why my python code can't get return value?

You are never adding anything to count, so it's basically like calling while 0 < 1:.

At the end of your while loop, add count+=1. Change your code to this:

def similarStrings(str1,str2):

 length=len(str1)
 count=0
 i=0

 while (count<length):
     if str1[count]==str2[count]:
         i=i+1
     else:
         i=i
     count+=1


 if i>=length-1:
    return True
 else:
    return False

As I stated in my comments, your algorithm does not work because you do not increment count in the while loop; while (count<lenth) will always be True and you therefor have an infinite loop. Similar to while True.

You asked how to compare two strings when one might be longer than the other. There are many ways, but one classic algorithm is the Levenshtein Distance. If the strings are the same, it returns 0. Any amount greater than 0 is the number of edits to get the to get one string to be the same as the other.

Here is the Rosetta Code version of Levenshtein distance:

def LevDist(s1,s2):
    if len(s1) > len(s2):
        s1,s2 = s2,s1
    distances = range(len(s1) + 1)
    for index2,char2 in enumerate(s2):
        newDistances = [index2+1]
        for index1,char1 in enumerate(s1):
            if char1 == char2:
                newDistances.append(distances[index1])
            else:
                newDistances.append(1 + min((distances[index1],
                                             distances[index1+1],
                                             newDistances[-1])))
        distances = newDistances
    return distances[-1]

Testing it:

>>> LevDist('kitten', 'kitten')
0
>>> LevDist('kittens', 'kitten')  
1
>>> LevDist('kitten', 'cat')  
5

Another method is to use the tools in difflib.

Good luck.

There are some issues in your code:

1- You have to increment count.

2- You are setting different names to the same variable "lenth" / "length".

3- str1[count]==str2[count]: will raise error if str1 is bigger than str2 (index out of range).

However, if you want to check if the two strings are similar you may just use:

if str1 == str2:
   return something

Otherwise let me know what do you want to do exactly.

Nothing appears when you run your program because it runs in an infinite loop.

The while (count<lenth): statement tells python to keep running while count is less than "lenth", but since the value of count never changes, this is always true, and python continues running endlessly.

It's rare that you actually need indexes or counters like this in python. You fix it by getting rid of the index by using a for loop, like so:

def similarStrings(str1, str2):
    length = len(str1)
    i = 0 

    for char1, char2 in zip(str1, str2):
        if char1 == char2:
            i = i+1

    if i >= length-1:
        return True
    else:
        return False

I'd rewrite the function like so though:

def similarStrings(str1, str2, difference=1):
    from itertools import izip_longest
    return sum(
        char1 != char2
        for char1, char2 in izip_longest(str1, str2)
    ) <= difference

I wrote a test suite, of sorts:

def test(s1, s2):
    if similarStrings(s1, s2):
        print 'SIMILAR:'
    else:
        print 'NOT SIMILAR:'

    print '   ', s1
    print '   ', s2

def main():
    test('', '')
    test('abc', 'abc')
    test('abcd', 'abc')
    test('abc', 'abcd')
    test('abcd', 'abce')
    test('abcd', 'abcee')
    test('abcdd', 'abce')
    test('abcdd', 'abcee')

if __name__ == '__main__':
    exit(main())

Which gave me these results, using my final implementation:

SIMILAR:


SIMILAR:
    abc
    abc
SIMILAR:
    abcd
    abc
SIMILAR:
    abc
    abcd
SIMILAR:
    abcd
    abce
NOT SIMILAR:
    abcd
    abcee
NOT SIMILAR:
    abcdd
    abce
NOT SIMILAR:
    abcdd
    abcee