how to get the file with certain name in a directory which contains many files

I have to open a single file(to read data from it)  from a directory which hundreds of files which contains certain character like  "new" how to do that ?
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AndyAinscowFreelance programmer / ConsultantCommented:
Extract from the help files about Directory::GetFiles (with a minor modification)

using namespace System;
using namespace System::IO;
int main()
      // Only get files that begin with the letter "c."
      array<String^>^dirs = Directory::GetFiles( "c:\\Mydir\\MySubdir\\", "*new*.*" );  //Find all files with 'new' somewhere in the file name
      Console::WriteLine( "The number of files starting with c is {0}.", dirs->Length );
      Collections::IEnumerator^ myEnum = dirs->GetEnumerator();
      while ( myEnum->MoveNext() )
         Console::WriteLine( myEnum->Current );
   catch ( Exception^ e ) 
      Console::WriteLine( "The process failed: {0}", e );


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Note you need a path with double backslashes: c:\\Mydir\\MySubdir\\ because a single backslash instructs the next character is a special character.

Once you identify which file you want to read from then
FileStream^ fs = File::OpenRead( pathOfFileGoesHere );
can be used to open a filestream to read it - look at the OpenRead method in help for some example code

AndyAinscowFreelance programmer / ConsultantCommented:
  FileStream^ fs = File::OpenRead( pathOfIleGoesHere );
panthiAuthor Commented:
i need to open the file(from a directory with many files)  which contains the name new in the file name? can you please give code example?
no of files in it are 100 and one of the file name is
"te****_new.csv" how do i pick that up
thanks for your help
Please try:

String^ directory="c:\test\";
array<String^>^files = Directory::GetFiles( directory, "te*_new.csv" ); // or "te????_new.csv", if exactly 4 chars

if (files->Length > 0)
	FileStream^ fs = File::OpenRead( files[0] );

	// read the file here

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Here is how you can get all the files in a particular directory.   You would just need some logic on what to do with the name of the file (i.e. something to compare it to)
On Error Resume Next

Set objFSO = CreateObject("Scripting.FileSystemObject")
objStartFolder = "C:\test"

Set objFolder = objFSO.GetFolder(objStartFolder)

Set colFiles = objFolder.Files
For Each objFile in colFiles
	Wscript.Echo objFile.Name

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