驱动控制浏览器 和排程算法

#从selenium框架中导入浏览器驱动
"""
我们就是使用浏览器驱动去控制我们浏览器
"""
from selenium import webdriver#自动化测试框架 ,  去对接scrapy框架去做爬虫
#WebDriverWait:是让浏览器等待加载
from selenium.webdriver.support.ui import WebDriverWait
import time
import log.settings


class taobao_infos:
    #对象初始化
    def __init__(self):
        #让浏览器指定的页面  淘宝登录页
        url="https://login.taobao.com/member/login.jhtml?"

        self.url=url
        self.brower=webdriver.Chrome(executable_path="G:\PY\chromedriver_win32\chromedriver.exe")
        # self.brower=webdriver.phantomjs #无界面浏览器
        self.wait=WebDriverWait(self.brower,10)

    #登录淘宝
    def login(self):
        #打开淘宝登录页
        self.brower.get(self.url)
        if self.brower.find_element_by_link_text("密码登录"):
            self.brower.find_element_by_link_text("密码登录").click()

        #等待微博登录选项出现
        if self.brower.find_element_by_xpath('//*[@id="J_OtherLogin"]/a[1]'):
            self.brower.find_element_by_xpath('//*[@id="J_OtherLogin"]/a[1]').click()
            time.sleep(1)
        #等待微博账号出现
        if self.brower.find_element_by_xpath('//*[@id="pl_login_logged"]/div/div[2]/div/input'):
            weibo_user=self.brower.find_element_by_xpath('//*[@id="pl_login_logged"]/div/div[2]/div/input')
            #send_keys是针对输入框自动填入你想要的值
            weibo_user.send_keys(weibo_username)

        #微博密码
        if self.brower.find_element_by_xpath('//*[@id="pl_login_logged"]/div/div[3]/div/span'):
            weibo_pwd=self.brower.find_element_by_xpath('//*[@id="pl_login_logged"]/div/div[3]/div/input')
            weibo_pwd.send_keys(weibo_password)

        #点击登录
        if self.brower.find_ele4ment_by_link_text("登录"):
            submit=self.brower.find_element_by_link_text("登录")
            submit.click()


if __name__=="__main__":
    weibo_username=log.settings.username
    weibo_password=log.settings.password

    #登录
    a=taobao_infos()
    a.login()

 排程

Python代码

def main(list1,list2,num):
    rows=zip(list1,list2)
    num=int(num)
    rows=list(rows)
    rows.sort(key=lambda x:x[1])
    rows.reverse()
    total=[]
    lists = [[] for i in range(num)]
    for m in range(0,num):
        total.append(8)  
    for  row in rows:
        print(row)
        min_time = total[0]
        k = 0
        for j in range(1,num):
            if min_time > total[j]:
                k = j
                min_time = total[j]      
        total[k]+=int(row[1])
        lists[k].append(row[0])
    return lists

C#代码

List<string> l1 = new List<string> { "k2", "k1", "k3","k4","k5" };
            List<double> l2 = new List<double> { 3,5,2,1,0.5};
            int num = 2;//2间手术室

            double temp = 0;
            string temps = "";
            for (int i = 0; i < l2.Count-1 ; i++)
            {
                for (int j = i+1; j < l2.Count; j++)
                {
                    if(l2[i]<l2[j])
                    {
                        temp = l2[i];
                        l2[i] = l2[j];
                        l2[j] = temp;

                        temps = l1[i];
                        l1[i] = l1[j];
                        l1[j] = temps;
                    }
                }
            }
           
            string[,] bString = new string[num, l1.Count()];
            int max = l1.Count() * num;
            int te = 0;
            int count = 0;
            int x = 0;
            for (int i = 0; i < max; i++)
            {
                bString[x, te] = l1[count];
                count += 1;
                x += 1;
                if (count>= l1.Count)
                {
                    break;
                }
                if (x == num)
                {
                    x = 0;
                    te += 1;
                }
            }

            for (int i = 0; i < num; i++)
            {
               
                for (int j = 0; j < l1.Count; j++)
                {
                    if (bString[i, j] != null)
                    {
                        string id = bString[i, j];
                    }
                }
            }
原文地址:https://www.cnblogs.com/ningxinjie/p/11013048.html