+ 7

Opening files in python

A bit of a stupid question, but what does the full file name look like when opening from a Mac? I have been trying this code: myfile = open("/Users/myname/Downloads/filename.docx") And as expected I got error: FileNotFoundError: [Errno 2] No such file or directory Because I did not use the full file path name I guess? But how do I find out the full path name??

20th Apr 2020, 8:16 AM
Julia Spencer
Julia Spencer - avatar
15 Antworten
+ 5
Code Crasher very good! "Automate the boring stuff " Is nice too but my version is a bit outdated.
20th Apr 2020, 9:50 AM
Oma Falk
Oma Falk - avatar
+ 4
Yeah so that's my question Oma Falk.. any idea how I find out what the full path name is?
20th Apr 2020, 8:38 AM
Julia Spencer
Julia Spencer - avatar
+ 3
For beginners, whenever you are working with files, I suggst to do the following: Put the files into the same directory like your code. That way you can access it with open("./filename.xyz") If you would like to specify specific paths and you aren't sure about the way a complete path should be written as a string (depends on the operating system), run this code: https://code.sololearn.com/ci0M98JQlj2l (Your code needs to saved to a file for this to work). This will give you the complete path to your current working directory as your operating system would stylize it. Then, just adapt it to your needs. Or, if you are lazy, copy the file with the above code into the same folder in which the file you want to open is located and run it.
20th Apr 2020, 6:12 PM
Froggy
Froggy - avatar
+ 2
Maybe your guess is true. Not full path. Which os by the way?
20th Apr 2020, 8:29 AM
Oma Falk
Oma Falk - avatar
+ 2
https://docs.python.org/3/library/os.path.html U can look in Browser or play with relative paths. I would recommend to use os and os.path
20th Apr 2020, 8:48 AM
Oma Falk
Oma Falk - avatar
+ 1
myfile = open("//Users//myname//Downloads//filename.docx")
20th Apr 2020, 8:22 AM
Oma Falk
Oma Falk - avatar
+ 1
That did not work either..
20th Apr 2020, 8:26 AM
Julia Spencer
Julia Spencer - avatar
+ 1
I have a windows and had the same problem with python 3.6. Try updating it to a newer version and enter the full file source. The with open() command ought to work.
21st Apr 2020, 11:43 AM
Evans
Evans - avatar
+ 1
First of all use with statement. open function is context manager so using with statement is more secure. Second, use os.path.join to build your path. The easiest way is to define it like below: ``` from os import path from functools import reduce path_parts = ['C:', 'Users', 'MyUser', 'test.txt'] # Windows example final_path = reduce(path.join, path_parts) with open(final_path, 'r') as file: content = file.readlines() ```
21st Apr 2020, 10:41 PM
Piotr Szajer
Piotr Szajer - avatar
+ 1
To work with files I make use of pathlib.Path. Path(yourfile).exists(). If the path is correct it will return True otherwise False. Also I prefer pathlib over os as you work with objects.
21st Apr 2020, 11:10 PM
GeraltdeRivia
0
I tried in dcode in mobile even it is saying "no such file". Why doesn't it work on Android mobile?
22nd Apr 2020, 6:50 AM
ajaynath reddy
ajaynath reddy - avatar
0
Does mobile has cmd?
22nd Apr 2020, 6:51 AM
ajaynath reddy
ajaynath reddy - avatar
0
It doesn't work in Android. At least not in SoloLearn
22nd Apr 2020, 7:08 AM
Evans
Evans - avatar
0
Y it doesn't work in android
22nd Apr 2020, 7:43 AM
ajaynath reddy
ajaynath reddy - avatar
0
Python File Open ❮ Previous Next ❯ f = open("demofile.txt", "r") print(f.read()) Open a file in a different location: Return the 5 first characters of the file: Read one line of the file: Read two lines of the file: Loop through the file line by line: Close the file when you are finished with it: Regards, Will
16th Jan 2023, 8:27 AM
william joe
william joe - avatar