Concatenates a specified separator String between each element of a specified String array, yielding a single concatenated string. Parameters specify the first array element and number of elements to use.
(in mscorlib.dll)
Visual Basic (Declaration)
Public Shared Function Join ( _
separator As String, _
value As String(), _
startIndex As Integer, _
count As Integer _
) As String
Dim separator As String
Dim value As String()
Dim startIndex As Integer
Dim count As Integer
Dim returnValue As String
returnValue = String.Join(separator, _
value, startIndex, count)
public static string Join(
string separator,
string[] value,
int startIndex,
int count
)
public:
static String^ Join(
String^ separator,
array<String^>^ value,
int startIndex,
int count
)
public static function Join(
separator : String,
value : String[],
startIndex : int,
count : int
) : String
Return Value
Type:
System..::.StringA String object consisting of the strings in value joined by separator. Or, Empty if count is zero, value has no elements, or separator and all the elements of value are Empty.
For example if separator is ", " and the elements of value are "apple", "orange", "grape", and "pear", Join(separator, value, 1, 2) returns "orange, grape".
If separator is nullNothingnullptra null reference (Nothing in Visual Basic), the empty string (Empty) is used instead.
The following example concatenates two elements from an array of names of fruit.
' Sample for String.Join(String, String[], int int)
Imports System
_
Class Sample
Public Shared Sub Main()
Dim val As [String]() = {"apple", "orange", "grape", "pear"}
Dim sep As [String] = ", "
Dim result As [String]
Console.WriteLine("sep = '{0}'", sep)
Console.WriteLine("val() = {{'{0}' '{1}' '{2}' '{3}'}}", val(0), val(1), val(2), val(3))
result = [String].Join(sep, val, 1, 2)
Console.WriteLine("String.Join(sep, val, 1, 2) = '{0}'", result)
End Sub 'Main
End Class 'Sample
'
'This example produces the following results:
'sep = ', '
'val() = {'apple' 'orange' 'grape' 'pear'}
'String.Join(sep, val, 1, 2) = 'orange, grape'
'
// Sample for String.Join(String, String[], int int)
using System;
class Sample {
public static void Main() {
String[] val = {"apple", "orange", "grape", "pear"};
String sep = ", ";
String result;
Console.WriteLine("sep = '{0}'", sep);
Console.WriteLine("val[] = {{'{0}' '{1}' '{2}' '{3}'}}", val[0], val[1], val[2], val[3]);
result = String.Join(sep, val, 1, 2);
Console.WriteLine("String.Join(sep, val, 1, 2) = '{0}'", result);
}
}
/*
This example produces the following results:
sep = ', '
val[] = {'apple' 'orange' 'grape' 'pear'}
String.Join(sep, val, 1, 2) = 'orange, grape'
*/
// Sample for String::Join(String, String[], int int)
using namespace System;
int main()
{
array<String^>^val = {"apple","orange","grape","pear"};
String^ sep = ", ";
String^ result;
Console::WriteLine( "sep = '{0}'", sep );
Console::WriteLine( "val[] = {{'{0}' '{1}' '{2}' '{3}'}}", val[ 0 ], val[ 1 ], val[ 2 ], val[ 3 ] );
result = String::Join( sep, val, 1, 2 );
Console::WriteLine( "String::Join(sep, val, 1, 2) = '{0}'", result );
}
/*
This example produces the following results:
sep = ', '
val[] = {'apple' 'orange' 'grape' 'pear'}
String::Join(sep, val, 1, 2) = 'orange, grape'
*/
Windows 7, Windows Vista, Windows XP SP2, Windows XP Media Center Edition, Windows XP Professional x64 Edition, Windows XP Starter Edition, Windows Server 2008 R2, Windows Server 2008, Windows Server 2003, Windows Server 2000 SP4, Windows Millennium Edition, Windows 98, Windows CE, Windows Mobile for Smartphone, Windows Mobile for Pocket PC, Xbox 360, Zune
The .NET Framework and .NET Compact Framework do not support all versions of every platform. For a list of the supported versions, see .NET Framework System Requirements.
.NET Framework
Supported in: 3.5, 3.0, 2.0, 1.1, 1.0
.NET Compact Framework
Supported in: 3.5, 2.0, 1.0
XNA Framework
Supported in: 3.0, 2.0, 1.0
Reference