Converter String para Duplo - VB

Question

Existe um método eficiente em VB para verificar se uma string pode ser convertido para um casal?

Atualmente estou fazendo isso tentando converter a string para um casal e, em seguida, ver se ele lança uma exceção. Mas isso parece estar a abrandar o meu pedido.

Try
    ' if number then format it.
    current = CDbl(x)
    current = Math.Round(current, d)
    Return current
Catch ex As System.InvalidCastException
    ' item is not a number, do not format... leave as a string
    Return x
End Try

Answer 1

Tente olhar para Double.TryParse () se você estiver usando o .NET 1,1 / 2,0 / 3,0 / 3,5 / 4,0 / 4,5

Answer 2

VB.NET Código de exemplo

Dim A as String = "5.3"
Dim B as Double

B = CDbl(Val(A)) '// Val do hard work

'// Get output 
MsgBox (B) '// Output is 5,3 Without Val result is 53.0

Answer 3

Dim text As String = "123.45"
Dim value As Double
If Double.TryParse(text, value) Then
    ' text is convertible to Double, and value contains the Double value now
Else
    ' Cannot convert text to Double
End If

Answer 4

As versões internacionais:

    Public Shared Function GetDouble(ByVal doublestring As String) As Double
        Dim retval As Double
        Dim sep As String = CultureInfo.CurrentCulture.NumberFormat.NumberDecimalSeparator

        Double.TryParse(Replace(Replace(doublestring, ".", sep), ",", sep), retval)
        Return retval
    End Function

    ' NULLABLE VERSION:
    Public Shared Function GetDoubleNullable(ByVal doublestring As String) As Double?
        Dim retval As Double
        Dim sep As String = CultureInfo.CurrentCulture.NumberFormat.NumberDecimalSeparator

        If Double.TryParse(Replace(Replace(doublestring, ".", sep), ",", sep), retval) Then
            Return retval
        Else
            Return Nothing
        End If
    End Function

Resultado:

        ' HUNGARIAN REGIONAL SETTINGS (NumberDecimalSeparator: ,)

        ' Clean Double.TryParse
        ' -------------------------------------------------
        Double.TryParse("1.12", d1)     ' Type: DOUBLE     Value: d1 = 0.0
        Double.TryParse("1,12", d2)     ' Type: DOUBLE     Value: d2 = 1.12
        Double.TryParse("abcd", d3)     ' Type: DOUBLE     Value: d3 = 0.0

        ' GetDouble() method
        ' -------------------------------------------------
        d1 = GetDouble("1.12")          ' Type: DOUBLE     Value: d1 = 1.12
        d2 = GetDouble("1,12")          ' Type: DOUBLE     Value: d2 = 1.12
        d3 = GetDouble("abcd")          ' Type: DOUBLE     Value: d3 = 0.0

        ' Nullable version - GetDoubleNullable() method
        ' -------------------------------------------------
        d1n = GetDoubleNullable("1.12") ' Type: DOUBLE?    Value: d1n = 1.12
        d2n = GetDoubleNullable("1,12") ' Type: DOUBLE?    Value: d2n = 1.12
        d3n = GetDoubleNullable("abcd") ' Type: DOUBLE?    Value: d3n = Nothing

Answer 5

I simples Eval(string) usado e avaliado como duplo.