在 VB.NET 中将字符串转换为整数的 TryCast
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TryCast for string to integer in VB.NET
提问by James Yeoman
I do not understand at all how to use TryCast in my code, but it is something I need to use for validating user input. I have done various searches and looked at various questions on here, but no one seems to actually say how to use it, and the MSDN website doesn't seem to help at all.
我完全不明白如何在我的代码中使用 TryCast，但我需要使用它来验证用户输入。我已经在这里进行了各种搜索并查看了各种问题，但似乎没有人真正说明如何使用它，而且 MSDN 网站似乎根本没有帮助。
Function ValidateInput(Var_In As String) As Integer
If TryCast(Var_In, Integer) = Nothing Then
Return vbNull
Else
Return Var_In
End If
End Function
The error says that
错误说
The operand must be of reference type but Integer is of value type
操作数必须是引用类型，但整数是值类型
What is the explanation of what I have done wrong?
我做错了什么的解释是什么？
TryParse doesn't accept more than 10 digits so for example, an input of "12345678901" won't be accepted. How do I fix this?
TryParse 不接受超过 10 位的数字，例如，“12345678901”的输入将不被接受。我该如何解决？
回答by Martin Verjans
Let's try to understand the differences between TryCast, Convert and TryParse.
让我们尝试了解TryCast、Convert 和 TryParse之间的区别。
TryCast
试播
This function will attempt to convert one object into another type, as long as it is a reference type.
此函数将尝试将一个对象转换为另一种类型，只要它是引用类型。
Dim MyNewObject = TryCast(MyObject, MyReferenceClass)
If IsNothing(MyNewObject) Then
MessageBox.Show("Impossible to cast")
End If
Since Integer is a value type, it will not work, so we have to figure something out...
由于Integer是值类型，所以行不通，所以我们得想办法……
Convert
转变
From MSDN:
来自 MSDN：
Converts a base data type to another base data type.
将基本数据类型转换为另一种基本数据类型。
So we can try:
所以我们可以试试：
Dim myInt = Convert.ToInt32(MyObject)
The problem is that it will generate an exception InvalidCastExceptionif it's impossible to do the conversion.
问题在于，InvalidCastException如果无法进行转换，则会产生异常。
TryParse
尝试解析
This function is trying to convert a Stringinto something you want. And it will not generate an exception:
这个函数试图将 a 转换String成你想要的东西。它不会产生异常：
Dim myInt As Integer = 0
If Not Integer.TryParse(MyString, myInt) Then
MessageBox.show("This is not an integer")
End If
Limitation
局限性
Converting a String into a Integer can sometimes be tricky... If the String represents a number that is greater or lesser than Integer.MaxValueand Integer.MinValue, you will end up with no conversion...
将字符串转换为整数有时会很棘手......如果字符串表示一个大于或小于Integer.MaxValueand 的数字Integer.MinValue，你最终将无法进行转换......
So you can go with a Double:
所以你可以使用Double：
Double.TryParse(MyString, MyDouble)
Or personally, if you know that it will be a number, use Decimal:
或者就个人而言，如果您知道它将是一个数字，请使用十进制：
Decimal.TryParse(MyString, MyDecimal)
Decimal still has a Max and Min value, according to MSDN:
根据 MSDN，十进制仍然有最大值和最小值：
The Decimal value type represents decimal numbers ranging from positive 79,228,162,514,264,337,593,543,950,335 to negative 79,228,162,514,264,337,593,543,950,335. The Decimal value type is appropriate for financial calculations that require large numbers of significant integral and fractional digits and no round-off errors.
Decimal 值类型表示从正 79,228,162,514,264,337,593,543,950,335 到负 79,228,162,514,264,337,593,543,950,335 的十进制数。Decimal 值类型适用于需要大量有效整数和小数位且没有舍入错误的财务计算。
Convert.ChangeType
转换.ChangeType
This oneis also interesting, but is a bit weird...
这个也很有意思，就是有点奇怪……
回答by David W
You are attempting to perform TryCastagainst an Integer, which is a value type. TryCastworks only on referencetypes, such as (but not limited to) a Class, Object, or Stringtype.
您正在尝试TryCast对 Integer执行，这是一个值类型。TryCast只对参考类型，诸如（但不限于）一个Class，Object或String类型。
If you are trying to convert the input parameter to an Integer, you might try one of the methods in the Convertclass, such as Convert.ToInt32()or Integer.TryParse.
如果您尝试将输入参数转换为整数，您可以尝试Convert类中的方法之一，例如Convert.ToInt32()或Integer.TryParse。
回答by Dr. Aaron Dishno
Instead of TryCast, use TryParse:
使用 TryParse 代替 TryCast：
Function ValidateInput(Var_In As String) As Integer
Dim iNum As Integer
If (Integer.TryParse(Var_In, iNum)) Then
Return iNum
Else
Return vbNull
End If
End Function
回答by Claudius
Much better is to use TryParse:
更好的是使用 TryParse：
Function ValidateInput(Var_In As String) As Integer
Dim num as Integer
If Not Integer.TryParse(Var_In, num) Then
Return vbNull
Else
Return num
End If
End Function
