I am having issues with the open() function in Python 3.2.3. The following code works well using 2.7.3, but not with Python 3:
file = open("text.txt", 'r')
In Python3, it gives me a standard IOError:
IOError: [Errno 2] No such file or directory: 'text.txt'
Note that the text.txt file that is referenced is in the same directory as the python file.
Any ideas?
The file name is not relative to the directory of the file, but your current working directory (which you can find out with os.getcwd()).
If you want to open a file whose name is relative to your Python file, you can use the magic variable __file__, like this:
import os.path
fn = os.path.join(os.path.dirname(__file__), 'text.txt')
with open(fn, 'r') as file:
# Do something, like ...
print(file.read())
You are trying to open a file in read mode, and this file has to exist.
Perhaps problem is that the file just doesn't exist in your python3 path and therefore the open command fails, but 'text.txt' exist on your python2.7 lib (or somewhere in python2.7 path) and this is why python is able to locate the file and open it.
You can just try this code (this will assure you the file exist since you create it):
f = open('text.txt','w')
f.close()
f.open('text.txt','r')
I was using Eclipse with Pydev, and had the text.txt file within the package instead of on the project level. To access a file within the package, you need to use:
file = open("[package]/text.txt", 'r')
os.getcwd()return?