逻辑错误：即使条件没有相遇，程序也会写入

Question

我正在尝试读写文件。在此示例中，我正在尝试读取文件user_admin.txt，并检查文件中要求注册的用户名是否存在。如果存在，它将通过（在此处显示一条消息），如果没有，则以给定格式注册用户（in str （self）），

"""
    ***user_admin.txt contains(separate lines)***
    111124;;;zerox;;;32@3432#4$$
    344377;;;zer;;;32@3432#4$$
    424554;;;zero;;;32@3432#4$$
    544334;;;zeronda;;;32@8986#5$$
    655334;;;zIT;;;32@0000#8$$
    113324;;;Bazer;;;32@0000#4$$
"""

#jupyter cell 1

import os
import re

class Admin:

    path = "./data/result/user_admin.txt" 

    def __init__(self, id=-1, username="", password=""): # constructor
        self.id = id
        self.username = username
        self.password = password

    def __str__(self):
        return f"{self.id};;;{self.username};;;{self.password}"
    
    def register_admin(self):
        path = "./data/result/user_admin.txt"  # Destination of the user_admin.txt file
        
        with open(path, "r", encoding="utf-8") as r:  # r means read
            r_contents = r.read()
            user_match = re.findall(r'\d+;;;\w+;;;.*\n?', r_contents)
            
            with open(path, "a+", encoding="utf-8") as f:  # f means append and read
                if os.stat(path).st_size == 0:
                    f.write(f"{self.id};;;{self.username};;;{self.password}\n")
                elif os.stat(path).st_size != 0:
                    for i in range(len(user_match)):
                        if self.username not in user_match[i]:
                            f.write(f"{self.id};;;{self.username};;;{self.password}\n")
                        else:
                            print("You are not welcome here") # I will use "pass" in actual code
        


#jupyter cell 2
admin_two = Admin(113324,'Bazer','32@0000#4$$')
admin_two.register_admin()

#Juppyter cell 3
 with open(path, "r", encoding="utf-8") as r:  # f means append and read
 r_contents = r.read()
 print(r_contents)
 

单元3的结果是：

11124;;;zerox;;;32@3432#4$$
344377;;;zer;;;32@3432#4$$ 
424554;;;zero;;;32@3432#4$$
544334;;;zeronda;;;32@8986#5$$ 
655334;;;zIT;;;32@0000#8$$
113324;;;Bazer;;;32@0000#4$$
113324;;;Bazer;;;32@0000#4$$
113324;;;Bazer;;;32@0000#4$$
113324;;;Bazer;;;32@0000#4$$
113324;;;Bazer;;;32@0000#4$$
113324;;;Bazer;;;32@0000#4$$

i不了解导致此问题的逻辑错误是什么。非常感谢您的帮助:)

Answer 1

Divide and Conquer

会将register_admin将拆分为3个函数：

1. 解析文件以使用正则示意用户
2. 查找所需的用户
3. 注册用户

在此处求解是 或单个职责原理（SRP）：

第一次搜索（是否找到），然后在分离的逻辑单元（函数或方法）中注册。

在循环中搜索

您可以使用for-loop在所有元素中搜索。
但是，请跟踪您的发现并在需要时尽早退出。

在此处求解是 boolean flag 与早期exit break break一起找到。

代码建议

# first decompose the methods in Cell 1 (split into smaller problem-solutions)

# (1) parse users
# class method, thus no self
def parse_users(path): 
    user_matches = []
    with open(path, "r", encoding="utf-8") as r:  # r means read
        r_contents = r.read()
        user_matches = re.findall(r'\d+;;;\w+;;;.*\n?', r_contents)
    return user_matches

# (2) find in users  (the search logic in the loop was the issue)
def find_in_users(self, user_matches):
    # search in all elements of user_matches
    found = False  # use a flag, which will turn True as soon as was found
    for i in range(len(user_matches)):
        if self.username in user_match[i]:  # remove `not` to find 
            found = True
            break  # as soon as found we must exit the loop
    # out of the loop: test if not in elements of user_matches at all
    return found

# (3) register at file
def register_at(self, path)
    user_matches = parse_users(path) 
    found = find_in_users(self, user_matches)
    if not found:  # may have two reasons: (a) no user in file or file size 0, (b) searched user not found   
        # append it to the file            
        with open(path, "a+", encoding="utf-8") as f:  # f means append and read
            if os.stat(path).st_size == 0:
                f.write(f"{self.id};;;{self.username};;;{self.password}\n")
            elif os.stat(path).st_size != 0:        
                f.write(f"{self.id};;;{self.username};;;{self.password}\n")
    else:
        print("You are not welcome here") # I will use "pass" in actual code


# then later in Cell 2
admin_two = Admin(113324,'Bazer','32@0000#4$')
admin_two.register_at(path)

# finally in Cell 3

进一步的问题并优化

重要的是：文件大小或发现？

如果文件为空，是否真的很重要（os.stat（path）.st_size）？

注册重要的结果是：

• 没有找到管理员，

因此我们可以简化这些行：

            if os.stat(path).st_size == 0:
                f.write(f"{self.id};;;{self.username};;;{self.password}\n")
            elif os.stat(path).st_size != 0:        
                f.write(f"{self.id};;;{self.username};;;{self.password}\n")

to：

            f.write(f"{self.id};;;{self.username};;;{self.password}\n")

在整个CSV线中找到单词？！

如果文件中的任何用户选择admin的名称bazer作为密码怎么办？

考虑以下行：

424554;;;zero;;;Bazer

尽管实际用户名是Zero，但它将发现用户被搜索为Bazer。

您可以通过使用捕获组调整正则表达式来提取用户名。因此，user_matches将是一个用户名的列表，只有例如eg ['zer'，'Zero']。

甚至更好地将真实的CSV-Parser模块用于Python。