第二模块第一章作业

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"