Develop Reference

Tkinter FileDialog.askopenfilename failing on already open files

I can not find a way using the the Tkinter tkFileDialog.askopenfilename / Open() method to get a file name of an already opened file. At least on Windows 7 anyway.
Simply need to return the filename selected, but when I select an 'opened' file I get the "This file is in use" popup. Any searches for getting filenames seems to always point back to using FileDialog.
There seems to be the same question but using c# here:
Open File Which is already open in window explorer
Are there any ways not listed that can be used to arrive at this? A file selection popup that doesn't fail on already opened files on Windows 7?
Python 2.7.11
Windows 7_64
Selection dialog:
opts = {}
opts['title'] = 'Select file.'
filename = tkFileDialog.Open(**opts).show()
or
filename = tkFileDialog.askopenfilename(**opts)
Which are the same as askopenfilename calls Open().show() which in turn is calling command = "tk_getOpenFile".
Searching tk_getOpenFile shows nothing to override this behavior.
Any pointers here would be much appreciated.
Update:
So as to not reiterate everything that's been gone over here's the break down of the Tkinter 8.5 FileDialog.
def askopenfile(mode = "r", **options):
"Ask for a filename to open, and returned the opened file"
filename = Open(**options).show()
if filename:
return open(filename, mode)
return None
def askopenfilename(**options):
"Ask for a filename to open"
return Open(**options).show()
Both call the Open class but askopenfile follows up with an open(file), here:
class Open(_Dialog):
"Ask for a filename to open"
command = "tk_getOpenFile"
def _fixresult(self, widget, result):
if isinstance(result, tuple):
# multiple results:
result = tuple([getattr(r, "string", r) for r in result])
if result:
import os
path, file = os.path.split(result[0])
self.options["initialdir"] = path
# don't set initialfile or filename, as we have multiple of these
return result
if not widget.tk.wantobjects() and "multiple" in self.options:
# Need to split result explicitly
return self._fixresult(widget, widget.tk.splitlist(result))
return _Dialog._fixresult(self, widget, result)
UPDATE2(answerish):
Seems the windows trc/tk tk_getOpenFile ultimately calls
OPENFILENAME ofn;
which apparently can not, at times, return a filename if the file is in use.
Yet to find the reason why but regardless.
In my case, needing to reference a running QuickBooks file, it will not work using askopenfilename().
Again the link above references someone having the same issue but using c# .

File upload at web2py

I am using the web2py framework.
I have uploaded txt a file via SQLFORM and the file is stored in the "upload folder", now I need to read this txt file from the controller, what is the file path I should use in the function defined in the default.py ?
def readthefile(uploaded_file):
file = open(uploaded_file, "rb")
file.read()
....

You can do join of application directory and upload folder to build path to file.
Do something like this:
import os
filepath = os.path.join(request.folder, 'uploads', uploaded_file_name)
file = open(filepath, "rb")
request.folder: the application directory. For example if the
application is "welcome", request.folder is set to the absolute path
"/path/to/welcome". In your programs, you should always use this
variable and the os.path.join function to build paths to the files you
need to access.
Read request.folder

The transformed name of the uploaded file is stored in the upload field of your database table, so you need a way to query the specific record that was inserted via the SQLFORM submission in order to get the name of the stored file. Here is how it would look assuming you know the record ID:
stored_filename = db.mytable(record_id).my_upload_field
original_filename, stream = db.mytable.my_upload_field.retrieve(stored_filename)
stream.read()
When you pass a filename to the .retrieve method of an upload field, it will return a tuple containing the original filename as well as the open file object (called stream in the code above).

wxPython FileOutputStream error

I'm trying to learn wxPython/python and I want to save text in a file. I found this example
def OnSaveAs(self, e):
saveFileDialog = wx.FileDialog(self, "SAVE txt file", "", "", "Textdocument (*.txt)|*.txt", wx.FD_SAVE | wx.FD_OVERWRITE_PROMPT)
if saveFileDialog.ShowModal() == wx.ID_CANCEL:
return # User canceled
# save the current contents in the file
# this can be done with e.g. wxPython output streams:
output_stream = wx.FileOutputStream(saveFileDialog.GetPath())
#My question: Insert what to write to output_stream here?
if not out_stream.IsOk():
wx.LogError("Cannot save current contents in file '%s'."%saveFileDialog.GetPath())
return
I get the error
in OnSaveAs output_stream = wx.FileOutputStream(saveFileDialog.GetPath()) AttributeError 'module' object has no attribute 'FileOutputStream'
Shouldnt output_stream contain the path to the file i want to save. And then I write to output_stream to save text in the file?
Thanks in advance!

Just use the Python functions to open and write content to the file. Something like this:
output = open(saveFileDialog.GetPath(), 'w')
ouput.write(stuff)
ouput.close()
In almost all cases wxPython only wraps the wxWidgets classes and functions which do not already have an equivallent in Python, and the AttributeError is telling you that there is no wx.FileOutputStream class available.

Save Dialog only appear once?

I am currently making a desktop widget and what I want to do is create a file in which a user can edit and then save. However, if you guys are familiar with Microsoft word or any other text editors, I want it so that after you hit File -> save, a save dialog appears in which you can choose where to save the file and the file's name. However, after the first time, if the file name stays the same, the save dialog will not come up--rather it will just automatically save over what was previously written. This is what I want to implement but I am having trouble trying to do this. Following is my method for saving files using the save dialog but I am unsure on how to save without having the save dialog pop up.
def saveFile(self):
filename = QtGui.QFileDialog.getSaveFileName(None, 'Save File', os.path.expanduser("~/Desktop/Calendar Data/"+self.dateString), ".txt")
f = open(filename, 'w')
filedata = self.text.toPlainText()
f.write(filedata)
f.close()
Anyone have any idea how to do this? If so that would be great! Thanks for helping.

You should make filename an instance attribute, so you can just check if it was already set:
class Spam:
...
def __init__(self):
self.filename = None
def saveFile(self):
if not self.filename:
self.filename = QtGui.QFileDialog.getSaveFileName(...)
# here you should check if the dialog wasn't cancelled
with open(filename, 'w') as f:
f.write(self.text.toPlainText())
...

File management

I am working on python and biopython right now. I have a file upload form and whatever file is uploaded suppose(abc.fasta) then i want to pass same name in execute (abc.fasta) function parameter and display function parameter (abc.aln). Right now i am changing file name manually, but i want to have it automatically.
Workflow goes like this.
----If submit is not true then display only header and form part
--- if submit is true then call execute() and get file name from form input
--- Then displaying result file name is same as executed file name but only change in extension
My raw code is here -- http://pastebin.com/FPUgZSSe
Any suggestions, changes and algorithm is appreciated
Thanks

You need to read the uploaded file out of the cgi.FieldStorage() and save it onto the server. Ususally a temp directory (/tmp on Linux) is used for this. You should remove these files after processing or on some schedule to clean up the drive.
def main():
import cgi
import cgitb; cgitb.enable()
f1 = cgi.FieldStorage()
if "dfile" in f1:
fileitem = f1["dfile"]
pathtoTmpFile = os.path.join("path/to/temp/directory", fileitem.filename)
fout = file(pathtoTmpFile, 'wb')
while 1:
chunk = fileitem.file.read(100000)
if not chunk: break
fout.write (chunk)
fout.close()
execute(pathtoTmpFile)
os.remove(pathtoTmpFile)
else:
header()
form()
This modified the execute to take the path to the newly saved file.
cline = ClustalwCommandline("clustalw", infile=pathToFile)
For the result file, you could also stream it back so the user gets a "Save as..." dialog. That might be a little more usable than displaying it in HTML.