I have vtk example which can read vtk file if it present, but I don't know where I to put vtp file?
If only the file name is passed to a reader, it should look in the current working directory (the same directory that your executable is created in).
To read and print to video a .vtp file use a piece of code like this:
filename = 'filename'
reader = vtk.vtkXMLPolyDataReader()
reader.setFileName(filename)
mapper = vtk.vtkPolyDataMapper()
mapper.SetInputConnection(reader.GetOutputPort())
actor = vtkActor()
actor.SetMapper(mapper)
renderer = vtkRenderer()
renderer.AddActor(actor)
renderWindow = vtk.vtkRenderWindow()
renderWindow.AddRenderer(renderer)
renderWindowInteractor = vtk.vtkRenderWindowInteractor()
renderWindowInteractor.SetRenderWindow(renderWindow)
renderer.ResetCamera()
renderWindow.Render()
renderWindowInteractor.Initialize()
renderWindowInteractor.Start()
Obviously substitute the 'filename' with the name of your .vtp file, like this
filename = 'example.vip'
Related
I'm using a shipping api, that has a method that spits out a pdf in this format (ascii, I believe):
%PDF-1.3\n1 0 obj\n
<<\n
/Type /Pages\n
/Count 1\n
many lines removed here
startxref\n
982\n
%%EOF\n
How do I..
Convert this code into a downloadable .pdf file?
Create a link to download the file?
Note - I do not need to store the file.
ADDENDUM
So, this is moving the correct file to the correct place, but is giving me an error "Call to a member function move() on string".
$pdf_raw = $this->create_label2($data->label_url);
$filename = 'label'.auth()->user()->id.'.pdf';
file_put_contents($filename, $pdf_raw);
$filename->move(public_path().'/img/pdf', $filename);
How can that be?
Here is the solution:
use Illuminate\Support\Facades\Storage;
$pdf_raw = $this->create_label2($data->label_url); //this is raw ascii data
$filename = 'label'.auth()->user()->id.'.pdf'; //can be any name .pdf
Storage::disk('pdf')->put($filename,$pdf_raw); //creates file and moves to correct path
$pdf_link = "/mypath/$filename"; //url to file
I am practicing Ruby, and I am trying to copy contents from file "from" to file "to". can you tell me where I did it wrong?
thanks !
from = "1.txt"
to = "2.txt"
data = open(from).read
out = open(to, 'w')
out.write(data)
out.close
data.close
Maybe I am missing the point, but I think writing it like so is more 'ruby'
from = "1.txt"
to = "2.txt"
contents = File.open(from, 'r').read
File.open(to, 'w').write(contents)
Personally, however, I like to use the Operating systems terminal to do File operations like so. Here is an example on linux.
from = "1.txt"
to = "2.txt"
system("cp #{from} #{to}")
And for Windows I believe you would use..
from = "1.txt"
to = "2.txt"
system("copy #{from} #{to}")
Finally, if you were needing the output of the command for some sort of logging or other reason, I would use backticks.
#A nice one liner
`cp 1.txt 2.txt`
Here is the system and backtick methods documentation.
http://ruby-doc.org/core-1.9.3/Kernel.html
You can't perform data.close — data.class would show you that you have a String, and .close is not a valid String method. By opening from the way you chose to, you lost the File reference after using it with your read. One way to fix that would be:
from = "1.txt"
to = "2.txt"
infile = open(from) # Retain the File reference
data = infile.read # Use it to do the read
out = open(to, 'w')
out.write(data)
out.close
infile.close # And finally, close it
i need to get the full address of a file which needs to be uploaded by the user using browse button. i tried getAbsolutePath, getAbsoluteFile, getCanonicalPath but they all are giving tomcat/bin location. i need the full path of the file which is to be uploaded.
MultipartFile doc_file = studentInfoBean.getUploadedDocument();
String fileName = doc_file.getOriginalFilename();
String fileExtension = FilenameUtils.getExtension(fileName);
File file = new File(fileName);
File path = file.getAbsoluteFile();
//String path = path.toString()
thank you
You'll probably want to use MultipartFile.transferTo(File dest) to save the uploaded file locally. You can then do your conversion, and whatever you need to do with your .csv file (store it somewhere, send it back to the client, etc.) So the full code might be:
MultipartFile doc_file = studentInfoBean.getUploadedDocument();
File temp_file = new File(doc_file.getOriginalFilename());
doc_file.transferTo(temp_file);
//convert doc_file to .csv
//store locally permanently or return to client
I have downloaded a set of html files and saved the file paths which I saved them to in a .txt file. It has each path on a new line. I wanted to look at the first file in the list and then itterate through the whole list, opening the files and extracting data before going on to the next file.
My code works fine with a single path put in directly (for the first file) as:
path = r'C:\path\to\file.html'
and works if I itterate through the text file using:
file_list_fp = r'C:\path\to\file_with_pathlist.txt'
with open(file_list_fp, 'r') as file_list:
for filepath in file_list:
pathend = filepath.find('\n')
path = file[:pathend]
q = open(path, 'r').read()
but it fails when I try getting a single path using either:
with open(file_list_fp, 'r') as file_list:
path_n = file_list.readline()
end = path_n.find('\n')
path_bad1 = path_n[:end]
or:
with open(file_list_fp, 'r') as file_list:
path_bad2 = file_list.readline().split('\n')[0]
With these two my code exits just after that point. I can't figure out why. Any pointers very welcome. (I'm using Python 3.3.1 on windows.)
I am having a problem in matlab and the problem is described as follows:
When i try to read an image ( I have several images) and write them to a specific folder, the matlab triggers an error saying
Error using ==> imwrite at 394
Can't open file "\Temp\\inim735282.4716703009300000.jpg" for writing.
You may not have write permission.
May I know why this is happening?
this is the code where the problem occurs
mkdir('.\Temp');
temp_name = sprintf('%.16f',now);
corner_file = ['\Temp\corners', temp_name,'.in'];
image_file = ['\Temp\inim', temp_name,'.jpg'];
out_file = ['\Temp\out', temp_name,'.desc'];
out_imname = ['\Temp\out', temp_name,'.desc.jpg'];
I tried to change it by omitting
mkdir('.\Temp');
moreoever, i direct the path in the folder to the folder by doing this
binary_path = 'C:\Users\cool\Documents\MATLAB\Experment\experiments\bag_of_words\Temp';
to read and and write in and out of the folder.
Can someone please help me figure out this problem?
Thank you guys
Open MatLAB with admin privileges.
A few suggestions:
To generate a temporary output name use the command tempname.
temp_name = tempname();
To concatenate paths and file names use fullfile.
conrner_file = fullfile( '\', 'Temp', 'corners', [temp_name, '.in'] );
You should be careful not to mix '\Temp' and '.\Temp': as the first is an absolute path, while the second is a relative path to cwd.
EDIT:
How about:
temp_name = tempname(); % temp name + folder name in TEMP
corner_file = [ temp_name,'.in'];
image_file = [ temp_name,'.jpg'];
out_file = [temp_name,'.desc'];
out_imname = [temp_name,'.desc.jpg'];
Is it working now?