
#include <iostream>
#include <ctime>
int *initialize(int []);
bool check(int random_num, int []);
int *draw(int []);
int *printout(int user_input, int []);
int entry(int user_input);
using namespace::std;
int wins[10]; // sets global array
int random_num; //stores global var for rando
在mysql8.0中有没有在JSON_ARRAYAGG或GROUP_CONCAT中进行限制的方法?例如: WITH season AS (
select 'DAL' as team, 4 as wins, 2000 as season UNION
select 'DAL' as team, 10 as wins, 2001 as season UNION
select 'DAL' as team, 9 as wins, 2002 as season UNION
select 'GB' as
我正在阅读一本Python编程入门书,作者在书中说,您希望避免直接访问类的属性,而是创建返回属性值的方法。然而,我发现有一些时候,这似乎有点多余,我想知道,如果只是采取了错误的方式。
这方面的一个例子是战斗函数,其中我直接访问player1.hand。这主意不好吗?还有,其他一切看上去都还好吗,还是有什么不好的做法,我应该扼杀在萌芽阶段?
#a game of rock paper scissors for two computers
#imports
import random
class Hand(object):
"""A hand with whic
假设我有以下数据集和查询: WITH results as (
SELECT 'DAL' as team, 2010 as season, 6 as wins union
SELECT 'DET' as team, 2010 as season, 6 as wins union
SELECT 'DET' as team, 2011 as season, 10 as wins union
SELECT 'DET' as team, 2012 as season, 4 as wins union
很抱歉,如果这是一个超级菜鸟的问题,但我拼命地尝试学习Backbone.js作为一份工作,我花了整整一周的时间阅读和观看视频在codeschool.com上,我仍然无法弄清楚。
我试图在模板中将json文件呈现为一个表,但我不知道自己做错了什么,而且每次尝试时都会做得越来越糟。
这是我的JS:
Game = Backbone.Model.extend({});
Games = Backbone.Collection.extend({
model: Game
})
window.AllGames = ne