Convertir Chaîne à Double - VB

Question

Y at-il une méthode efficace en VB pour vérifier si une chaîne peut être converti en un double?

Je suis en train de le faire en essayant de convertir la chaîne en un double, puis voir si elle lance une exception. Mais cela semble ralentir ma demande.

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

Essayez de regarder Double.TryParse () si vous utilisez .NET 1.1 / 2.0 / 3.0 / 3.5 / 4.0 / 4.5

Answer 2

  

VB.NET Exemple de code

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

Les versions internationales:

    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

Résultats:

        ' 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 simple, Eval(string) utilisé et évalué comme Double.