Re: file path [modified] - C# Discussion Boards

AhsanS19-Aug-08 0:01

AhsanS

19-Aug-08 0:01

Right click it and resolve assembly name. Thats it.

or add reference to System.Reflection

Ahsan Ullah
Senior Software Engineer

bassjuh18-Aug-08 23:54

bassjuh

18-Aug-08 23:54

try
Environment.CurrentDirectory

arkiboys19-Aug-08 0:11

19-Aug-08 0:11

Environment.CurrentDirectory gives the following path:
"C:\\Work\\Developments\\SQLchecks\\SQLchecks\\SQLchecks\\bin\\Debug"
Please note the .xml file is sitting under:
C:\\Work\\Developments\\SQLchecks\\SQLchecks\\DataAccess\\text.xml
Any thoughts please?

sumit703419-Aug-08 0:22

sumit7034

19-Aug-08 0:22

Replace
SQLchecks\\bin\\Debug

with

SQLchecks\\DataAccess\\text.xml

using string functions to replace string

arkiboys19-Aug-08 1:25

19-Aug-08 1:25

Hi,
I now get this error:
Access to the path 'C:\Documents and Settings\jodowe\Local Settings\Apps\2.0\JXW1XBXW.7LE\414E3W43.NMO\sqlc..tion_3237500c67d360be_0001.0000_06a11946b45767e8' is denied.

bassjuh19-Aug-08 0:25

bassjuh

19-Aug-08 0:25

I guess the data directory is always inside the the project. Just choose to copy the file into the exe location with every build(with subpath dataaccess) in the designer. Would be reallife behaviour.

sumit703419-Aug-08 0:30

sumit7034

19-Aug-08 0:30

you can replace like this

string str = "C:\\Work\\Developments\\SQLchecks\\SQLchecks\\SQLchecks\\bin\\Debug";

str=str.Substring(0,str.IndexOf("\\SQLchecks\\bin\\Debug"));
str = str + "\\DataAccess\\text.xml";

Re: file path [modified]

Csharp_Raja19-Aug-08 1:17

19-Aug-08 1:17

This following code will help you to access the file which is added to the project

Add the file into the project and then set the property as embedded file then you will be able to access with the following code. or you can add the file in the resource file and you will be able to access the file.

Assembly embeddedFile = Assembly.GetExecutingAssembly();
Stream StreamEmbeddedFile;
StreamEmbeddedFile = embeddedFile.GetManifestResourceStream ("NameSpace.Test.xml");
XmlDocument myStylesheet = new XmlDocument();
myStylesheet.Load(StreamEmbeddedFile);

Here, "NameSpace" is default namespace.

if you have any other problem on this pls let me know

Thanks
Raja.S

modified on Tuesday, August 19, 2008 7:25 AM

arkiboys19-Aug-08 1:33

19-Aug-08 1:33

Hi Raja.S,
This code does not seem to work as I need to run it from the class inside a class library project. i.e. Assembly is not recognised in intellisense

Csharp_Raja19-Aug-08 2:16

19-Aug-08 2:16

You have to include "using System.Reflection;" it is available in Class Library as well. you mean to say the color of the code is not changed in the IDE that means you haven’t included the name space.

This is getting the running assembly if it is class library then class library(dll) will be loaded to the running process on the demand , when it executes the code from class library the running assembly will be the class library(dll) .

Thanks
Raja.S

arkiboys19-Aug-08 2:28

19-Aug-08 2:28

Correction to my previous post.

I have created a windows application.
There are forms inside the windows form application project. There is a class library project too.

The .sln is in the following path:
C:\Work\Developments\SQLchecks\SQLchecks

The windows forms are inside the windows application project as below:
C:\Work\Developments\SQLchecks\SQLchecks\SQLchecks

There is a class library project called DataAccess which has classes and an xml file called Settings.xml :
C:\Work\Developments\SQLchecks\SQLchecks\DataAccess

Question:
from within a class inside the DataAccess project which is a class library project, how do I refer to the Settings.xml file which is also in the same directory?

Thanks

Csharp_Raja19-Aug-08 2:54

19-Aug-08 2:54

Got that !! your class library and the setting.xml in the same folder you want to access that setting.xml from the class library

simple:

using System.Reflection;

string path = Assembly.GetExecutingAssembly().Location;
path = path + "setting.xml";

let me know the result.

Thanks
Raja.S

arkiboys19-Aug-08 3:02

19-Aug-08 3:02

This seems to be IN-correct because it returns:
"C:\\Work\\Developments\\SQLchecks\\SQLchecks\\SQLchecks\\bin\\Debug\\DataAccess.dllSettings.xml"
Please note the Settings.xml file is inside C:\\Work\\Developments\\SQLchecks\\SQLchecks\\DataAccess
Where DataAccess is a classLibrary Project which sits inside a dataAccess folder.

I then modified the above code and placed "\Settings.xml"

The new error is:
Could not find a part of the path 'C:\\Work\\Developments\\SQLchecks\\SQLchecks\\SQLchecks\\bin\\Debug\\DataAccess.dll\\Settings.xml'."

Any thoughts please?

Csharp_Raja19-Aug-08 3:08

19-Aug-08 3:08

Sorry it should be

path = path + "\\setting.xml";

this should work

Thanks
Raja.S

arkiboys19-Aug-08 3:11

19-Aug-08 3:11

The new error is:
Could not find a part of the path 'C:\\Work\\Developments\\SQLchecks\\SQLchecks\\SQLchecks\\bin\\Debug\\DataAccess.dll\\Settings.xml'."

Thanks

Csharp_Raja19-Aug-08 3:14

19-Aug-08 3:14

Have checked is that setting.xml is there in that path ?

can i get your hotmail id if possible we can chat ?

Thanks
Raja.S

Csharp_Raja19-Aug-08 3:17

19-Aug-08 3:17

hey man your getting the path with the assembly name discard assembly name from the path and put the setting.xml every thing should work fine

it should be

'C:\\Work\\Developments\\SQLchecks\\SQLchecks\\SQLchecks\\bin\\Debug\\Settings.xml'."

Thanks
Raja.S

arkiboys19-Aug-08 3:19

19-Aug-08 3:19

Hi,
These files are present in:
C:\Work\Developments\SQLchecks\SQLchecks\DataAccess\bin\Debug

DataAccess.dll
Settings.xml
...
...

Csharp_Raja19-Aug-08 3:24

19-Aug-08 3:24

Find the code here

// get current path for the assembly

string path = Assembly.GetExecutingAssembly().Location;
path = Path.GetFullPath(Path.Combine(Path.GetDirectoryName(path),".."));
path = path + @"\setting.xml";

Thanks
Raja.S

Csharp_Raja19-Aug-08 3:30

19-Aug-08 3:30

final code here

using System.IO;
using System.Reflection;

// get current path for the assembly

string path = Assembly.GetExecutingAssembly().Location;
path = Path.GetFullPath(Path.Combine(Path.GetDirectoryName(path),".."));
path = path + @"\setting.xml";

pls let me know the result

Thanks
Raja.S

arkiboys19-Aug-08 3:33

19-Aug-08 3:33

Hi,
Error is:
Message: "Value cannot be null.\r\nParameter name: inputUri"

Csharp_Raja19-Aug-08 4:53

19-Aug-08 4:53

this excemption seems to be some problem in your application code , code which i have provided you, i have tested, it is getting the running assembly path and then with .. its discaring the assembly name and get the setting.xml file.

Thanks
Raja.S

arkiboys19-Aug-08 20:59

19-Aug-08 20:59

Hi,
Using your code the path is:
"C:\\Work\\Developments\\SQLchecks\\SQLchecks\\SQLchecks\\bin\\Settings.xml"

But after publishing, the Settings.xml is sitting inside the debug so somehow I need to access:
"C:\\Work\\Developments\\SQLchecks\\SQLchecks\\SQLchecks\\bin\\Debug\\Settings.xml"
So I used the following code:
path = AppDomain.CurrentDomain.BaseDirectory + "Settings.xml";
This seems to be working fine on my Development machine but after publishing it to users machines then this does not seem to get the Settings.xml file.

Not sure what I am doing wrong?
Thanks

chaiguy133719-Aug-08 16:02

chaiguy1337

19-Aug-08 16:02

Here's a chunk of code that might help you with this. If you set "test.xml" as an "embedded resource", rather than "content", then it will actually reside inside the compiled library file, and you can access it like this:

using ( Stream prototype = System.Reflection.Assembly.GetExecutingAssembly().GetManifestResourceStream( "Namespace.test.xml" ) ) {
	byte[] fileData = new byte[ prototype.Length ];
	prototype.Read( fileData, 0, (int)prototype.Length );
	File.WriteAllBytes( filePath, fileData );
}

Be sure to use the namespace when identifying the file name. I think there are other overloads of this function you can use as well.

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arkiboys19-Aug-08 21:23