Question

I m having serious trouble sorting out scoping error in my project. My project looks something like this:

这份主要文件载有这一进程,并用警棍打造了全球调查。

interfaceAndProcess.pyw
import lib1
...
#do something with tkinter
filePath = askopenfilename(filetypes=(("All files", "*.*")))
...
lib1.checkSomeDocument(filePath)

然后,我有“其他平衡”职能。

moalibs.py

def parseSomething(lookForStringX)
  position = line.index(lookForStringX, 0)
  return(position)

def bla():
  ...

def blabla():
  ...

这里是使用分子方法的许多平衡之一。 y

lib1.py
from moalibs import *
def checkSomeDocument(filePath)
global line
  fileContent = open(filePath,  r )
  for line in fileContent:
    tmpVar = parseSomething(lookForStringX)
    ...
    tmpVar = bla()
    ...
    tmpVar = blabla()
    ...
    tmpVar = bla()
    # In any of my many libs the methods from moalib are called
    # serveral times in different orders, that s why this part
    # is pretty "hard coded"

我的问题是,实施接口。 pyw向以下线投掷了名词:“名称线没有界定”。

为什么可以打上<条码>。见《瓦尔<条码>。

当我把<代码>parseSomething列入moalibs.py。正在做的是罚款。

I m sorry for this question to be pretty specific but I m searching and trying for for more than two hours now. Been playing with global line inside the methods, been defining line inside interfaceAndProcess.pyw, nothing...

任何建议?

Edit: Ok, I understood that what I have tried can t work as I expected. How would I achieve this without passing the variable as an argument?

Answer 1

对于开端人来说,global keyword isn t do any for a changing have been used in the high-level namespace (as is line in this case). The Stack Overflow question Discussiones why.



您重新获得姓名是因为lib1.py 进口<代码>moalibs.py,而不是副面。 页: 1 页: 1

To see what I mean, you can ask each module to tell you its current namespace contents by using the dir() function. You ll find that lib1.py is fully aware of all of moalibs.py s identifiers, but moalibs.py is not aware of lib1.py s identifiers.

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