[Tutor] n.isalnum() is failing

[Terry]

Alan,

Wow! Those changes radically change the program! 

Is there a way to get the try-exception to cause the end user questions 
to be repeated until acceptable answers are given? Like a kind of
"return" 
or "repeat" command that could be added to the except side of things?

try:
    start = int(raw_input("Enter yyyy for beginning year : "))
    end = int(raw_input("Enter yyyy for ending year : "))
except ValueError: print "Must be a integer number as YEAR."


I also noticed that I had to make the statement "start = end = 0" prior 
to the 'try'. To my eye, it looks to me that 'start' and 'end' are
being 
declared inside the try-exception, but evidently python does not agree.

Thanks,
Terry












On Tue, 2007-07-03 at 21:42 +0100, Alan Gauld wrote:

> "Terry" <terry.kemmerer at gmail.com> wrote
> 
> > trapping for possible user  errors.
> > I am using x.isalnum() to check that I have a number
> 
> First, alnum() checks for alpha-numeric characters so
> will allow both alphabetic characters and numerics.
> 
> BUT in python its better to ask forgiveness that permission
> so don't bother doing the check but catch the errors when
> they occur:
> 
> try:        start = int(start); end = int(end)
> except: ValueError: start = None  # or whatever you prefer
> 
> > def leap(yyyy):
> >    answer = 0
> >    t1 = yyyy / 4
> >    if t1 == int(t1):
> 
> For integer division this will always be true.
> 
> >        t2 = yyyy / 100
> >        t3 = yyyy / 400
> >        if t2 <> int(t2) or t3 == int(t3):
> 
> and similarly the second test will always be true.
> 
> >            answer = "-- leap year!"
> 
> thus this will always be true.
> 
> > start = raw_input("Enter yyyy for beginning year : ")
> > end = raw_input("Enter yyyy for ending year : ")
> 
> The easiest way to solve your conversion problems
> is to do it at source so wrap raw_input in a call to int()
> and wrap both prompts in a try/except as shown above.
> 
> try:
>    start = int(raw_input....)
>    end = int(raw_input....)
> except ValueError: pass # do something with error
> 
> > if len(start) == 4 and len(end) == 4:
> 
> that way you don't need these lines but can instead
> test if the numeric value is greater than some lower
> limit
> 
> >    if start.isalnum() == 1 and end.isalnum() == 1:
> > #<----------fails to detect 'aaaa' as not a number
> >        start = int(start); end = int(end)
> > #<----------commits suicide here
> >        if start > 0 and end > start:
> 
> As you are doing it start can never be less than 1000
> (unless the user enters 0001 I suppose...)
> 
> Also in Python you can use the neater style of
> 
> if 0 < start < end:
> 
> to test whether start lies between 0 and end.
> 
> >            for i in range(start, end + 1):
> >                answer = leap(i)
> >                if answer != 0:
> 
> Its best not to mix return types in a function. If it is a
> leap year you return a string. If it's not, you return a
> number - 0. It would probably be better for the function
> to return a boolean: True if it is a leap year and False
> if not. That style of function is often called a predicate
> and is usually named with an 'is' in front so it would
> become isLeapYear() Then this test becomes the more
> readable
> 
> for y in range(start, end+1):  # use y for year instead of meaningless 
> i
>       if isLeapYear(y):
>          print i, 'is a leap year'
> 
> HTH,
> 
> 
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