Writing into a File using a Try Except Loop always results in the except block

The function should open the file and if an exception occurs the function should display an error message and the function should return None. If an exception does not occur while opening, the code should write into the file text.

def write_string(fname, text):
    try:
        f = open(fname, 'r')
        f.read()
        l = open(fname, 'w')
        return l.write(text)
    except:
        print(f'Error occurred when opening {fname} to read')
        return None

However, it seems to always return an error and never actually write the file. Any advice? I'm still new to Python.

When testing it I get this:
Full printing (excluding input prompts/lines and empty lines) should look something like:

"""
The following (excluding input prompts/lines and empty lines) was printed:
"""
Error occurred when opening basic_io.write_string_0.txt to read
Error occurred when opening basic_io.write_string_0.txt to read

In general it's bad practice to catch all exceptions -- you should catch the specific exceptions that you know how to handle (for example, FileNotFoundError is a specific exception you might encounter when trying to open a file), and let others raise. That will make your program easier to debug when you encounter unexpected exceptions, because Python will automatically halt the program and print an error message that tells you exactly what exception was raised and where, instead of putting you in the situation where you're getting an error message but have no idea what's causing it.

If you are going to broadly catch exceptions and print an error message, you should strongly consider printing the error message that's contained in the exception itself:

def write_string(fname, text):
    try:
        f = open(fname, 'r')
        f.read()
        l = open(fname, 'w')
        return l.write(text)
    except Exception as e:
        print(f'Error occurred when opening {fname} to read: {e}')
        return None

This will probably give you something to go on.

Note that the except will catch any error that happens in the try block, not just the open(fname, 'r') call! So the error you're printing is misleading -- it might not necessarily be something that's happening because you opened the file to read. For example, if I run the above version of the function with a read-only file, I get:

>>> write_string("foo.txt", "asdfasdf")
Error occurred when opening foo.txt to read: [Errno 13] Permission denied: 'foo.txt'

If I simply remove the try/except the error is more informative:

Traceback (most recent call last):
  File "test.py", line 7, in <module>
    write_string("foo.txt", "asdfasdf")
  File "test.py", line 4, in write_string
    l = open(fname, 'w')
PermissionError: [Errno 13] Permission denied: 'foo.txt'

because now I can see that I'm getting a PermissionError and that it happened on the open(fname, 'w') call, i.e. my attempt to open the file for write, not read.

You are not passing any argument to the write function which expects a single argument.

def write_string(fname, text):
    try:
        f = open(fname, 'r')
        f.read()
        l = open(fname, 'w')
        return l.write(text)
    except:
        print(f'Error occurred when opening {fname} to read')
        return None

should work.

Also, IN YOUR CASE, I would recommend using:-

try:
    # Do something
except Exception as e:
    print(e)

to print the error message.

But generally, as @michael-butscher pointed out, it is better to "catch specific expected expections".