CODEWITHSUNDEEP
pythonpathfilepath
nina_berlini
nina_berlini

Reputation: 195

Concatenate path and filename

I have to build the full path together in python. I tried this:

filename= "myfile.odt"

subprocess.call(['C:\Program Files (x86)\LibreOffice 5\program\soffice.exe',
                    '--headless',
                    '--convert-to',
                    'pdf', '--outdir',
                    r'C:\Users\A\Desktop\Repo\',
                    r'C:\Users\A\Desktop\Repo\'+filename])

But I get this error

SyntaxError: EOL while scanning string literal.

Upvotes: 12

Views: 84564

Answers (7)

Ethan
Ethan

Reputation: 171

You can also simply add the strings together. Personally I like this more.

filename = r"{}{}{}".format(dir, foldername, filename)

Upvotes: 1

mamal
mamal

Reputation: 1986

add library to code :

from pathlib import Path

when u want get current path without filename use this method :

print("Directory Path:", Path().absolute())

now you just need to add the file name to it :for example

   mylink = str(Path().absolute())+"/"+"filename.etc" #str(Path().absolute())+"/"+"hello.txt"

If normally addes to the first path "r" character for example: r"c://..."

You do not need to do here

Upvotes: 1

Armin
Armin

Reputation: 168

To anyone else stumbling across this question, you can use \ to concatenate a Path object and str.

Use path.Path for paths compatible with both Unix and Windows (you can use it the same way as I've used pathlib.PureWindowsPath).

The only reason I'm using pathlib.PureWindowsPath is that the question asked specifically about Windows paths.

For example:

import pathlib
# PureWindowsPath enforces Windows path style
# for paths that work on both Unix and Windows use path.Path
base_dir = pathlib.PureWindowsPath(r'C:\Program Files (x86)\LibreOffice 5\program')
# elegant path concatenation
myfile = base_dir / "myfile.odt"

print(myfile)
>>> C:\Program Files (x86)\LibreOffice 5\program\myfile.odt

Upvotes: 1

zanseb
zanseb

Reputation: 1425

Try:

import os
os.path.join('C:\Users\A\Desktop\Repo', filename)

The os module contains many useful methods for directory and path manipulation

Upvotes: 38

zvone
zvone

Reputation: 19352

Backslash character (\) has to be escaped in string literals.

  • This is wrong: '\'
  • This is correct: '\\' - this is a string containing one backslash

Therefore, this is wrong:

'C:\Program Files (x86)\LibreOffice 5\program\soffice.exe'

There is a trick!

String literals prefixed by r are meant for easier writing of regular expressions. One of their features is that backslash characters do not have to be escaped. So, this would be OK:

r'C:\Program Files (x86)\LibreOffice 5\program\soffice.exe'

However, that wont work for a string ending in backslash:

  • r'\' - this is a syntax error

So, this is also wrong:

r'C:\Users\A\Desktop\Repo\'

So, I would do the following:

import os
import subprocess


soffice = 'C:\\Program Files (x86)\\LibreOffice 5\\program\\soffice.exe'
outdir = 'C:\\Users\\A\\Desktop\\Repo\\'
full_path = os.path.join(outdir, filename)

subprocess.call([soffice,
                 '--headless',
                 '--convert-to', 'pdf',
                 '--outdir', outdir,
                 full_path])

Upvotes: 7

Aaron
Aaron

Reputation: 2073

To build on what zanseb said, use the os.path.join, but also \ is an escape character, so your string literal can't end with a \ as it would escape the ending quote.

import os
os.path.join(r'C:\Users\A\Desktop\Repo', filename)

Upvotes: 2

Simon Callan
Simon Callan

Reputation: 3130

The problem you have is that your raw string is ending with a single backslash. For reason I don't understand, this is not allowed. You can either double up the slash at the end:

r'C:\Users\A\Desktop\Repo\\'+filename

or use os.path.join(), which is the preferred method:

os.path.join(r'C:\Users\A\Desktop\Repo', filename)

Upvotes: 2

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