Using Resources With FileOpen

Question

4.00/5 (1 vote)

See more:

Hi,

I have some .txt files in the resources of my vb.net project. I am trying to use the command:

VB

FileOpen(1, "FILE NAME", OpenMode.Input)

How do I use my resource location as the file name?

Thanks

Posted 26-Nov-11 3:01am

RoomTen

Add a Solution

2 solutions

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)

**Manoj K Bhoir** · Answer 1 · 2011-11-26T03:24:00

Solution 1

You place your .Txt file in Debug folder of your Project insted of Resourses.Then you can get this path dynamically using following code :

VB

Dim Path As String = My.Application.Info.DirectoryPath
'It returns the Path of the Debug folder where your Exe file created.

Code to get Resource folder path :

VB

Dim Path As String = My.Application.Info.DirectoryPath
Path = Path.Substring(0, Path.LastIndexOf("\"))
Path = Paht.Substring(0, Path.LastIndexOf("\")) & "\Resources"
MessageBox.Show(Path)

I hope it will help you.

Posted 26-Nov-11 3:24am

Manoj K Bhoir

Updated 26-Nov-11 3:40am

v2

Comments

RoomTen 26-Nov-11 9:33am

I would like it to still be able to work self standing of the debug folder for when I build it into a .exe . Do I need to use something else?

Sergey Alexandrovich Kryukov 26-Nov-11 22:38pm

Forget it. You don't need anything like that if you use resource. Please see my solution.
--SA

Manoj K Bhoir 26-Nov-11 9:40am

Check out the updated solution!

Manoj K Bhoir 26-Nov-11 9:42am

I think Path is not required because all resources are added into the .exe while building the project.

Sergey Alexandrovich Kryukov · Answer 2 · 2011-11-26T16:37:00

If you put your text file in the .resx resource, you don't need anything to read it. As soon as you create a resource, an auto-generated source file is created. Look at this file: it has a static class (module) and a static ("Shared") property in it; the name of this property will be the same or close to the name of your original file. Use this property directly in your code: its value will be equal to the content of the source text file.

You can use this rule for many other cases. Suppose you have a PNG file and added it to some resource file. The property in the auto-generated file will be of the type Bitmap; and you can use it directly. You won't need to read resource stream or anything like that.

—SA