How to use the output from glob.glob to read the outputted file?

Question

I'm writing a program that scans for specific file types and reads them for certain 'commands'. I have the code filenameglob = (glob.glob("*.doremi")) but upon returning the variable with print it comes out as ['test.doremi'] instead of the optimal test.doremi. The problem with this is I can not use it to read the file. How would I go about using glob to search for the file, output the name and then read the file with that name?

Answer 1

Basic Answer

glob.glob returns a list of items matching the search criteria passed at runtime- or an empty list if no matches are found. This is why you are getting ['test.doremi'] instead of test.doremi.

The filenameglob variable in your example is a list containing 1 element - the string test.doremi - which can be accessed using it's index within the list i.e filenameglob[0] is test.doremi.

The below code provides an appropriate answer/solution to your original question assuming that there will be 0 or 1 .doremi files in the target directory.

import glob

filenameglob = glob.glob("*.doremi")

if filenameglob: # If glob.glob returned list of matches
    filename = filenameglob[0] # Take first item from filename glob
    print(filename)
    with open(filename, "r") as f: # Open file in read mode
        contents = f.read() # Read contents from file to variable
    # Do something with contents read from .doremi file
    
else: # Else if glob.glob returned empty list
    print("No .doremi file found!")
    # Define behaviour in case of no .doremi file found

Processing multiple .doremi files from directory

The above code assumes 0 or 1 files with the .doremi extension within the execution directory as per your original question.

As described above, glob.glob returns a list of strings representing matches based on the arguments passed to the function at runtime - in this case '*.doremi'.

Therefore it is probably worth considering how you will handle multiple .doremi files should they be found within the execution directory by glob.glob.

One solution may be to write a function that defines how you will handle contents read from each .doremi file and call this within a loop for each item returned by glob.glob. See below for an example of how you might do this.

import glob

def process_contents(contents):
    # Process file contents here

filenameglob = glob.glob("*.txt")

if filenameglob: # If glob.glob returned list of matches
    for filename in filenameglob: # For each file in filenameglob list
        print(filename)
        with open(filename, "r") as f: # Open current file in read mode
            contents = f.read() # Read contents from current file to variable
            process_contents(contents) # Send contents from file to function for processing
            
else: # Else if glob.glob returned empty list
    print("No .doremi files found!")
    # Define behaviour in case of no .doremi files found

The above code will work for any number (0...n) of files returned by glob.glob, as such I would recommend taking this approach as to better mitigate the risk of your code processing the "wrong" .doremi file - as may happen in the Basic Answer as it will only process the first in the list, even if multiple are present.