Vb.net 仅从带整数的字符串中获取整数

Question

我有这个字符串123abc123我怎样才能从这个字符串中只得到整数？

例如，将123abc123转换为123123 。

我尝试了什么：

Integer.Parse(abc)

Answer 1

您可以使用Char.IsDigit

Dim str = "123abc123"
Dim onlyDigits = New String(str.Where(Function(c) Char.IsDigit(c)).ToArray())
Dim num = Int32.Parse(onlyDigits)

Answer 2

提取整数的正确方法是使用isNumbric函数：

Dim str As String = "123abc123"
Dim Res As String
For Each c As Char In str
    If IsNumeric(c) Then
        Res = Res & c
    End If
Next
MessageBox.Show(Res)

其他方式：

Private Shared Function GetIntOnly(ByVal value As String) As Integer
    Dim returnVal As String = String.Empty
    Dim collection As MatchCollection = Regex.Matches(value, "\d+")
    For Each m As Match In collection
        returnVal += m.ToString()
    Next
    Return Convert.ToInt32(returnVal)
End Function

Answer 3

    Dim input As String = "123abc456"
    Dim reg As New Regex("[^0-9]")
    input = reg.Replace(input, "")
    Dim output As Integer
    Integer.TryParse(input, output)

Answer 4

您可以使用带有\\D模式的正则表达式来匹配非数字字符并删除它们，然后解析其余的字符串：

Dim input As String = "123abc123"

Dim n As Integer = Int32.Parse(Regex.Replace(input, "\D", ""))

Answer 5

您还可以使用FindAll提取所需的内容。 我们还应该考虑使用Val函数来处理空字符串。

    Dim str As String = "123abc123"
    Dim i As Integer = Integer.Parse(Val(New String(Array.FindAll(str.ToArray, Function(c) "0123456789".Contains(c)))))

Answer 6

在这种情况下，使用 Val() function of vb.net. https://www.hscripts.com/tutorials/v.net/val-function.html 。

 Dim str As String = "123abc123"
 Dim result As Integer =Val(str)

Answer 7

您还可以使用以下代码：

Public Class Form1

Private Sub Form1_OnLoad() Handles Me.Load
    Dim str1 As String = "2,147,4abc83,648" '"123abc123"
    FindingNumbers(str1)
End Sub

Private Sub FindingNumbers(ByRef str1 As String)
    Dim str2 As String = ""
    For i = 0 To str1.Length - 1 Step 1
        Dim chr As Char = Mid(str1, i + 1, 1)
        Select Case chr
            Case "0" : str2 &= "0"
            Case "1" : str2 &= "1"
            Case "2" : str2 &= "2"
            Case "3" : str2 &= "3"
            Case "4" : str2 &= "4"
            Case "5" : str2 &= "5"
            Case "6" : str2 &= "6"
            Case "7" : str2 &= "7"
            Case "8" : str2 &= "8"
            Case "9" : str2 &= "9"
        End Select
    Next
    Try
        Dim num As Integer = 0 '(-2,147,483,648 through 2,147,483,647.)
        num = CType(str2, Integer)
        MsgBox("There was found the following number: " & Str(num), MsgBoxStyle.Information, "Success")
    Catch ex As Exception
        MsgBox("There was a error in conversion to int from string.", MsgBoxStyle.Exclamation, "Error")
        MsgBox("There was found the following number: " & str2, MsgBoxStyle.Information, "Success")
    End Try
End Sub

End Class