web-dev-qa-db-fra.com

Convertir une chaîne en double - VB

Existe-t-il une méthode efficace dans VB pour vérifier si une chaîne peut être convertie en double?

Je fais actuellement cela en essayant de convertir la chaîne en double, puis de voir si elle lève une exception. Mais cela semble ralentir ma candidature.

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
vb.netstringdoubletry-catch
21
mscccc

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

23
Kane

Exemple de code VB.NET

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
20
Pooky
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
11
Patrick McDonald

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
3
SZL

J'ai simplement utilisé Eval(string) et il a été évalué comme Double.

0
gcrav