Python 我的代码运行得很好，但当我尝试搜索已经包含的字符时，它不是'；不匹配，一直说找不到

这是我的输出：
输入要输入的字母数3
输入字母表A
输入字母B
输入字母C
您输入的字母表是数组（'u'，'ABC'）
A
B
C
输入要搜索的字母是否可用b
找不到

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此外，我还附加了我的输出图像

删除

中断

，它将找到它。下面是一个稍微好一点的代码。我对你的做了些改变

from array import *

alphabets = array('u', [])

lettersRange = int(input("Enter how many letters do you want to enter "))
for i in range(lettersRange):
    addLetter = input("Enter Alphabet ")
    alphabets.append(addLetter)

print("alphabets that you've entered are ", alphabets)

for i in range(len(alphabets)):
    print(alphabets[i])

searchLetter = input("enter the letter to search if its is available or not")
count = 0
for i in alphabets:
    if i == searchLetter:
        print('The Letter "',searchLetter,'" is available and it\'s position is',count )

    else:
        print("Letter Not found")
        break

    count += 1

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检查最后一段代码：

from array import *

alphabets = array('u', [])

lettersRange = int(input("Enter how many letters do you want to enter "))
for i in range(lettersRange):
    addLetter = input("Enter Alphabet ")
    alphabets.append(addLetter)

print("alphabets that you've entered are ", alphabets)

for i in range(len(alphabets)):
    print(alphabets[i])

searchLetter = input("enter the letter to search if its is available or not")
count = 0
for i in alphabets:
    if i == searchLetter:
        print('The Letter "',searchLetter,'" is available and it\'s position is',count )
    break
    count += 1

if(count==len(alphabets)):
    print("Not Found")

分解：该代码表示，对于字母表中的每个字母，如果该字母是搜索字母，则打印（“xyz”）。如果该字母不在字母表中，请打印（“未找到字母”）。然后打破循环。这是一个好主意，但不幸的是，循环只测试第一个字母a。然后，如果它不是searchLetter，它将打破循环。我建议：

searchLetter = input("enter the letter to search if its is available or not")
count = 0
for i in alphabets:
    if i == searchLetter:
        print('The Letter "',searchLetter,'" is available and it\'s position is',count )

    else:
        print("Letter Not found")
        break

    count += 1

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如果（count==len（alphabets））：print（“Not Found”）////你能解释一下我编辑代码的原因吗。本质上，如果找不到元素，那么循环将完成相当于字母数的迭代次数，因此计数将等于字母数。在这种情况下，无法找到打印，明白了！！谢谢…-）

found = False
for i in alphabets:
    if i == searchLetter:
        print("xyz")
        found = True
if not found:
    print("Letter not found")