How to get clean name from a linked file?

Developers
1

Hello,

i got the name from my linked instance but for any reason, the name includes other stuff

import clr
import sys
clr.AddReference('RevitAPI')
from Autodesk.Revit.DB import *
from Autodesk.Revit.DB.Structure import *
clr.AddReference('RevitServices')
import RevitServices
from RevitServices.Persistence import DocumentManager
from RevitServices.Transactions import TransactionManager

doc = DocumentManager.Instance.CurrentDBDocument
uidoc = DocumentManager.Instance.CurrentUIApplication.ActiveUIDocument

rvt_links = FilteredElementCollector(doc).OfCategory(BuiltInCategory.OST_RvtLinks).WhereElementIsNotElementType()

names = []

for i in rvt_links:
	names.append(i.Name)


OUT = rvt_links, names

How do i call it correctly?

2023-02-23_14h22_23
2023-02-23_14h22_23664×716 54.3 KB

grafik

KR
Andreas

2

You are looking at link instances not link types. What would you expect for the name if you had multiple instances of one of the links?

If you want the file name you can try working with the link types (via collector or by getting the instances link type), or pull each instance’s document and get that document’s title.

1 Like
3

@jacob.small ,

based on time-presure. i solved it this way…

grafik
grafik1174×492 110 KB

KR
Andreas

4

Try the following:

import clr
clr.AddReference('RevitAPI')
from Autodesk.Revit.DB import *
clr.AddReference('RevitServices')
import RevitServices
from RevitServices.Persistence import DocumentManager

doc = DocumentManager.Instance.CurrentDBDocument

rvt_links = FilteredElementCollector(doc).OfCategory(BuiltInCategory.OST_RvtLinks).WhereElementIsNotElementType()

OUT = []

for i in rvt_links:
	rvtFileName = i.Name.split(".rvt")[0]
	OUT.append([i,rvtFileName])

OUT = zip(*OUT)
2 Likes
5

A non - string manipulation method:

for i in rvt_links:
         lnkDoc = I. GetLinkDocument()
         names.append(lnkDoc.Title)
1 Like
6

Beat me to it, i was just about to post a version including the GetLinkDocument() as i just remembered about that :rofl:

import clr
clr.AddReference('RevitAPI')
from Autodesk.Revit.DB import *
clr.AddReference('RevitServices')
import RevitServices
from RevitServices.Persistence import DocumentManager

doc = DocumentManager.Instance.CurrentDBDocument

#Gets a list of Revit Link elements
rvt_links = FilteredElementCollector(doc).OfCategory(BuiltInCategory.OST_RvtLinks).WhereElementIsNotElementType()

#Sets OUT to be a list
OUT = []

for i in rvt_links:
	# Outputs the revit link and the title of the link document(eg file name without format)
	# If you need more data from the link Document than just the title, then you should get 
	# the link document first as a variable then get what you need before putting into the list append.
	# eg   rvtDoc = i.GetLinkDocument()
	#  rvtTitle = rvtDoc.Title
	#  rvtPath = rvtDoc.PathName
	OUT.append([i,i.GetLinkDocument().Title])

# Transposes the list and outputs it
OUT = zip(*OUT)
2 Likes
7

Personally when working with documents I always return the link, as it allows you to get not just the title (aka, name which is certainly useful in it’s own right), but also path which in my experience has been the instant follow up to the file name ion 80% of uses, and a whole bunch more.

2 Likes
8

Good point and I have added that as a comment on my code that i posted :slight_smile:

1 Like
9

Hello this is Gulshan Negi
There are lots of ways to do this. To get a clean name from a linked file in Revit, you can also use the Document.GetElement() method and the RevitLinkType.Name property.
Thanks

1 Like