第二模块第一章作业

电玩女神 2021-12-21 07:07 247阅读 0赞

知识：以后还要再回顾，到时再写

str.find()实际返回的是索引值，如果需对查找字符串作条件判断，第一个字符串刚好是索引0，也就是布尔值False，而找不到返回的是-1或者索引非0，则返回true。
    　　　　　---》用以下方法可对查找字符串作判断，根据.find 找不到字符串返回-1的特点，把返回的索引值+1再判断即可，找不到返回-1，+1后刚好变成0，为false，而找得到就算+1也是True

#! /usr/bin/env python
    #.-*- coding:utf-8 -*-
    #.__author__.='J"
    
    # 删除信息和第4题只操作一遍，所以不计数n了
    
    
    def file_to_D(): # 把文件内容变成以索引为key的字典，每个索引指向一条个人信息
        global D
        D={}
        f = open('E:test.txt','r')
        d = f.readlines()
        for i in d:
            i = i.split(',')
            i[0] = int(i[0])
            D[i[0]] = i
        print('总表',D)
        # print(len(D))
    
    def save_to_file():
        f = open('E:test.txt','w')
        for i in D.values():
            i = ','.join(i)
            f.write(i)
        f.flush()
        f.close()
    
    
    def question1_1(s):
        l1 = []
        n = 0
        for i in D.values():
            if int(i[2]) > s:
                l1.append((i[1],i[2]))
                n += 1
        print(l1)
        print('记录: ',n)
    
    
    def question1_2(s):
        global p
        p = []
        n = 0
        for i in D.values():
            if i[4] == s:
                p.append(i)
                n += 1
        print(p)
        print('记录: ', n)
    
    def question1_3(s):
        l3 = []
        n = 0
        for i in D.values():
            if　i[5].find(s)+1:  # 为什么这里找到2013返回的是false，百度上都说返回索引值---实际返回的是索引值，因为2013在日期字符串排第一位，也即是索引位为0，返回布尔值False。而如果搜索非首位的关键字，则这个代码是无效的，因为找不到返回的是-1，也就是true。---》根据这个特点，把返回值+1再判断即可，找不到返回-1，+1后刚好变成0，为false，而找得到就算+1也是True
                l3.append(i)
                n += 1
        print(l3)
        print('记录: ', n)
    
    
    def question2(name,age,phone,dept,birth):
        global n2
        global staff_id
        n2 = 1
        staff_id = next(a)  # next 冻结生成器，不是冻结函数，yield才会
        birth = str(birth)
        staff_id = str(staff_id)
        D[phone] = {staff_id,name,age,phone,dept,birth}
        print(D)
        save_to_file()  # 实际没写进去，为什么
        print('记录 ',n2)
    
    def question3(id):
        if 0 < id < len(D):
            D.pop(id)
            for i in D.values():
                i[0] = str(i[0])
        else:
            print('输错了')
    
    
        print(D)
        save_to_file()
    
    def question4(name='handsome', age = '666', dept= 'super man', birth = '1'):  # 默认了这几个参数不让输入
        for index,i in D.items():
            if dept == 'IT':
                i[4] = 'market'
                print('修改部门后的：',i)
            if name == 'Alex Li':
                i[2] = '25'
                print('修改年龄后的：',i)
            D[index] = i
        print('总表:', D)
        save_to_file()
    
    file_to_D()
    a = (i for i in range(1, 1000000001))
    
    
    syntax = input('syntax:>>')
    
    def judge_str(syntax):
        l = syntax.split(' ')  # 把命令变成一个个字符串的列表
        print('命令： ',l)
        for i in l:
            i.strip()
        # print(l)
        # if l[0] == 'find' and l[1] == 'name'  # 1_1 name,age这里判断不了，先不做
        if l[0] == 'find':  # 这里判断却不用对s进行eval（s)
            if l[1] == '*' and l[2] == 'from' and l[3] == 'staff_table' and l[4] == 'where' and l[5] == 'dept' and l[6] == '=' and len(l) == 8:
                s = l[7]
                # print(s)
                s = eval(s)   # 为什么加了这句后S就能识别到了
                if s in ['IT',"Sales",'HR','Marketing','Operation','Administration']:
                    question1_2(s)  # 为什么question1_2(s)返回的是一个空列表，函数读不到s变量字符串的值？
                else:
                    print('输入错误') #第一题第二问
    
            elif l[1] == 'name,age' and l[2] == 'from' and l[3] == 'staff_table' and l[4] == 'where' and l[5] == 'age' and l[6] == '>' and len(l) == 8:
                s = l[7]
                s = eval(s)
                print(s,type(s))
                question1_1(s)  #  第一题第一问
    
            elif l[1] == '*' and l[2] == 'from' and l[3] == 'staff_table' and l[4] == 'where' and l[5] == 'enroll_date' and l[6] == 'like' and len(l) == 8:
                s = l[7]
                # s = eval(s)
                print(s, type(s))
                question1_3(s)   
    　　　　　　# 第3题第3问
        elif l[0] == 'add':  # 第二题
            if l[1] == 'staff_table' and len(l) == 7:
                question2(l[2],str(l[3]),str(l[4]),l[5],l[6])
        elif l[0] == 'del':  # 第三题
            if l[1] == 'from' and l[2] == 'staff' and l[3] == 'where' and l[4] == 'id' and l[5] == '=' and len(l) == 7:
                s = int(l[6])
                question3(s)
        # 第四题不用写了，复制粘贴，一样套路
    
    judge_str(syntax)  # 实现根据任意部门查找个人信息find * from staff_table where dept = "IT"

转载于:https://www.cnblogs.com/jackfree/p/9671657.html