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SourceReader.py
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87 lines (79 loc) · 4.2 KB
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# -*- coding: utf8 -*-
import sqlite3
""" 下面是用來取得一天有幾個班的函數 """
def CountJobQuantityInOneDay(c):
c.execute('select count(job_name) from job where enable = 1')
intJobQuantity = c.fetchall()[0][0]
return intJobQuantity
def ReturnJobsList(c):
c.execute('select * from job where enable = 1 order by job_order')
listJobsInOneDay = c.fetchall()
return listJobsInOneDay
def ShowForArrangeMemberTable(listMemberForArrange):
i = 0
for tupleMember in listMemberForArrange:
print(tupleMember, end = '\t')
if i % 4 == 3:
print()
i += 1
print()
def ReturnRegularMemberName(c, intMemberId):
c.execute('select name from member_array where ID = ' + str(intMemberId))
strMemberName = c.fetchall()[0][0]
return strMemberName
def ReturnMemberChange(conn):
c = conn.cursor()
c.execute('select * from MemberChange')
listMemberChange = c.fetchall()
#抓完要改變的人員清單後把表格清掉
conn.execute('delete from MemberChange')
c.execute('vacuum;')
conn.commit()
return listMemberChange
def RecordNamesAndPattern(listMemberForArrange, intStartMemberId):
listNowAndNextMemberNamesAndPattern = []
#迴圈跑十次，保留十個人名
for i in range(0,10,1):
#先看看排到的這個人在listMemberForArrange中出現幾次
listIdsAndNames = list(filter(lambda Names: Names[1] == listMemberForArrange[intStartMemberId - 1][1], listMemberForArrange))
#如果抓出來的list中有兩個元素，從第一個元素開始，如果ID = 現在排到的人的ID，則把listPattern的list塞入一個1。當然另外一個就是塞入0，如果抓出的元素只有一個，那麼listPattern只會有一個1。這可以告訴我們，如果是明旭，是前面的明旭還是後面的明旭，也就是製造pattern
listPattern=[]
for IDandName in listIdsAndNames:
if IDandName[0] == intStartMemberId:
listPattern.append(1)
else:
listPattern.append(0)
listNowAndNextMemberNamesAndPattern.append((listMemberForArrange[intStartMemberId - 1][1],) + (listPattern,))
intStartMemberId += 1
if intStartMemberId > len(listMemberForArrange):
intStartMemberId = 1
#listNowAndNextMemberNamesAndPattern裡面大概是這樣 [('明旭', [1,0]), ('慧枚', [1,0])....]
return listNowAndNextMemberNamesAndPattern
def CheckPattern(listZipped):
#為了避免這個問題 -> 就是萬一這個人本來沒有折扣，如果pattern是[0,1]那就ok。但是萬一他在listMemberForArrange加加減減的過程中變成有折扣，這時候如果pattern還是[0,1]，這就糟了。zip後可能會產生[0, (145, '詠馨')]....於是在下面的程式碼中，就變成最後回傳的intStartMemberId == 0
if (len(listZipped) == 1 and listZipped[0][0] == 0) or len(listZipped) == 0:
return 0
else:
return 1
def GetBackStarterID(listNowAndNextMemberNamesAndPattern, listMemberForArrange):
listIdsAndNames = []
i = 0
listZipped = []
#有保留十個人名，萬一一個個被抽走，還可以有十個機會。如果超過十個就會出現錯誤
#為了避免這個問題 -> 就是萬一這個人本來沒有折扣，如果pattern是[0,1]那就ok。但是萬一他在listMemberForArrange加加減減的過程中變成有折扣，這時候如果pattern還是[0,1]，這就糟了。zip後可能會產生[0, (145, '詠馨')]....於是在下面的程式碼中，就變成最後回傳的intStartMemberId == 0
while len(listIdsAndNames) == 0 or CheckPattern(listZipped) == 0:
listIdsAndNames = list(filter(lambda Names: Names[1] == listNowAndNextMemberNamesAndPattern[i][0], listMemberForArrange))
listZipped = list(zip(listNowAndNextMemberNamesAndPattern[i][1], listIdsAndNames))
i += 1
intStartMemberId = 0
for item in listZipped:
if item[0] == 1:
intStartMemberId = item[1][0]
return intStartMemberId
def ReturnHolidays(c):
c.execute('select date from Holidays')
list_of_tuplesHolidays = c.fetchall()
listHolidays = []
for Holiday in list_of_tuplesHolidays:
listHolidays.append(Holiday[0])
return listHolidays