Hoe haal je het workspace-pad op van een laag met arcpy?

Question

<\/HEAD>Jason Camerano<\/A> gaf me deze python snippet:<\/P>head, <\/SPAN>sep<\/SPAN>, <\/SPAN>tail = arcpy.Describe(parcelFabricLayer).Path.partition(".gdb"<\/SPAN>) workspacePath = head + sep<\/SPAN><\/PRE>uitleg:<\/P>Describe().Path retourneert het volledige pad dat ook de string ".gdb" bevat<\/P>partition scheidt dit met behulp van de opgegeven ".gdb" <\/SPAN>string om:<\/P>head - vóór de ".gdb" string<\/SPAN><\/LI>sep - de ".gdb" string<\/SPAN><\/LI>tail - na de ".gdb" string<\/SPAN><\/LI><\/UL><\/P>Iemand een betere methode?<\/P><\/BODY><\/HTML>

MattHowe · Answer

For Julian's path example I'd normally use os.

a = r"D:\Data\GISData\dataeceived\Transport.gdb\Routing\AccessRoad"
os.path.dirname(os.path.dirname(a))
'D:\Data\GISData\data\received\Transport.gdb'

MarkDonnelly · Answer

In Pro you can use the following:

layer.connectionProperties["connection_info"]["database"]

JulianInskip · Answer

Hi Amir. I do mine a little different. I don't use arcpy to get my workspace path. I find mine takes a lot quicker as it does not have to call arcpy every time, saving a few seconds. With this method you can also use any geodatabase type (.sde, .gdb or .mdb). accessRoadFc = r"D:\Data\GISData\data eceived\Transport.gdb\Routing\AccessRoad" wkspSplit = accessRoadFc.split(".") wkspExt = wkspSplit[1].split("\")[0] workspace = "{}.{}".format(wkspSplit[0], wkspExt)‍‍‍‍‍‍‍