Python程序不接受十进制原始输入

Question

I am working on small python payroll project where you enter employee name, wage, and hours worked. When I enter decimals for the wage input, I am getting "invalid entry" because of my exception handling. Why are decimals being returned as invalid? Also, how can I loop this program so that it keeps the same 3 questions until the user types "Done"? Any help will be greatly appreciated! Thanks!

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

Here's the core of it:

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

You can't convert a string to an integer if it's not an integer. So when you do intWage=int(raw_input("Enter employee's wage: ")) it throws the ValueError . Perhaps you should convert it directly to float .

Answer 2

因为您要将输入转换为int ：

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

Answer 3

You are assuming that the wage is an integer, which by definition does not have a decimal place. Try this:

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

Answer 4

Try using

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

This will accept a decimal number as input.

Answer 5

Error w/ Floats

As others said before, you are assuming the input will be a float. Either use float() or eval() instead of int() :

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

Or use input() instead of int(raw_input()) :

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

Both will accomplish the same thing. Oh, and change intWage to floatWage atleast, or even better, don't use Hungarian Notation.

The Code

As for your code, I did a couple of things:

• Used break and/or return to terminate loops instead of keeping track of booleans (that's the whole purpose of the break and continue statements)
• Changed intWage to floatWage
• Rewrote number comparisons in a more concise way ( x <= y and x >= z can be written as z >= x >= y )
• Added return statements. I don't get why you didn't put them yourself, unless you wanted to assign None to strUserName , strWage and strHours )
• Added a loop as you requested when asking for an employee's details.
• Modified getDone() to work w/ the loop.

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() 

An Example of break and continue

The break statements inside a loop ( for and while ) terminate the loop and skip all 'else' clauses (if there are any).

The continue statements skips the rest of the code in the loop and the continues the loop as if nothing happened.

The else clause in a for...else construct, executes its code block when the loop exhausted all the items and exited normally, ie, when it's not terminated by break or something.

for no in range(2, 10):
    for factor in range(2, no):
        if no % factor == 0:
            if factor == 2:
                print "%d is even" % no
                continue  
                # we want to skip the rest of the code in this for loop for now
                # as we've already done the printing
            print "%d = %d * %d" % (no, factor, n/x)
            break 
           # We've asserted that the no. isn't prime, 
           # we don't need to test the other factors
    else:
        # if 'break' wasn't called
        # i.e., if the loop fell through w/o finding any factor
        print no, 'is a prime number'