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"