Reputation: 201
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?
Upvotes: 20
Views: 119944
Reputation: 27
Instead of this D:\path\0.html
try this D:/path/0.html
Reason of error is python interpret \0 instead of path string.
Upvotes: 1
Reputation: 11
I was also getting the same error with the following code:
with zipfile.ZipFile("C:\local_files\REPORT.zip",mode='w') as z:
z.writestr(data)
It was happening because I was passing the bytestring i.e. data in writestr()
method without specifying the name of file i.e. Report.zip
where it should be saved.
So I changed my code and it worked.
with zipfile.ZipFile("C:\local_files\REPORT.zip",mode='w') as z:
z.writestr('Report.zip', data)
Upvotes: 1
Reputation: 51
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
Upvotes: 5
Reputation: 191
Try putting r (raw format).
r'D:\python_projects\templates\0.html'
Upvotes: 7
Reputation: 3561
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'
Upvotes: 2
Reputation: 152
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.
Upvotes: 2
Reputation: 179
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())
Upvotes: 1
Reputation: 5075
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")
Upvotes: -1
Reputation: 339
It seems that you have problems with characters "\" and "/". If you use them in input - try to change one to another...
Upvotes: 22
Reputation: 1191
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()
Upvotes: 8