"ValueError: embedded null character" when using open()

I am taking python at my college and I am stuck with my current assignment. We are supposed to take 2 files and compare them. I am simply trying to open the files so I can use them but I keep getting the error "ValueError: embedded null character"

file1 = input("Enter the name of the first file: ") file1_open = open(file1) file1_content = file1_open.read() 

What does this error mean?

8 Answers

It seems that you have problems with characters "\" and "/". If you use them in input - try to change one to another...

Default encoding of files for Python 3.5 is 'utf-8'.

Default encoding of files for Windows tends to be something else.

If you intend to open two text files, you may try this:

import locale locale.getdefaultlocale() file1 = input("Enter the name of the first file: ") file1_open = open(file1, encoding=locale.getdefaultlocale()[1]) file1_content = file1_open.read() 

There should be some automatic detection in the standard library.

Otherwise you may create your own:

def guess_encoding(csv_file): """guess the encoding of the given file""" import io import locale with io.open(csv_file, "rb") as f: data = f.read(5) if data.startswith(b"\xEF\xBB\xBF"): # UTF-8 with a "BOM" return "utf-8-sig" elif data.startswith(b"\xFF\xFE") or data.startswith(b"\xFE\xFF"): return "utf-16" else: # in Windows, guessing utf-8 doesn't work, so we have to try try: with io.open(csv_file, encoding="utf-8") as f: preview = f.read(222222) return "utf-8" except: return locale.getdefaultlocale()[1] 

and then

file1 = input("Enter the name of the first file: ") file1_open = open(file1, encoding=guess_encoding(file1)) file1_content = file1_open.read() 

I got this error when copying a file to a folder that starts with a number. If you write the folder path with the double \ sign before the number, the problem will be solved.

On Windows while specifying the full path of the file name, we should use double backward slash as the seperator and not single backward slash. For instance, C:\\FileName.txt instead of C:\FileName.txt

The problem is due to bytes data that needs to be decoded.

When you insert a variable into the interpreter, it displays it's repr attribute whereas print() takes the str (which are the same in this scenario) and ignores all unprintable characters such as: \x00, \x01 and replaces them with something else.

A solution is to "decode" file1_content (ignore bytes):

file1_content = ''.join(x for x in file1_content if x.isprintable()) 

The first slash of the file path name throws the error.

Need Raw, r
Raw string

FileHandle = open(r'..', encoding='utf8') 

FilePath='C://FileName.txt'
FilePath=r'C:/FileName.txt'

Try putting r (raw format).

r'D:\python_projects\templates\0.html'

If you are trying to open a file then you should use the path generated by os, like so:

import os os.path.join("path","to","the","file")