Open a file that has a dynamically generated file name with date

Question

0.00/5 (No votes)

See more:

All,
How would I open a dynamically generated file?
For instance here is code:

StreamReader reader = File.OpenText(@"C:\Users\Iman\Downloads\tandem-win32-10-12-01-1\tandem-win32-10-12-01-1\bin\output.xml");

but the output.xml is created every time with a previous process with date and time like: output.2011_07_06_23_34_53.t

Posted 9-Jul-11 9:44am

shekarchee

Updated 9-Jul-11 16:43pm

thatraja

v2

Add a Solution

Comments

Emrah Taylan 9-Jul-11 16:14pm

Do you want to read file same second?

2 solutions

Solution 2

Prefer using special format for string representation of time called "sortable".

C#

string time = System.DateTime.Now.ToString("s").Replace(':', '-');
// it will be formatted like 2011-06-15T13-45-30
// ':' should be removed to comply with
// the requirements for file system names

// just for example:
string fileName = string.Format("output.{0}.xml", time);

The format "s" is specially designed for such things. Is you sort your files by alphanumeric order, it will be the same as the order by time.

See:
http://msdn.microsoft.com/en-us/library/1k1skd40.aspx[^],
http://msdn.microsoft.com/en-us/library/az4se3k1.aspx[^].

—SA

Posted 9-Jul-11 21:44pm

Sergey Alexandrovich Kryukov

Updated 10-Jul-11 1:33am

v3

Comments

Espen Harlinn 10-Jul-11 7:34am

Good point, my 5

Sergey Alexandrovich Kryukov 10-Jul-11 13:34pm

Thank you, Espen.
--SA

Sergey Alexandrovich Kryukov 12-Jul-11 16:25pm

Thank you for fixing my text (I did not see where, I think I can trust you :-).
--SA

Espen Harlinn 12-Jul-11 16:51pm

I just removed a single character :)

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Dr.Walt Fair, PE · Accepted Answer · 2011-07-09T10:34:00

Solution 1

I would probably use something like

DirectoryInfo dir_info = new DirectoryInfo(my_file_path);
FileSystemInfo[] my_files = dir_info.GetFileSystemInfo("output.*");

to put the file information in an array, then use Array.Sort() to find the newest file and go from there.

Posted 9-Jul-11 10:34am

Dr.Walt Fair, PE

Comments

Espen Harlinn 10-Jul-11 7:33am

That would work, nice and simple - my 5