Tracubik wrote:
********************************
error message of variable lenght
********************************
to print the asterisks line i do this:
def StringOfAsterisks(myString):
asterisksString = "*"
for i in range(1,len(myString):
asterisksString += "*"
print asterisksString
so i create the asterisksString of the lenght i need
There is a better way of creating asterisksString?
well, to make it more pythonic (ignoring camel-case
variable-names for now), I'd just use
print '*' * len(myString)
print myString
print '*' * len(myString)
possibly stashing the resulting asterisksString once:
asterisks_string = '*' * len(my_string)
print asterisks_string
print my_string
print asterisks_string
If I used it in multiple places, I might wrap it in a function
and/or define a "DIVIDER_CHARACTER" constant, something like
DIVIDER_CHARACTER = '*'
def surround(s, divider=DIVIDER_CHARACTER):
d = divider[0]
print d * len(s)
print s
print d * len(s)
surround('hello')
surround('world', '-')
surround('foo', 'xo')
depending on the sort of behavior you want
the code seem to work for me, but it doesn't work properly if in errorMsg
there is some "esotic" char like euro char (€):
euro = "€"
len(euro)
3
I suspect you're seeing the byte-representation of your string
with a particular-yet-undisclosed encoding. If it was a unicode
string (a good idea to specify the encoding at the top of your
file), the length should be accurate, so what happens if you
>>> euro = u"€"
>>> len(euro)
? (I don't have ready access to a terminal where I can enter
unicode characters, and you don't display the representation of
the string with
print repr(euro)
so I can steal the byte values to recreate it; but I suspect the
result will correctly be 1).
-tkc
--
http://mail.python.org/mailman/listinfo/python-list
