How to confirm a file is a valid csv file

Question

I have a program that reads csv files. Sometimes users mistakenly try to pass Excel files into the program and I would like to detect this.

Currently, I get a nasty error and I can't seem to catch it.

A simple example:

from csv import DictReader
with open("an_excel_file_pretending_to_be_a_csv_file.csv", "rb") as f:
    reader = csv.DictReader(f)
    for row in reader:
        print(row)

This produces the following error message.

    for row in reader:
  File "c:\Python27\Lib\csv.py", line 107, in next
    self.fieldnames
  File "c:\Python27\Lib\csv.py", line 90, in fieldnames
    self._fieldnames = self.reader.next()
Error: line contains NULL byte

This is happening when the DictReader tries to read the first line of data.

What I would like to do is to catch this as an exception. Unfortunately, this is not an exception , it just borks out of my code with the error message.

from csv import DictReader
with open("an_excel_file_pretending_to_be_a_csv_file.csv", "rb") as f:
    reader = csv.DictReader(f)
    try:
        for row in reader:
            print(row)
    except Exception:
        print "error"

Nothing is captured, the error still breaks out of my code which is in an infinite loop processing many files.

There is a similar question here: Python CSV error: line contains NULL byte

But I would like to catch the error gracefully in my code. How can I do this?

Answer 1

You could use a general try-except to capture ALL exceptions (be careful because this may be dangerous, hiding errors in your code).

try:
    # your code that can raise exceptions here
except Exception as e:
    # handle ANY exception
else:
    # this executes if no exception occurred (optional)

Answer 2

readline function can be used to detect the type before using the csv.DictReader

with open("file1.csv", "rb") as f:
   print f.readline().split(",")