blocks|key|1496505|text|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()|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1496506|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Try looking at Double.TryParse() if you are using .NET 1.1/2.0/3.0/3.5/4.0/4.5

blocks|key|1496526|text|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|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1496527|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>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
</code></pre>

blocks|key|1496541|text|国际版本：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1496542|++++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|code-block|syntax|javascript|1496543|结果：|1496544|++++++++'+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|1496545|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

The international versions:

<pre><code> 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
</code></pre>

Results:

<pre><code> ' 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
</code></pre>

blocks|key|1281811|text|我简单地使用了Eval(string)，它的计算结果是双精度的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1281812|entityMap^0|7|C|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

I simple used <code>Eval(string)</code> and it evaluated as Double.

Is there an efficient method in VB to check if a string can be converted to a 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.

<pre><code>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
</code></pre>

Convert String to Double - VB

在VB中有没有一种有效的方法来检查字符串是否可以转换为双精度？我目前正在尝试将字符串转换为双精度值，然后查看它是否会抛出异常。但这似乎减慢了我的应用程序。Try    ' if number then format it.    current = CDbl(x)    current = Math.Round(cur...

问将字符串转换为Double - VB
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