[Scons-users] Getting directory in which Sconscript is running

[Bill Deegan]

Tom,

Not sure what you're really trying to do, but it sounds ugly.
I've found if it's hard to do, you're probably doing it wrong..

Perhaps you can expand on the what you're trying to do, rather than the
'how you're trying to do it'..

-Bill


On Thu, May 1, 2014 at 9:23 AM, Tom Tanner (BLOOMBERG/ LONDON) <
ttanner2 at bloomberg.net> wrote:


> I've found half the answer. Dir('.').path gives me A/B

>

> Now I need to find the path of the actual directory that contains the

> SConscript (I know I can do os.path.dirname(File('SConscript').rstr()) but

> - that is not pretty)

>

>

> ----- Original Message -----

> From: ttanner2 at bloomberg.net

> To: scons-users at scons.org

> At: May 1 2014 17:19:59

>

> But I don't want to do #A/B. What happens if for some reason, I want to

> move B (or A). Or want to share code between 2 SConscripts.

>

> ----- Original Message -----

> From: wblevins001 at gmail.com

> To: Tom Tanner (BLOOMBERG/ LONDON) <ttanner2 at bloomberg.net>,

> scons-users at scons.org

> At: May 1 2014 17:00:00

>

> See scons 2.3.1 user guide 14.3 for more.

> On May 1, 2014 11:55 AM, "William Blevins" <wblevins001 at gmail.com> wrote:

>

>> The '#' symbol always refers to the SConstruct dir, so you can refer as

>> '#A' join 'B' and files stay relative to the root src dir.

>>

>> V/R,

>> William

>> On May 1, 2014 11:48 AM, "Tom Tanner (BLOOMBERG/ LONDON)" <

>> ttanner2 at bloomberg.net> wrote:

>>

>>> I'd really like to be able to get hold of the current SConscript

>>> directory relative to the SConstruct path, so with a directory structure in

>>> a repository a little like this:

>>>

>>> +-SConstruct

>>> |

>>> +-A

>>> |

>>> +-B

>>> |

>>> +-SConscript

>>>

>>> There will be something I can do in the SConscript that will get me

>>> "A/B" (i.e. the relative directory in which things will be built), and

>>> <wherever the source is>/A/B, which is the sort of thing I need to stuff in

>>> environment variable (for instance for getting the right PYTHONPATH for

>>> running python scripts which live in a subdirectory of B).

>>>

>>> Using Dir('.').rstr() or .rdir() will get you '.'. Throwing

>>> srcdir().abspath into the mix will get me the full path to the build

>>> directory, not the full path to the source directory.

>>>

>>>

>>>

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