Php – Create a folder if it doesn’t already exist

TFS users: If you are using source control that requires you to warn it before your rename files/folders then look at this answer instead which covers the extra steps required.

To rename a project's folder, file (.*proj) and display name in Visual Studio:

• Close the solution.
• Rename the folder(s) outside Visual Studio. (Rename in TFS if using source control)
• Open the solution, ignoring the warnings (answer "no" if asked to load a project from source control).
• Go through all the unavailable projects and...
  • Open the properties window for the project (highlight the project and press Alt+Enter or F4, or right-click > properties).
  • Set the property 'File Path' to the new location.
    • If the property is not editable (as in Visual Studio 2012), then open the .sln file directly in another editor such as Notepad++ and update the paths there instead. (You may need to check-out the solution first in TFS, etc.)
  • Reload the project - right-click > reload project.
  • Change the display name of the project, by highlighting it and pressing F2, or right-click > rename.

Note: Other suggested solutions that involve removing and then re-adding the project to the solution will break project references.

If you perform these steps then you might also consider renaming the following to match:

1. Assembly
2. Default/Root Namespace
3. Namespace of existing files (use the refactor tools in Visual Studio or ReSharper's inconsistent namespaces tool)

Also consider modifying the values of the following assembly attributes:

1. AssemblyProductAttribute
2. AssemblyDescriptionAttribute
3. AssemblyTitleAttribute

On Python ≥ 3.5, use pathlib.Path.mkdir:

from pathlib import Path
Path("/my/directory").mkdir(parents=True, exist_ok=True)

For older versions of Python, I see two answers with good qualities, each with a small flaw, so I will give my take on it:

Try os.path.exists, and consider os.makedirs for the creation.

import os
if not os.path.exists(directory):
    os.makedirs(directory)

As noted in comments and elsewhere, there's a race condition – if the directory is created between the os.path.exists and the os.makedirs calls, the os.makedirs will fail with an OSError. Unfortunately, blanket-catching OSError and continuing is not foolproof, as it will ignore a failure to create the directory due to other factors, such as insufficient permissions, full disk, etc.

One option would be to trap the OSError and examine the embedded error code (see Is there a cross-platform way of getting information from Python’s OSError):

import os, errno

try:
    os.makedirs(directory)
except OSError as e:
    if e.errno != errno.EEXIST:
        raise

Alternatively, there could be a second os.path.exists, but suppose another created the directory after the first check, then removed it before the second one – we could still be fooled.

Depending on the application, the danger of concurrent operations may be more or less than the danger posed by other factors such as file permissions. The developer would have to know more about the particular application being developed and its expected environment before choosing an implementation.

Modern versions of Python improve this code quite a bit, both by exposing FileExistsError (in 3.3+)...

try:
    os.makedirs("path/to/directory")
except FileExistsError:
    # directory already exists
    pass

...and by allowing a keyword argument to os.makedirs called exist_ok (in 3.2+).

os.makedirs("path/to/directory", exist_ok=True)  # succeeds even if directory exists.