Question

我遇到了一个问题，即python没有为代码中的输入调用输入输入。它只是跳过获取输入并将一个空字符串放入变量源，这会导致程序在尝试打开名为source（空字符串）的文件时失败。

可能导致它跳过一行输入的原因是什么？

我有这段代码我写的只是几行要求用户输入。

def main():
    source = input('please enter source file name: ')
    file_name = input('please enter file name you want script to be saved as: ')
    language = input('please enter language to generate script in: ')
    args = input('please enter comma delimited list of attributes from source file (if any): ')
    print('\nGenerating script...\n')
    generate_script(source,file_name,args,language)
    return

当我运行程序时，它正常工作并输出

please enter source file name: Tree.txt

please enter file name you want script to be saved as: t

please enter language to generate script in: matlab

please enter comma delimited list of attributes from source file (if any): 

Generating script...

Generation successful

当我将此代码交给其他人并且他们在计算机上运行时发生此错误

please enter source file name: /* SHOULD TAKE USER INPUT HERE BUT IMMEDIATELY PRINTS NEXT INPUT LINE INSTEAD */ please enter file name you want script to be saved as: 'C:\Users\pmade\Desktop\DecisionTreeGenerator-master\DecisionTreeGenerator-master\T.txt'
please enter language to generate script in: 'matlab'
please enter comma delimited list of attributes from source file (if any): 'PCI0, FREEZE_THAW_YR'

Generating script...
  File "<stdin>", line 1, in <module>
  File "C:\Program Files\Anaconda3\lib\site-packages\spyder\utils\site\sitecustomize.py", line 866, in runfile
    execfile(filename, namespace)
  File "C:\Program Files\Anaconda3\lib\site-packages\spyder\utils\site\sitecustomize.py", line 102, in execfile
    exec(compile(f.read(), filename, 'exec'), namespace)
  File "C:/Users/pmade/Desktop/DecisionTreeGenerator-master/DecisionTreeGenerator-master/Text2Code.py", line 162, in <module>
    main()
  File "C:/Users/pmade/Desktop/DecisionTreeGenerator-master/DecisionTreeGenerator-master/Text2Code.py", line 9, in main
    generate_script(source,file_name,args,language)
  File "C:/Users/pmade/Desktop/DecisionTreeGenerator-master/DecisionTreeGenerator-master/Text2Code.py", line 32, in generate_script
    with io.open(source,'r') as f_r:
FileNotFoundError: [Errno 2] No such file or directory: "runfile('C:/Users/pmade/Desktop/DecisionTreeGenerator-master/DecisionTreeGenerator-master/Text2Code.py', wdir='C:/Users/pmade/Desktop/DecisionTreeGenerator-master/DecisionTreeGenerator-master')"
>>> Traceback (most recent call last):

似乎解释器在main中打印出第一个输入调用的消息，然后在等待输入之前直接进入第二个输入调用，导致它抱怨该文件名为＆＃34;＆＃34;在当前目录中不存在。任何人都知道发生了什么？

编辑：抱歉忘记解释正确，函数generate_script使用从跳过的输入中取出的字符串来打开要读取的源文件。这是代码：

def generate_script(source,file_name,args='',language='python'):

    with io.open(source,'r') as f_r:
        do work with file f_r....

由于错误，解释器会跳过通常放入源变量的输入，导致io尝试打开名为＆＃34;＆＃34;的文件。由于来源为空白，导致上述错误。

Answer 1

错误消息的最后一行几乎说明了一切：

FileNotFoundError：[Errno 2]没有这样的文件或目录：＆＃34; runfile（＆＃39; C：/Users/pmade/Desktop/DecisionTreeGenerator-master/DecisionTreeGenerator-master/Text2Code.py' ;, wdir =＆＃39; C：/用户/ pmade /桌面/ DecisionTreeGenerator主/ DecisionTreeGenerator主＆＃39;）＆＃34;

错误发生在函数generate_script中，该函数不是您发布的代码的一部分。

Answer 2

看起来错误发生在第7行，您调用＆＃34; generate_script＆＃34;。根据具体情况以及您可能调用的模块，其他人可能没有安装这些模块，或者没有正确调用它们。

Python3跳过第一个输入调用并继续第二个输入

2 个答案: