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Python打开文件按钮

[英]Python Open File Button

I am having trouble, trying to get my open button to open a text file. 我遇到了麻烦,试图让我的打开按钮打开文本文件。 I want it to open the text file into the not pad, when I click the open button. 当我单击打开按钮时,我希望它将文本文件打开到not pad中。 If some one could help me or tell me what I am doing wrong, I would appreciate it. 如果有人可以帮助我或告诉我我做错了什么,我将不胜感激。

def _open(self):
        open("../Address.txt","r").close()
        with open("../Address.txt", "a") as file:
            self._outputArea.insert("1.0", file.read)
            file.read()
  • Why open and close the file first? 为什么要先打开和关闭文件? Just use the with line. 只需使用with行。
  • Don't use file as a variable name, it's also a type. 不要将file用作变量名,它也是一种类型。
  • You're not calling read . 您不是在打电话给 read
  • 'a' is a flag for appending a file, use 'r' (for read) instead. 'a'是用于附加文件的标志,请改用'r' (供读取)。

Try something like: 尝试类似:

def _open(self):
    with open("../Address.txt", "r") as the_file:
        self._outputArea.insert("1.0", the_file.read())

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