string 将字符串转换为双精度 - VB
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Convert String to Double - VB
提问by mscccc
Is there an efficient method in VB to check if a string can be converted to a double?
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？
I'm currently doing this by trying to convert the string to a double and then seeing if it throws an exception. But this seems to be slowing down my application.
我目前正在通过尝试将字符串转换为双精度然后查看它是否引发异常来执行此操作。但这似乎减慢了我的申请速度。
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
回答by Kane
Try looking at Double.TryParse() if you are using .NET 1.1/2.0/3.0/3.5/4.0/4.5
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()
回答by Kane
VB.NET Sample Code
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
回答by Patrick McDonald
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
回答by SZL
The international versions:
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
Results:
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
回答by gcrav
I simple used Eval(string)and it evaluated as Double.
我简单地使用过Eval(string)，它被评估为 Double。
