excel-vba
Tablice

Wypełnianie tablic (dodawanie wartości)

Istnieje wiele sposobów zapełniania tablicy.

Bezpośrednio

'one-dimensional
Dim arrayDirect1D(2) As String
arrayDirect(0) = "A"
arrayDirect(1) = "B"
arrayDirect(2) = "C"

'multi-dimensional (in this case 3D)
Dim arrayDirectMulti(1, 1, 2)
arrayDirectMulti(0, 0, 0) = "A"
arrayDirectMulti(0, 0, 1) = "B"
arrayDirectMulti(0, 0, 2) = "C"
arrayDirectMulti(0, 1, 0) = "D"
'...

Korzystanie z funkcji Array ()

'one-dimensional only
Dim array1D As Variant 'has to be type variant
array1D = Array(1, 2, "A")
'-> array1D(0) = 1, array1D(1) = 2, array1D(2) = "A"

Z zakresu

Dim arrayRange As Variant 'has to be type variant
    
'putting ranges in an array always creates a 2D array (even if only 1 row or column)
'starting at 1 and not 0, first dimension is the row and the second the column
arrayRange = Range("A1:C10").Value
'-> arrayRange(1,1) = value in A1
'-> arrayRange(1,2) = value in B1
'-> arrayRange(5,3) = value in C5
'...
    
'Yoo can get an one-dimensional array from a range (row or column)
'by using the worksheet functions index and transpose:

'one row from range into 1D-Array:
arrayRange = Application.WorksheetFunction.Index(Range("A1:C10").Value, 3, 0)
'-> row 3 of range into 1D-Array
'-> arrayRange(1) = value in A3, arrayRange(2) = value in B3, arrayRange(3) = value in C3

'one column into 1D-Array:
'limited to 65536 rows in the column, reason: limit of .Transpose
arrayRange = Application.WorksheetFunction.Index( _
Application.WorksheetFunction.Transpose(Range("A1:C10").Value), 2, 0)
'-> column 2 of range into 1D-Array
'-> arrayRange(1) = value in B1, arrayRange(2) = value in B2, arrayRange(3) = value in B3
'...

'By using Evaluate() - shorthand [] - you can transfer the
'range to an array and change the values at the same time.
'This is equivalent to an array formula in the sheet:
arrayRange = [(A1:C10*3)]
arrayRange = [(A1:C10&"_test")]
arrayRange = [(A1:B10*C1:C10)]
'...

2D with Evaluate ()

Dim array2D As Variant
'[] ist a shorthand for evaluate()
'Arrays defined with evaluate start at 1 not 0
array2D = [{"1A","1B","1C";"2A","2B","3B"}]
'-> array2D(1,1) = "1A", array2D(1,2) = "1B", array2D(2,1) = "2A" ...

'if you want to use a string to fill the 2D-Array:
Dim strValues As String
strValues = "{""1A"",""1B"",""1C"";""2A"",""2B"",""2C""}"
array2D = Evaluate(strValues)

Korzystanie z funkcji Split ()

Dim arraySplit As Variant 'has to be type variant
arraySplit = Split("a,b,c", ",")
'-> arraySplit(0) = "a", arraySplit(1) = "b", arraySplit(2) = "c"

Tablice dynamiczne (zmiana rozmiaru i obsługa dynamiczna)

Ze względu na to, że nie jest zawartością wyłączną Excel-VBA, ten przykład został przeniesiony do dokumentacji VBA.

Link: Tablice dynamiczne (zmiana rozmiaru i obsługa dynamiczna)

Jagged Arrays (Arrays of Arrays)

Ze względu na to, że nie jest zawartością wyłączną Excel-VBA, ten przykład został przeniesiony do dokumentacji VBA.

Link: Jagged Arrays (Arrays of Arrays)

Sprawdź, czy tablica jest zainicjowana (jeśli zawiera elementy, czy nie).

Częstym problemem może być próba iteracji w tablicy, która nie zawiera żadnych wartości. Na przykład:

Dim myArray() As Integer
For i = 0 To UBound(myArray) 'Will result in a "Subscript Out of Range" error

Aby uniknąć tego problemu i sprawdzić, czy tablica zawiera elementy, użyj tego narzędzia oneliner :

If Not Not myArray Then MsgBox UBound(myArray) Else MsgBox "myArray not initialised"

Tablice dynamiczne [Deklaracja tablicy, zmiana rozmiaru]

Sub Array_clarity()

Dim arr() As Variant  'creates an empty array
Dim x As Long
Dim y As Long

x = Range("A1", Range("A1").End(xlDown)).Cells.Count
y = Range("A1", Range("A1").End(xlToRight)).Cells.Count

ReDim arr(0 To x, 0 To y) 'fixing the size of the array

For x = LBound(arr, 1) To UBound(arr, 1)
    For y = LBound(arr, 2) To UBound(arr, 2)
        arr(x, y) = Range("A1").Offset(x, y) 'storing the value of Range("A1:E10") from activesheet in x and y variables
    Next
Next

'Put it on the same sheet according to the declaration:
Range("A14").Resize(UBound(arr, 1), UBound(arr, 2)).Value = arr

End Sub