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苏州质量展厅大概费用

发布于:2023-10-08 20:12:56     

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There are a few different ways to approach this problem, but one possible solution is to use a loop to iterate through each character in the string and check if it is a vowel. If it is, increment a counter variable. Here's an example in Python:

```python

苏州质量展厅大概费用

def count_vowels(string):

vowels = ['a', 'e', 'i', 'o', 'u'] count = 0 for char in string: if char.lower() in vowels: count += 1 return count

# Example usage:

string = "Hello, World!"

print(count_vowels(string)) # Output: 3

```

In this example, we define a function `count_vowels` that takes a string as input. We initialize a counter variable `count` to 0. Then, we iterate through each character in the string using a for loop. Inside the loop, we check if the lowercase version of the character is in the list of vowels. If it is, we increment the `count` variable by 1. Finally, we return the value of `count`.

Note that this solution treats both uppercase and lowercase vowels as equivalent by converting all characters to lowercase before checking for vowel status.

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