vb.net - significa - funcion cint visual basic

Convertir cadena a doble-VB

Question

¿Existe un método eficiente en VB para verificar si una cadena se puede convertir a doble?

Actualmente estoy haciendo esto al tratar de convertir la cadena a un doble y luego ver si se produce una excepción. Pero esto parece estar desacelerando mi aplicación.

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

Código de ejemplo 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

Answer 2

Intente buscar en Double.TryParse () si está utilizando .NET 1.1 / 2.0 / 3.0 / 3.5 / 4.0 / 4.5

Answer 3

Las versiones internacionales:

    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

Resultados:

        '' 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 4

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