Getting exception details in Python

Question

I have to open & write to about 10 different files all within the same loop. e.g:

for i in range(0,10):
    try:
        a=5
        file1 = open(\"file1.txt         


        

Answer 1

You mention using a loop, but you're not actually using a loop. Use a loop. That way you can write each file, one at a time, inside a single try block. You don't appear to be doing anything with the files except write one value to each, so you don't need to keep them all open.

for filename in ['file1.txt', 'file2.txt', ...]:
    try:
        with open(filename, 'w+') as f:
            f.write(str(a)+"whatever")
    except IOError:
        print("Error occurred with", filename)

Edit: If you have wildly different things to write to the different files, create a dictionary or other data structure ahead of time storing the mapping between files and data, then use that in the loop.

data = {'file1.txt': str(a), 'file2.txt': 'something else', 'file3.txt': str(a)+str(b)}

for filename, output in data.items():
    try:
        with open(filename, 'w+') as f:
            f.write(output)
    except IOError:
        print("Error occurred with", filename)