从单个文本框中获取多个输入

Question

我是vb的新手，并且试图从文本框中获取多个输入，我正在制作一个二次求解器，其中我打算使用文本框执行以下操作。 当一个数字在空的文本框中输入时，它被存储在一个变量'a'中，然后如果某个操作符被按下，例如（*， - ，/，+，=），则下面的数字被存储在另一个变量中。

功能如图所示;

Dim a As Long 'coefficient of x^2 
Dim b As Long 'coeffecient of x 
Dim c As Long 'constant 
Dim o As String 
Dim ansa As Long 
Dim ansb As Long 
Dim ans As String 
Dim detrmnt As Long 'discriminant 
Private Function solve() 
    detrmnt = (b^2 - 4 * a * c) 
    ansa = (-b + Math.Sqrt(dscmnt))/2 * a ' quadratic formula root1 
    ansb = (-b - Math.Sqrt(dscmnt))/2 * a ' quadratic formula root2 
    ans = ansa & ", " & ansb 
    Return ans 
End Function 

Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click 
    solve() 
    TextBox2.Text = solve() 
End Sub 

我该怎么办;

TextBox1的需要二次方程和上面所说的变量分配号码，

！ - textbox2给出了答案。

编辑;我用的公式为二次公式是用于（AX^2 + BX + C = 0）

检查它here on wikipedia

注释;

（-b + Math.Sqrt（dscmnt））是在根部中的一个和2分子*一个是分母。

这里check this formula on wikipedia

“鉴别”，上面dscmnt = B^2 - 4 * A * C

Answer 1

与往常一样，我得到了忘乎所以的实施可能的解决方案，所以让我们现在先用碱性溶液中，然后用测试场景:)

分析文本的“全面”解决方案，您将需要确定：

• 您的输入有效
• 如果它们是正数或负数
• 哪里是A，B的位置，C

一个可能的解决方案可能是以下（我敢肯定有还有更多的方法可以做到这一点，并且我对+/-的解析不是100％满意，但它确实有效，并让你知道如何解析文本）。

注意，这仅仅是类的解决方法，全面过火的实施遵循以下

当不能得到解决，此方法将抛出错误，所以，无论是附上尝试在此方法中/ Catch块或使用的TryParse方法，你可以在下面找到:)

Public Overrides Sub Resolve(input As String) 
    ' lets say 2x^2 + 4x - 4 = 0 
    ' lets say 5x^2 + 3x + 1 = 0 
    If String.IsNullOrWhiteSpace(input) Then 
     Throw New ArgumentException("Input string cannot be null", "input") 
    End If 
    If input.IndexOf("=") >= 0 Then 
     input = input.Split("=")(0).Trim() 
     If String.IsNullOrWhiteSpace(input) Then 
      Throw New FormatException("'=' is at the beginning of the equation") 
     End If 
    End If 
    If input.Contains(" ") Then 
     input = input.Replace(" ", "") 
    End If 
    Dim lstGroup As New List(Of String) 
    Dim position As Integer = 0 
    Dim delimiters() As String = {"+", "-"} 
    Dim minIndex As Integer 
    Dim curDel As String 
    Dim lastDel As String = "+" 
    Dim part As String 

    While position < input.Length 
     minIndex = input.Length 
     curDel = "+" 
     For Each del In delimiters 
      Dim targetIndex As Integer = input.IndexOf(del, position) 
      If targetIndex > 0 AndAlso targetIndex < minIndex Then 
       minIndex = targetIndex 
       curDel = del 
      End If 
     Next 
     part = input.Substring(position, minIndex - position) 
     lstGroup.Add(lastDel + part.ToLower()) 
     position = minIndex + 1 
     lastDel = curDel 
    End While 

    CoefficientA = 0 
    CoefficientB = 0 
    CoefficientC = 0 
    For Each group In lstGroup 
     If group.Contains("x^2") Then 
      If CoefficientA <> 0 Then 
       Throw New FormatException("'x^2' was already determined!") 
      End If 
      If Not Integer.TryParse(group.Replace("x^2", ""), CoefficientA) Then 
       ' it is there so set it to 1 
       CoefficientA = 1 
      End If 
      Continue For 
     End If 
     If group.Contains("x") Then 
      If CoefficientB <> 0 Then 
       Throw New FormatException("'x' was already determined!") 
      End If 
      If Not Integer.TryParse(group.Replace("x", ""), CoefficientB) Then 
       CoefficientB = 1 
      End If 
      Continue For 
     End If 
     If CoefficientC <> 0 Then 
      Throw New FormatException("CoefficientC was already determined!") 
     End If 
     CoefficientC = Convert.ToInt32(group) 
    Next 
End Sub 

上述方法将设置你的职业等级，然后你可以解决所有必要的真实的值，并显示你觉得应该文本（见测试用例场景）

正如我所提到的那样，它在顶层实现方面有点过分，但如果你有更多的这个方程需要解决，你可以采用更灵活的方法来解析你的新类（它们只需要一个Solve/Resolve方法它可以节省你在未来的一些可能的麻烦：d）

走，接口IEquation，具有解决和解决方法，以及解决参数

Public Interface IEquation 
    Sub Resolve(input As String) 
    Sub Solve() 
    ReadOnly Property Solved As Boolean 
End Interface 

而那类的为MustInherit实施

''' <summary>Abstract implementation of IEquation, offers TryParse & Parse methods,that in their turn refer to the IEquation.Resolve() to do the actual parsing</summary> 
''' <remarks>Any implementation of IEquation must have a paramless constructor</remarks> 
Public MustInherit Class Equation 
    Implements IEquation 

    Public Shared Function TryParse(Of T As {New, IEquation})(input As String, ByRef equation As T) As Boolean 
     Dim succeeded As Boolean = True 

     Try 
      If String.IsNullOrWhiteSpace(input) Then 
       Throw New ArgumentException("Input string cannot be empty or nothing!", "input") 
      End If 
      equation = Parse(Of T)(input) 
     Catch ex As Exception 
      equation = Nothing 
      succeeded = False 
     End Try 
     Return succeeded 
    End Function 

    Public Shared Function Parse(Of T As {New, IEquation})(input As String) As T 
     Dim equation As New T() 
     equation.Resolve(input) 
     Return equation 
    End Function 

    Private _solved As Boolean = False 
    Public Property Solved As Boolean 
     Get 
      Return _solved 
     End Get 
     Protected Set(value As Boolean) 
      _solved = value 
     End Set 
    End Property 

    Public ReadOnly Property SolvedExplicit As Boolean Implements IEquation.Solved 
     Get 
      Return Solved 
     End Get 
    End Property 

    Public MustOverride Sub Resolve(input As String) Implements IEquation.Resolve 

    Public MustOverride Sub Solve() Implements IEquation.Solve 
End Class 

然后个

这个类可以作为你想创建，因为这样的一元二次方程的所有其他可解析式底座（再次，稍稍过了头实施）

Public Class QuadraticEquation 
    Inherits Equation 

    Private _cA As Integer 
    Public Property CoefficientA As Integer 
     Get 
      Return _cA 
     End Get 
     Set(value As Integer) 
      If _cA = value Then 
       Return 
      End If 
      _cA = value 
      Solved = False 
     End Set 
    End Property 

    Private _cB As Integer 
    Public Property CoefficientB As Integer 
     Get 
      Return _cB 
     End Get 
     Set(value As Integer) 
      If _cB = value Then 
       Return 
      End If 
      _cB = value 
      Solved = False 
     End Set 
    End Property 

    Private _cC As Integer 
    Public Property CoefficientC As Integer 
     Get 
      Return _cC 
     End Get 
     Set(value As Integer) 
      If _cC = value Then 
       Return 
      End If 
      _cC = value 
      Solved = False 
     End Set 
    End Property 

    Private _positiveRoot As Decimal 
    Public Property PositiveRoot As Decimal 
     Get 
      Return _positiveRoot 
     End Get 
     Protected Set(value As Decimal) 
      _positiveRoot = value 
     End Set 
    End Property 

    Private _negativeRoot As Decimal 
    Public Property NegativeRoot As Decimal 
     Get 
      Return _negativeRoot 
     End Get 
     Protected Set(value As Decimal) 
      _negativeRoot = value 
     End Set 
    End Property 

    Public Overrides Sub Resolve(input As String) 
     ' lets say 2x^2 + 4x - 4 = 0 
     ' lets say 5x^2 + 3x + 1 = 0 
     If String.IsNullOrWhiteSpace(input) Then 
      Throw New ArgumentException("Input string cannot be null", "input") 
     End If 
     If input.IndexOf("=") >= 0 Then 
      input = input.Split("=")(0).Trim() 
      If String.IsNullOrWhiteSpace(input) Then 
       Throw New FormatException("'=' is at the beginning of the equation") 
      End If 
     End If 
     If input.Contains(" ") Then 
      input = input.Replace(" ", "") 
     End If 
     Dim lstGroup As New List(Of String) 
     Dim position As Integer = 0 
     Dim delimiters() As String = {"+", "-"} 
     Dim minIndex As Integer 
     Dim curDel As String 
     Dim lastDel As String = "+" 
     Dim part As String 

     While position < input.Length 
      minIndex = input.Length 
      curDel = "+" 
      For Each del In delimiters 
       Dim targetIndex As Integer = input.IndexOf(del, position) 
       If targetIndex > 0 AndAlso targetIndex < minIndex Then 
        minIndex = targetIndex 
        curDel = del 
       End If 
      Next 
      part = input.Substring(position, minIndex - position) 
      lstGroup.Add(lastDel + part.ToLower()) 
      position = minIndex + 1 
      lastDel = curDel 
     End While 

     CoefficientA = 0 
     CoefficientB = 0 
     CoefficientC = 0 
     For Each group In lstGroup 
      If group.Contains("x^2") Then 
       If CoefficientA <> 0 Then 
        Throw New FormatException("'x^2' was already determined!") 
       End If 
       If Not Integer.TryParse(group.Replace("x^2", ""), CoefficientA) Then 
        ' it is there so set it to 1 
        CoefficientA = 1 
       End If 
       Continue For 
      End If 
      If group.Contains("x") Then 
       If CoefficientB <> 0 Then 
        Throw New FormatException("'x' was already determined!") 
       End If 
       If Not Integer.TryParse(group.Replace("x", ""), CoefficientB) Then 
        CoefficientB = 1 
       End If 
       Continue For 
      End If 
      If CoefficientC <> 0 Then 
       Throw New FormatException("CoefficientC was already determined!") 
      End If 
      CoefficientC = Convert.ToInt32(group) 
     Next 
    End Sub 

    Public Sub New() 

    End Sub 

    Public Overrides Sub Solve() 
     Solved = False 

     Try 
      Dim determinant As Decimal 
      Dim squareRootDeterminant As Decimal 
      Dim doubleA As Decimal 

      determinant = ((CoefficientB^2) - (4 * CoefficientA * CoefficientC)) 
      If determinant >= 0 Then 
       squareRootDeterminant = Math.Sqrt(Math.Abs(determinant)) 
      Else 
       Throw New InvalidOperationException("Cannot get Square root of negative determinant " & determinant) 
      End If 
      doubleA = 2 * CoefficientA 

      PositiveRoot = (-CoefficientB + squareRootDeterminant)/doubleA ' quadratic formula root1 
      NegativeRoot = (-CoefficientB - squareRootDeterminant)/doubleA ' quadratic formula root2 
      Solved = True 
     Catch ex As Exception 
      Solved = False 
      Console.WriteLine("{0}", ex.Message) 
     End Try 
    End Sub 

End Class 

那么这将是检验的这样

Sub Main() 
    Dim test() As String = {"2x^2 + 4x - 4 = 0", "5x^2 + 3x + 1", "2x^2+5x", "x^2+5", "5x - 5 + 3x^2"} 
    Dim eq As IEquation = Nothing 

    For Each expression In test 
     Console.WriteLine("Trying to resolve: {0}", expression) 
     If Equation.TryParse(Of QuadraticEquation)(expression, eq) Then 
      eq.Solve() 
      If Not eq.Solved Then 
       Console.WriteLine(vbTab & "Although it could be read, the equation failed to be solved!") 
      Else 
       Dim qe As QuadraticEquation = DirectCast(eq, QuadraticEquation) 

       Console.WriteLine(vbTab & "Result: [{0}; {1}]", qe.NegativeRoot, qe.PositiveRoot) 
      End If 
     Else 
      Console.WriteLine("Couldn't be resolved!") 
     End If 
    Next 

    Console.ReadLine() 
End Sub