String.Join Method (String, array<String>[]()[], Int32, Int32)
[ This article is for Windows Phone 8 developers. If you’re developing for Windows 10, see the latest documentation. ]
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.
Namespace: System
Assembly: mscorlib (in mscorlib.dll)
Syntax
Public Shared Function Join ( _
separator As String, _
value As String(), _
startIndex As Integer, _
count As Integer _
) As String
public static string Join(
string separator,
string[] value,
int startIndex,
int count
)
Parameters
- separator
Type: System..::.String
The string to use as a separator.
- value
Type: array<System..::.String>[]()[]
An array that contains the elements to concatenate.
- startIndex
Type: System..::.Int32
The first element in value to use.
- count
Type: System..::.Int32
The number of elements in value to use.
Return Value
Type: System..::.String
A string that consists of the strings in value delimited by the separator string.
-or-
String..::.Empty if count is zero, value has no elements, or separator and all the elements of value are String..::.Empty.
Exceptions
| Exception | Condition |
|---|---|
| ArgumentNullException | value is nullNothingnullptra null reference (Nothing in Visual Basic). |
| ArgumentOutOfRangeException | startIndex or count is less than 0. -or- startIndex plus count is greater than the number of elements in value. |
| OutOfMemoryException | Out of memory. |
Remarks
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. If any element in value is nullNothingnullptra null reference (Nothing in Visual Basic), an empty string is used instead.
Examples
The following code example concatenates two elements from an array of names of fruit.
' Sample for String.Join(String, String[], int int)
_
Class Example
Public Shared Sub Demo(ByVal outputBlock As System.Windows.Controls.TextBlock)
Dim val As [String]() = {"apple", "orange", "grape", "pear"}
Dim sep As [String] = ", "
Dim result As [String]
outputBlock.Text += String.Format("sep = '{0}'", sep) & vbCrLf
outputBlock.Text += String.Format("val() = {{'{0}' '{1}' '{2}' '{3}'}}", val(0), val(1), val(2), val(3)) & vbCrLf
result = [String].Join(sep, val, 1, 2)
outputBlock.Text += String.Format("String.Join(sep, val, 1, 2) = '{0}'", result) & vbCrLf
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 Example
{
public static void Demo(System.Windows.Controls.TextBlock outputBlock)
{
String[] val = { "apple", "orange", "grape", "pear" };
String sep = ", ";
String result;
outputBlock.Text += String.Format("sep = '{0}'", sep) + "\n";
outputBlock.Text += String.Format("val[] = {{'{0}' '{1}' '{2}' '{3}'}}", val[0], val[1], val[2], val[3]) + "\n";
result = String.Join(sep, val, 1, 2);
outputBlock.Text += String.Format("String.Join(sep, val, 1, 2) = '{0}'", result) + "\n";
}
}
/*
This example produces the following results:
sep = ', '
val[] = {'apple' 'orange' 'grape' 'pear'}
String.Join(sep, val, 1, 2) = 'orange, grape'
*/
Version Information
Windows Phone OS
Supported in: 8.1, 8.0, 7.1, 7.0
Platforms
Windows Phone