错误w / Floats

Question

我正在开发小型python工资单项目，您可以在其中输入员工姓名，工资和工时。当我输入工资输入的小数时，由于我的异常处理，我得到“无效输入”。为什么小数被返回为无效？此外，如何循环此程序，以便它保持相同的3个问题，直到用户键入“完成”？任何帮助将不胜感激！谢谢！

import cPickle

def getName():
    strName="dummy"
    lstNames=[]
    strName=raw_input("Enter employee's Name: ")
    lstNames.append(strName.title() + " \n")


def getWage():
    lstWage=[]
    strNum="0"
    blnDone=False
    while blnDone==False: #loop to stay in program until valid data is entered
        try:
            intWage=int(raw_input("Enter employee's wage: "))
            if intWage >= 6.0 and intWage <=20.0:
                lstWage.append(float(strNum)) #convert to float
                blnDone=True
            else:
                print "Wage must be between $6.00 and $20.00"
        except(ValueError): #if you have Value Error exception.  Explicit on error type
            print "Invalid entry"


def getHours():
    lstHours=[]
    blnDone=False
    while blnDone==False: #loop to stay in program until valid data is entered
        try:
            intHrs=int(raw_input("Enter number of hours worked: "))
            if intHrs >= 1.0 and intHrs <=60.0:
                blnDone=True
            else:
                print "Hours worked must be 1 through 60."
        except(ValueError): #if you have Value Error exception.  Explicit on error type
            print "Invalid entry"

def getDone():
    strDone=""
    blnDone=False
    while blnDone==False:
        try:
            srtDone=raw_input("Type \"DONE\" if you are finished entering names, otherwise press enter: ")
            if strDone.lower()=="done":
                blnDone=True
            else:
                print "Type another empolyee name"
        except(ValueError): #if you have Value Error exception.  Explicit on error type
            print "Invalid entry"


##### Mainline ########

strUserName=getName()
strWage=getWage()
strHours=getHours()
srtDone1=getDone()

Answer 1

以下是它的核心：

>>> int("4.3")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: invalid literal for int() with base 10: '4.3'

如果字符串不是整数，则无法将字符串转换为整数。因此，当您执行intWage=int(raw_input("Enter employee's wage: "))时，它会抛出ValueError。也许您应该将其直接转换为float。

Answer 2

因为您要将输入转换为int：

intWage=int(raw_input("Enter employee's wage: "))

Answer 3

您假设工资是一个整数，根据定义，它没有小数位。试试这个：

intWage=float(raw_input("Enter employee's wage: "))

Answer 4

错误w / Floats

正如其他人之前所说，你假设输入将是一个浮点数。使用float()或eval()代替int()：

intWage = float(raw_input("Enter employee's wage: "))

或使用input()代替int(raw_input())：

intWage = input("Enter employee's wage:")

两者都将完成同样的事情。哦，至少将intWage更改为floatWage，甚至更好，不要使用匈牙利表示法。

代码

至于你的代码，我做了几件事：

使用break和/或return来终止循环，而不是跟踪布尔值（这是break和continue语句的全部目的）
将intWage更改为floatWage
以更简洁的方式重写数字比较（x <= y and x >= z可写为z >= x >= y）
添加了退货声明。除非你想None分配给strUserName，strWage和strHours）
在询问员工的详细信息时，根据您的要求添加了一个循环。
修改getDone()以使用循环。

import cPickle



def getName():
    strName = "dummy"
    lstNames = []
    strName = raw_input("Enter employee's Name: ")
    lstNames.append(strName.title() + " \n")
    return strName

def getWage():
    lstWage = []
    strNum = "0"
    while True:                                                #Loop to stay in program until valid data is entered
        try:
            floatWage = float(raw_input("Enter employee's wage: "))
            if 6.0 <= floatWage <= 20.0:
                lstWage.append(floatWage)
                return floatWage
            else:
                print "Wage must be between $6.00 and $20.00"
        except ValueError:                                     #Catches ValueErrors from conversion to float
            print "Invalid entry"

def getHours():
    lstHours = []
    while True:                                                #loop to stay in program until valid data is entered
        try:
            intHrs=int(raw_input("Enter number of hours worked: "))
            if 1.0 <= intHrs <= 60.0:
                return intHrs
            else:
                print "Hours worked must be 1 through 60."
        except ValueError:                                     #Catches ValueErrors from conversion to int
            print "Invalid entry"

def getDone():
    strDone = ""
    while True:
        srtDone = raw_input('Type "DONE" if you are finished entering names, otherwise press enter: ')
        if strDone.strip().lower() == "done":
            return True
        else:
            print "Type another empolyee name"

while not getDone():
    strUserName = getName()
    strWage = getWage()
    strHours = getHours()

def getName(): strName = "dummy" lstNames = [] strName = raw_input("Enter employee's Name: ") lstNames.append(strName.title() + " \n") return strName def getWage(): lstWage = [] strNum = "0" while True: #Loop to stay in program until valid data is entered try: floatWage = float(raw_input("Enter employee's wage: ")) if 6.0 <= floatWage <= 20.0: lstWage.append(floatWage) return floatWage else: print "Wage must be between $6.00 and $20.00" except ValueError: #Catches ValueErrors from conversion to float print "Invalid entry" def getHours(): lstHours = [] while True: #loop to stay in program until valid data is entered try: intHrs=int(raw_input("Enter number of hours worked: ")) if 1.0 <= intHrs <= 60.0: return intHrs else: print "Hours worked must be 1 through 60." except ValueError: #Catches ValueErrors from conversion to int print "Invalid entry" def getDone(): strDone = "" while True: srtDone = raw_input('Type "DONE" if you are finished entering names, otherwise press enter: ') if strDone.strip().lower() == "done": return True else: print "Type another empolyee name" while not getDone(): strUserName = getName() strWage = getWage() strHours = getHours()

和`break`

的示例

循环中的continue语句（break和for）终止循环并跳过所有'else'子句（如果有的话）。

while语句跳过循环中的其余代码，继续循环，就好像什么也没发生一样。

continue构造中的else子句在循环耗尽所有项目并正常退出时执行其代码块，即，当它没有被for...else或其他东西终止时。 / p>

break

Answer 5

尝试使用

intWage=int(float(raw_input("Enter employee's wage: ")))

这将接受十进制数作为输入。

Python程序不接受十进制原始输入

5 个答案:

错误w / Floats

代码

和`break`

Python程序不接受十进制原始输入

5 个答案:

错误w / Floats

代码

和break

和`break`