Python open()需要完整路径 [英] Python open() requires full path

问题描述

我正在编写一个脚本来读取一个csv文件. csv文件和脚本位于同一目录中.但是，当我尝试打开文件时，它会给我FileNotFoundError: [Errno 2] No such file or directory: 'zipcodes.csv'.我用来读取文件的代码是

I am writing a script to read a csv file. The csv file and script lies in the same directory. But when I tried to open the file it gives me FileNotFoundError: [Errno 2] No such file or directory: 'zipcodes.csv'. The code I used to read the file is

with open('zipcodes.csv', 'r') as zipcode_file:
    reader = csv.DictReader(zipcode_file)

如果我提供文件的完整路径，它将起作用.为什么open()需要文件的完整路径?

If I give the full path to the file, it will work. Why open() requires full path of the file ?

推荐答案

我已经确定了问题所在.我在Visual Studio代码调试器上运行代码.我打开的根目录高于文件级别.当我打开相同的目录时，它起作用了.

I have identified the problem. I was running my code on Visual Studio Code debugger. The root directory I have opened was above the level of my file. When I opened the same directory, it worked.

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