Re: File Path retrieving problem

2009-02-26 Thread music24by7 
Hi Dennis Thanks for your reply and also the detailed explanation.

 Or do you mean you want something that, given a bare name, searches
 your file system to find where a file with that name actually is?


Yes, this is what i exactly needed.
I have found something interesting to do this using the os.walk(path)
function, and it worked
I am pasting the code which i used for this:



testpath=rC:\\
logpath=rC:\Pathfinder.txt
csvpath=rC:\Pathfinder.csv
import os, csv

def logData(d={}, logfile=c://filename999.txt, separator=\n):
Takes a dictionary of values and writes them to the provided
file path
logstring=separator.join([str(key)+:  +d[key] for key in
d.keys()])+\n
f=open(logfile,'a')
f.write(logstring)
f.close()
return
def walker(topPath,csvpath):
fDict={}
logDict={}
limit=1000
freed_space=0
items=0
fileHandle = csv.writer(open(csvpath, 'w'))
fileHandle.writerow(['Index','Path'])
for root, dirs, files in os.walk(topPath):
for name in files:
fpath=os.path.join(root,name)
logDict[Name]=name
logDict[Path]=fpath
if name.find(NEWS):
continue
else:
logData(logDict, logpath, \t)
fDict[items] = name
items=len(fDict.keys())
fileHandle.writerow([items,fpath])
print Dict entry:  ,items,
print Path:  ,fpath
if items  limit:
break
if items  limit:
break
walker(testpath,csvpath)



Finally thank you all very much for the patience to listen to me.


Regards,
Sudhir

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File Path retrieving problem

2009-02-25 Thread music24by7 
Hi all,

I am new to Python, i have installed python 2.5.4 and it is my
requirement.

I need to retrieve the path of filename in python.

I have found some API's to get this:

from os.path import realpath
print realpath(NEWS.txt)  # here NEWS.txt exists and it shows the
path of the file as C:\Python25\WorkSpace\NEWS.txt
print realpath(abc.txt) # here abc.txt does not exist but still it
shows C:\Python25\WorkSpace\abc.txt

can anybody tell the reason why


Now took some safety measures:

found = lexists(realpath(filename))
if found == 0:
print Not Found
else:
print realpath(filename)

i have given the filename as NEWS.txt and abc.txt but i am always
getting the output as Not Found


Can anyone please tell me where am i doing wrong

also any suggestions to retrieve the filepath from a given filename is
highly appreciated.

Thanks in advance.


Regards,
Sudhir
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Re: File Path retrieving problem

2009-02-25 Thread music24by7 
On Feb 25, 8:57 pm, Steve Holden st...@holdenweb.com wrote:
 music24...@gmail.com wrote:
  Hi all,

  I am new to Python, i have installed python 2.5.4 and it is my
  requirement.

  I need to retrieve the path of filename in python.

  I have found some API's to get this:

  from os.path import realpath
  print realpath(NEWS.txt)  # here NEWS.txt exists and it shows the
  path of the file as C:\Python25\WorkSpace\NEWS.txt
  print realpath(abc.txt) # here abc.txt does not exist but still it
  shows C:\Python25\WorkSpace\abc.txt

  can anybody tell the reason why

  Now took some safety measures:

  found = lexists(realpath(filename))
          if found == 0:
              print Not Found
          else:
              print realpath(filename)

  i have given the filename as NEWS.txt and abc.txt but i am always
  getting the output as Not Found

  Can anyone please tell me where am i doing wrong

 It seems pretty apparent that lexists() nevert returns a true result.

 Why not just

 if os.path.exists(filename):
     print os.path.realpath(filename)
 else:
     print Not found

  also any suggestions to retrieve the filepath from a given filename is
  highly appreciated.

 Well, realpath returns the path of the file targeted after any symbolic
 links have been evaluated, which may or may not be what you want. Have
 you looked at os.path.abspath?

 regards
  Steve
 --
 Steve Holden        +1 571 484 6266   +1 800 494 3119
 Holden Web LLC              http://www.holdenweb.com/



Hi Steve,

I have tried your suggested code and also replaced os.path.realpath
with os.path.abspath but still getting the same result.
I want to know is there any workaround for retrieving the filepaths
given only filename

Regards,
Sudhir
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Re: File Path retrieving problem

2009-02-25 Thread music24by7 
On Feb 26, 2:35 am, Emile van Sebille em...@fenx.com wrote:
 Peter Otten wrote:
  Maybe it's about access rights?

  $ mkdir alpha
  $ touch alpha/beta
  $ python -cimport os; print os.path.exists('alpha/beta')
  True
  $ chmod u-x alpha
  $ python -cimport os; print os.path.exists('alpha/beta')
  False
  $

  I Don't know how this is handled on Windows...

 Here's one way

 Microsoft Windows XP [Version 5.1.2600]
 (C) Copyright 1985-2001 Microsoft Corp.
 C:\mkdir alpha
 C:\touch alpha\beta  # cygwin at work here...

 C:\python -cimport os; print os.path.exists('alpha/beta')
 True

 C:\cacls alpha /E /R Everyone
 processed dir: C:\alpha
 C:\cacls alpha /E /R Users
 processed dir: C:\alpha
 C:\cacls alpha /E /R Administrators
 processed dir: C:\alpha

 C:\python -cimport os; print os.path.exists('alpha/beta')
 False

Hi,

This is the following code which i have used to retrieve the pathname
of a filename given by the user

filename = self.enText.get() # get the text entered in the
text box
if os.path.exists(filename): # checking if the filename
exists , had replaced it with os.path.exists(os.path.abspath
(filename))
print os.path.abspath(filename)
else:
print Not found

actually i need to list all the files/folders with the given filename
alongwith path and store it in an CSV file with index.

But i am unable to get even a single filepath correctly.
Whatever filename i enter the output is as : C:\Python25\WorkSpace
\xyz
i.e the name i entered is being appended to my workspace and
displayed.
Guys, i am not able to understand what shall i do here...plz help me
out.


Regards,
Sudhir
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Re: File Path retrieving problem

2009-02-25 Thread music24by7 
On Feb 26, 9:03 am, Steve Holden st...@holdenweb.com wrote:
 music24...@gmail.com wrote:
  On Feb 26, 2:35 am, Emile van Sebille em...@fenx.com wrote:
  Peter Otten wrote:
  Maybe it's about access rights?
  $ mkdir alpha
  $ touch alpha/beta
  $ python -cimport os; print os.path.exists('alpha/beta')
  True
  $ chmod u-x alpha
  $ python -cimport os; print os.path.exists('alpha/beta')
  False
  $
  I Don't know how this is handled on Windows...
  Here's one way

  Microsoft Windows XP [Version 5.1.2600]
  (C) Copyright 1985-2001 Microsoft Corp.
  C:\mkdir alpha
  C:\touch alpha\beta  # cygwin at work here...

  C:\python -cimport os; print os.path.exists('alpha/beta')
  True

  C:\cacls alpha /E /R Everyone
  processed dir: C:\alpha
  C:\cacls alpha /E /R Users
  processed dir: C:\alpha
  C:\cacls alpha /E /R Administrators
  processed dir: C:\alpha

  C:\python -cimport os; print os.path.exists('alpha/beta')
  False

  Hi,

  This is the following code which i have used to retrieve the pathname
  of a filename given by the user

          filename = self.enText.get() # get the text entered in the
  text box
          if os.path.exists(filename): # checking if the filename
  exists , had replaced it with os.path.exists(os.path.abspath
  (filename))
              print os.path.abspath(filename)
          else:
              print Not found

  actually i need to list all the files/folders with the given filename
  alongwith path and store it in an CSV file with index.

  But i am unable to get even a single filepath correctly.
  Whatever filename i enter the output is as : C:\Python25\WorkSpace
  \xyz
  i.e the name i entered is being appended to my workspace and
  displayed.
  Guys, i am not able to understand what shall i do here...plz help me
  out.

 Look, we aren't psychic. We can't debug code and directories we can't
 see. Work with us here - give us the information we need to help you.

 The actual code. A directory listing? Any error messages?

 regards
  Steve
 --
 Steve Holden        +1 571 484 6266   +1 800 494 3119
 Holden Web LLC              http://www.holdenweb.com/





Hi Steve,

Following is the code which i have written to display a GUI textbox
and get any name entered by the user and search the path of this
filename and then list it into an CSV file.

For this i initially created an textbox widget and added an button to
it.
Now, i have entered a filename (like NEWS.txt) in the text boz,
as you can see when i press the button i retrieve the text and then
search its filepath.
currently i am printing this filepath on the shell.


from Tkinter import *
import os
from os.path import realpath, exists, abspath
import tkMessageBox
import Tkinter
#import filePath

class GUIFrameWork(Frame):
This is the GUI

def __init__(self,master=None):
Initialize yourself

Initialise the base class
Frame.__init__(self,master)

Set the Window Title
self.master.title(Enter Text to search)

Display the main window
with a little bit of padding
self.grid(padx=10,pady=10)
self.CreateWidgets()

def CreateWidgets(self):

self.enText = Entry(self)
self.enText.grid(row=0, column=1, columnspan=3)



Create the Button, set the text and the
command that will be called when the button is clicked
self.btnDisplay = Button(self, text=Display!,
command=self.Display)
self.btnDisplay.grid(row=0, column=4)

def Display(self):
Called when btnDisplay is clicked, displays the contents of
self.enText
filename = self.enText.get()
if os.path.exists(os.path.abspath(filename)):
print os.path.abspath(filename)
else:
print Not found
#tkMessageBox.showinfo(Text, You typed: %s %
os.path.abspath(os.path.dirname(filename)))

#filepath.mydir(self.enText.get())


if __name__ == __main__:
guiFrame = GUIFrameWork()
guiFrame.mainloop()





User Input: NEWS.txt
Output: Not Found (though this file is actually present)




Regards,
Sudhir
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