[IfcOpenShell] How to work with file content instead of file path

B by Borna_Molnar on 16 Jul 2021, edited 18 Jun 2022 #

I am trying to develop a Dash app for ifc files but I am stuck at the first step.

The dash upload component gives acces to the content of the uploaded file but not the uploaded path so I am not really sure how to work with that.

The ifc.open() method doesnt work without a path.

I looked around in the ifcopenshell folder and found that the ifcopenshell_wrapper has an read() method so I tried the following code:

import ifcopenshell as ifc

from ifcopenshell import ifcopenshell_wrapper

from ifcopenshell.ifcopenshell_wrapper import file

f = ifcopenshell_wrapper.read(contents)

ifc_file = file(f)

But I get this error:

Wrong number or type of arguments for overloaded function 'new_file'.

Possible C/C++ prototypes are:

IfcParse::IfcFile::IfcFile(std::string const &)

IfcParse::IfcFile::IfcFile(std::istream &,int)

IfcParse::IfcFile::IfcFile(void *,int)

IfcParse::IfcFile::IfcFile(IfcParse::IfcSpfStream *)

IfcParse::IfcFile::IfcFile(IfcParse::schema_definition const *)

IfcParse::IfcFile::IfcFile()

What did I do wrong? Or is there another way around?

M by Moult on 17 Jul 2021 # +1 votes

@Borna_Molnar try this:


 f = ifcopenshell.file.from_string(open('/home/dion/tmp.ifc').read())

K by keshavanarayan on 18 Jun 2022, edited 19 Jun 2022 #

I'm getting a file doesn't exist error if I do this. (open('/home/USER/tmp.ifc'))Please let me know if you have found a solution.