Seeing if a file exists and printing it

Question

So I'm trying to create a method that can search files and print them out.

def search():
    searched = input("... ")
    with open(searched + ".py", "r") as my_file:
        print(my_file)
search()

but the console prints out <_io.TextIOWrapper name='RandomPrac.py' mode='r' encoding='UTF-8'> What does this mean?

You printed the file object instead of the contents. Try `print(my_file.read())` — khelwood, May 13 '18 at 20:51
This is file handler object, maybe you need `my_file.read()` to read file data. — ᴀʀᴍᴀɴ, May 13 '18 at 20:51

score 3 · Accepted Answer · answered May 13 '18 at 20:51

3

you print my_file variable, which is of type _io.TextIOWrapper. What you actually need is

print(my_file.read())

answered May 13 '18 at 20:51

Ali Yılmaz

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abarnert · Answer 2 · 2018-05-13T21:11:30.980

As Ali Yilmaz's answer already nicely explains, you're printing the file object, not the contents of that file.

For smallish files, just read() the file to get the contents as a string, and print that, as in his answer. And, since you're searching for *.py files, you almost certainly do have smallish files.

But if the file might be gigantic, it's probably better to loop over it and print each line as you read it:

for line in my_file:
    print(line, end='')

Or, if you want to be more concise but possibly too clever:

print(*my_file, sep='')

Seeing if a file exists and printing it

2 Answers2