在python中查找字符串中子字符串的出现次数

Question

question:In this challenge, the user enters a string and a substring. You have to print the number of times that the substring occurs in the given string. String traversal will take place from left to right, not from right to left. NOTE: String letters are case-sensitive.

Sample Input

ABCDCDC CDC

Sample Output 2

This is my solution but i am not getting the correct output(i am getting 1 as output). What is the reason?

def count_subtring(string, sub_string):  
    len_string = int(len(string))
    len_substring = int(len(sub_string))
    l = len_string - len_substring + 1
    n = 0
    for i in range (l):
        if string[i:i + len_substring] == sub_string:
            n =+ 1
    return n

if __name__ == '__main__':
    string = input().strip()
    sub_string = input().strip()
    
    count = count_substring(string, sub_string)
    print(count)

and i found a solution but i am unable to understand it and i want a explanation for the below code

def count_substring(string, sub_string):
    count=0
    for i in range(len(string)):
        for j in range(len(sub_string)):
            if string[i+j]==sub_string[j] and j==(len(sub_string)-1):
                count=count+1  
            if string[i+j]!=sub_string[j]:
                break  
        if i==len(string)-len(sub_string):
            break            
    return count


if __name__ == '__main__':
    string = input().strip()
    sub_string = input().strip()
    
    count = count_substring(string, sub_string)
    print(count)

thanks

Answer 1

You are assigning n with +1 when you find a match instead of adding 1 to it ( n += 1 ).

n =+ 1 is the same as n = +1 or n = 1 .

So the result will never be greater than 1.