Python误差计算程序

Question

要求用户输入公司的工资信息。设置一个循环，继续询问信息，直到他们输入“完成”。对于每位员工询问三个问题：

姓名（名字和姓氏） 本周工作小时数（仅允许 1 到 60 小时） 小时工资（仅允许 6.00 至 20.00） 验证工作时间和小时工资，并确保输入姓名。

计算每个员工的工资，并将其写入顺序文件。请务必包含文件 I/O 错误处理逻辑。

仅包括每周工资 周工资计算方式： 对于（1-40 小时），它是小时费率 * 工作时间 对于（41-60 小时），其为（工作小时数 – 40）*（每小时费率 * 1.5） + 小时费率 * 40

输入所有员工后，将顺序文件读入名为 PAY 的列表中，作为每个员工的每周工资。对列表进行排序。现在打印本周的最低、最高和平均周薪。

我对这段代码有明显的问题

while len(eName)>0:
     eName=raw_input("\nPlease enter the employees' first and last name. ")
     hWork=raw_input("How many hours did they work this week? ")
     hoursWork=int(hWork)
     if hoursWork < 1 or hoursWork > 60:
         print "Employees' can't work less than 1 hour or more than 60 hours!"

     else:
         pRate=raw_input("What is their hourly rate? ")
         payRate=int(pRate)
         if payRate < 6 or payRate > 20:
              print "Employees' wages can't be lower than $6.00 or greater than $20.00!"
         if hoursWork <=40:
              grossPay=hoursWork*payRate
         else:
              grossPay=((hoursWork-40)*(payRate*1.5))+(40*payRate)
         lsthours.append(grossPay)
     print grossPay
     print lsthours



     ePass=raw_input("Type DONE when finished with employees' information. ")
     ePass.upper() == "DONE"
     if ePass == "DONE":
          break
     else:
          continue

Answer 1

这段代码有几个问题：

• 缩进到处都是。例如，while 循环在第一个 if 语句处结束。
• while 循环的测试几乎肯定是 false（因为 eName 未初始化），因此循环永远不会进入
• ePass.upper() == " 处的代码DONE”正在尝试设置 ePass 变量，这意味着测试将不起作用。您需要：

  如果 ePass.upper() == "DONE":
  中断

Answer 2

试试这个：

lsthours = list()
eName = "start" # initialize to something to start the loop
while eName:
    eName = raw_input("\nPlease enter the employees' first and last name. ")
    if not eName:
        break #loop will exit also when blank name is inserted
    hWork = raw_input("How many hours did they work this week? ")
    hoursWork = int(hWork)
    if hoursWork < 1 or hoursWork > 60:
        print "Employees' can't work less than 1 hour or more than 60 hours!"
        continue #skip

    pRate = raw_input("What is their hourly rate? ")
    payRate = int(pRate)
    if payRate < 6 or payRate > 20:
        print "Employees' wages can't be lower than $6.00 or greater than $20.00!"
        continue #skip
    if hoursWork <= 40:
        grossPay = hoursWork * payRate
    else:
        grossPay = ((hoursWork - 40) * (payRate * 1.5)) + (40 * payRate)
    lsthours.append(grossPay)
    print grossPay
    print lsthours
    ePass = raw_input("Type DONE when finished with employees' information. ")
    if ePass.upper() == "DONE":
        break

它仍然缺乏异常检查，但应该可以工作。

“数据错误”检查应该只是短路主循环，它更简单，但您可以有更复杂的代码并将它们放入自己的循环中。

Answer 3

已经指出的一些错误：

在Python中，缩进决定代码块

while循环：

while logic_test:
    # this is inside while loop
    ....
# this is outside while loop

字符串上的某些函数不会就地替换字符串，它们通过返回值返回另一个字符串< /strong>

upper：

>>> a = "done"
>>> a.upper()
'DONE'
>>> a
'done'
>>> 

始终初始化变量。

如果您使用序列方法，则变量应该早先定义为序列。

>>> t.append('ll')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 't' is not defined
>>> t = []
>>> t.append('ll')
>>> 

明确你的范围

lsthours = []
while len(eName)>0:
    ........
    lsthours.append(grossPay)

Answer 4

Yu 可以这样做：

grossPay = 0.0
lsthours = []

eName=raw_input("\nPlease enter the employees' first and last name (type 'PASS' to  exit): ") 

while eName.upper() != "PASS":      
   hWork=raw_input("How many hours did they work this week? ") 
   hoursWork=int(hWork)

   if hoursWork < 1 or hoursWork > 60: 
      print "Employees' can't work less than 1 hour or more than 60 hours!" 
   else: 
      pRate=raw_input("What is their hourly rate? ") 
      payRate=int(pRate) 

      if payRate < 6 or payRate > 20: 
         print "Employees' wages can't be lower than $6.00 or greater than $20.00!" 

      if hoursWork <=40: 
         grossPay=hoursWork*payRate 
      else: 
         grossPay=((hoursWork-40)*(payRate*1.5))+(40*payRate) 

      lsthours.append(grossPay) 

      print grossPay 
      print lsthours 

  eName=raw_input("\nPlease enter the employees' first and last name. (type 'PASS' to exit): ")