Convertire String Double - VB
Domanda
Esiste un metodo efficiente in VB per verificare se una stringa può essere convertito in un doppio?
Al momento sto facendo questo, cercando di convertire la stringa in un doppio e poi vedere se viene generata un'eccezione. Ma questo sembra rallentare la mia domanda.
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
Soluzione
Prova a guardare Double.TryParse () se si utilizza .NET 1.1 / 2.0 / 3.0 / 3.5 / 4.0 / 4.5
Altri suggerimenti
VB.NET Codice di esempio
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
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
Le versioni internazionali:
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
Risultati:
' 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
I semplice Eval(string)
utilizzato e valutato come doppio.
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