I am trying to create a file in the folder I running my .py script. This is the code I am using. The problem is that the open function requires / for directories. new_file_path uses \ instead. This is causing the open function to fail. How do I fix this?
import os
dir_path = os.path.dirname(os.path.realpath(__file__))
new_file_path = str(os.path.join(dir_path, 'mynewfile.txt'))
open(new_file_path, "x")
First of all, as @buran commented, there is no need to use str, the following suffices:
new_file_path = os.path.join(dir_path, 'mynewfile.txt')
There is a distinction between where the script exists, given by __file__, and the current working directory (usually the place from which the script was invoked), given by os.getcwd(). It is not entirely clear from the question wording which one was intended, although they are often the same. But not in the following case:
C:\Booboo>python3 test\my_script.py
But they are in the following case:
C:\Booboo>python3 my_script.py
But if you are trying to open a file in the current working directory, why should you even be bothering with making a call to os.getcwd()? Opening a file without specifying any directory should by definition place the file in the current working directory:
import os
with open('mynewfile.txt', "x") as f:
# do something with file f (it will be closed for you automatically when the block terminates
Another problem may be that you are opening the file with an invalid flag, "x" if the file already exists. Try, "w":
with open(new_file_path, "w") as f:
# do something with file f (it will be closed for you automatically when the block terminates
/home/levon) from some other directory (/root) . The script is creating a log file open('logfile.txt', a), but rather than creating the logfile in /home/levon where the script is located, it creates it in /root.os.getcwd(). I am saying that the current working directory by default (unless you change it) is the directory from which the script is invoked. In your example that would be /root. Am I missing something?have you tried simply
import os
dir_path = os.getcwd()
open(dir_path +'mynewfile.txt', "x")
Edit: Sorry for the last message, it saved incomplete
dir_path + os.path.sep + 'mynewfile1.txt' or this open can't even succeed. Second, by default if you do not specify a directory, the system will attempt to open the file in the current working directory by default (that is the definition of the current working directory), so that makes using os.getcwd() totally unnecessary. How can this be the accepted answer?/. In any other directory, such as /home/Roni, getcwd() would just return /home/Roni without a trailing path separator and your code would not work in the general case (as would blindly adding a path separator would not work if you are in the root directory). That is why you would want to use os.path.join, which will work for both cases. So my comment was based on your answer as *a general solution." My other point is that calling getcwd() is altogether unnecessary. Just open('mynewfile.txt', 'x')..__file__, which is where the file lives, but he likes your getcwd(), which is the current working directory, and again, they may not always be the same.you need to use os.getcwd to get the crrent working directory
import os
dir_path = os.getcwd()
new_file_path = str(os.path.join(dir_path, 'mynewfile.txt'))
open(new_file_path, "x")
open('mynewfile.txt', 'x'), what directory would the file it go into if not the current working directory?
os.getcwd() to put a file into the current working directory.
str()function on the result fromos.path.join(). And post full traceback you get. There is no problem with using forward slash in pathopen('mynewfile.txt', 'w')will do the job...