将字符串转换为Double - VB

时间:2009-07-23 14:49:21

标签: vb.net string double try-catch

VB中是否有一种有效的方法来检查字符串是否可以转换为double?

我目前正在尝试将字符串转换为double,然后查看是否会引发异常。但这似乎减缓了我的申请。

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

5 个答案:

答案 0 :(得分:21)

如果您使用的是.NET 1.1 / 2.0 / 3.0 / 3.5 / 4.0 / 4.5,请尝试查看Double.TryParse()

答案 1 :(得分:19)

  

VB.NET示例代码

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

答案 2 :(得分:11)

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

答案 3 :(得分:3)

国际版本:

    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

结果:

        ' 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

答案 4 :(得分:0)

我简单地使用Eval(string)并将其评估为Double。