blocks|key|1429884|text|acc_num[starting_index]+=+0++
//this+sets+0+..+In+the+last+while+pass
//acc_num+will+value+[0,0,0,0,0,0,0,0,,,,,,,]

int(acc_num[starting_index])+
//This+does+nothing

starting_index+%2B=+1+
search+=+int(input('Enter+Account+Number:'))
//+Above+search+is+set+as+int.
//+Bellow+search+should+be+a+string.+
if(search+in+acc_num):+|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1429885|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>acc_num[starting_index] = 0 
//this sets 0 .. In the last while pass
//acc_num will value [0,0,0,0,0,0,0,0,,,,,,,]

int(acc_num[starting_index]) 
//This does nothing

starting_index += 1 
search = int(input('Enter Account Number:'))
// Above search is set as int.
// Bellow search should be a string. 
if(search in acc_num): 
</code></pre>

blocks|key|5440445|text|def+main():

try:++++
++++file+=+open('charge_accounts.txt','r')+
++++search+=+input('Enter+Account+Number:')
++++acc_num+=+file.readlines()++++

++++for+x+in+acc_num:
++++++++if+search+in+x:
++++++++++++print+"Yup"

++++file.close()++++++++

except+ValueError:++
++++print('Unable+to+open+the+file')|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|5440446|main()|unstyled|5440447|如果您正在读取文本文件，则所有内容都将作为字符串读取。"+12345“不等同于12345。首先是一个字符串。如果每一行都是一个整数，那么可以写成var+=+int(input(.......))(在清理了非int之后)，然后与int进行比较，但对于您所描述的内容可能不是必需的。|5440448|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>def main():

try: 
 file = open('charge_accounts.txt','r') 
 search = input('Enter Account Number:')
 acc_num = file.readlines() 

 for x in acc_num:
 if search in x:
 print "Yup"

 file.close() 

except ValueError: 
 print('Unable to open the file')
</code></pre>

main()

If you are reading a text file everything is read as a string. "12345" is not the same as 12345. First is a string. If every line is just an integer, you could write var = int(input(.......)) (after sanitizing for non-ints) and then compare against an int, but likely not necessary for what you are describing.

blocks|key|1424692|text|假设您在charge_accounts.txt中有这些帐号|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1424693|12345
23456
34567
45678
56789|code-block|syntax|javascript|1424694|当python执行这一行时-->+acc_num+=+file.readlines()|offset|length|style|CODE|1424695|acc_num将拥有以下内容：|1424696|ipdb>+acc_num
['12345\n',+'23456\n',+'34567\n',+'45678\n',+'56789']|1424697|为什么将列表中的第一项设置为0？|1424698|acc_num[starting_index]+=+0

ipdb>+acc_num
[0,+'23456\n',+'34567\n',+'45678\n',+'56789']|1424699|根据mshsayem的最后一个提示：|1424700|你的列表是string，你的搜索是int。删除int()并为输入设置str。|1424701|Enter+Account+Number:56789
('56789',+':Yes,+the+account+number+is+VALID')|1424702|最后但同样重要的是，尝试使用pdb/ipdb进行调试。|1424703|entityMap^0|0|0|H|Q|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|16|8|@$I|17|J|18|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|19|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|1A|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|1B|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|1C|8|@]|9|@]|A|$E|F]]|$1|U|3|V|5|6|7|1D|8|@]|9|@]|A|$]]|$1|W|3|X|5|6|7|1E|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|D|7|1F|8|@]|9|@]|A|$E|F]]|$1|10|3|11|5|6|7|1G|8|@]|9|@]|A|$]]|$1|12|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]]|13|$]]

Let say you have these account number in charge_accounts.txt

<pre><code>12345
23456
34567
45678
56789
</code></pre>

When python execute this line --> <code>acc_num = file.readlines()</code>

acc_num will have these :-

<pre><code>ipdb&gt; acc_num
['12345\n', '23456\n', '34567\n', '45678\n', '56789']
</code></pre>

Why you set the first item in list to 0?

<pre><code>acc_num[starting_index] = 0

ipdb&gt; acc_num
[0, '23456\n', '34567\n', '45678\n', '56789']
</code></pre>

And the last hint according to mshsayem:

your list is string and your search is int. Remove the int() and set str for your input.

<pre><code>Enter Account Number:56789
('56789', ':Yes, the account number is VALID')
</code></pre>

Last but not least, try using pdb/ipdb to debug.

blocks|key|5440515|text|您可以考虑将您的程序更改为以下内容：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5440516|def+main():

++++try:++++
++++++++with+open('charge_accounts.txt','r')+as+f:
++++++++++++acc_num+=+f.read().splitlines()+
++++++++starting_index+=+0++++++

++++++++while(starting_index+!=+len(acc_num)):++++
++++++++++++search+=+raw_input('Enter+Account+Number:')
++++++++++++starting_index+%2B=+1+
++++++++++++if(search+in+acc_num):++
++++++++++++++++print(search,':Yes,+the+account+number+is+VALID')
++++++++++++else:+++
++++++++++++++++print(search,':No,+the+account+number+is+INVALID')
++++++++f.close()
++++++++return

++++except+ValueError:++
++++++++print('Unable+to+open+the+file')
++++++++return

main()|code-block|syntax|javascript|5440517|这消除了acc_num中的'\n‘字符，并使用raw_input读取字符串，现在您的搜索比较会找到帐号。|5440518|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You may consider changing your program to the following:

<pre><code>def main():

 try: 
 with open('charge_accounts.txt','r') as f:
 acc_num = f.read().splitlines() 
 starting_index = 0 

 while(starting_index != len(acc_num)): 
 search = raw_input('Enter Account Number:')
 starting_index += 1 
 if(search in acc_num): 
 print(search,':Yes, the account number is VALID')
 else: 
 print(search,':No, the account number is INVALID')
 f.close()
 return

 except ValueError: 
 print('Unable to open the file')
 return

main()
</code></pre>

This eliminates the '\n' characters from acc_num, and using raw_input reads string and now your search comparison finds the account number(s).

I'm trying to write a program that brings in a file and it asks to input an account number and if it matches a number on the file it will tell the user it is either valid or invalid.

The program runs but always says invalid no matter what. What's wrong?

Here's the code:

<pre><code>def main():

 try: 
 file = open('charge_accounts.txt','r') 

 acc_num = file.readlines() 
 starting_index = 0 

 while(starting_index != len(acc_num)): 
 acc_num[starting_index] = 0 
 int(acc_num[starting_index]) 
 starting_index += 1 
 search = int(input('Enter Account Number:')) 

 if(search in acc_num): 
 print(search,':Yes, the account number is VALID')
 else: 
 print(search,':No, the account number is INVALID')

 file.close() 

 except ValueError: 
 print('Unable to open the file')

main()
</code></pre>

Python Program homework

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

我正在尝试写一个程序，它引入一个文件，它要求输入一个帐号，如果它与文件上的一个数字匹配，它就会告诉用户它是有效的还是无效的。程序会运行，但无论如何都会显示无效。怎么了？代码如下：def main():    try:            file = open('charge_accounts.txt','r')  ...

问Python程序作业
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问Python程序作业EN