Vb.net 比较两个字符串,找出字母位置匹配的位置

Vb.net 比较两个字符串,找出字母位置匹配的位置,vb.net,Vb.net,我想对两个字符串执行按位and运算,以便: 鉴于: Dim word As String = "abcd" Dim temp As String = "a-d-" Dim word As String = "abcd" Dim temp As String = "a--d" 我只想返回“a” 鉴于: Dim word As String = "abcd" Dim temp As String = "a-d-" Dim word As String = "abcd" Dim

我想对两个字符串执行按位and运算,以便:

鉴于:

 Dim word As String = "abcd"
 Dim temp As String = "a-d-"
 Dim word As String = "abcd"
 Dim temp As String = "a--d"
我只想返回“a”

鉴于:

 Dim word As String = "abcd"
 Dim temp As String = "a-d-"
 Dim word As String = "abcd"
 Dim temp As String = "a--d"
我只想返回'a--d'

我尝试过intersect,但它只在一个字符串中找到与另一个字符串中的字符匹配的字符,而不管位置如何

我在这里用“-”来表示空格


任何建议都将不胜感激。

这将处理长度不匹配的字符串:

Public Function CheckMask(ByVal word As String, ByVal mask As String) As String
    Dim wordChars() As Char = word.ToCharArray()
    Dim maskChars() As Char = mask.ToCharArray()

    Dim i As Integer = 0       
    While i < wordChars.Length AndAlso i < maskChars.Length
        If wordChars(i) <> maskChars(i) Then wordChars(i) = " "c
        i = i + 1
    End While

    'If string lengths are equal or the mask is longer, we're done
    'If the word is longer, need to set remaining characters to " "
    While i < wordChars.Length
        wordChars(i) = " "c
    End While

    Return New String(wordChars)
End Function
公共函数检查掩码(ByVal字作为字符串,ByVal掩码作为字符串)作为字符串
Dim wordChars()作为Char=word.ToCharArray()
Dim maskChars()作为Char=mask.tocharray()显示
尺寸i为整数=0
而我
这基本上循环到两个字符串中较短的一个的末尾。如果给定位置的字母匹配,则将字母添加到结果中,否则将添加空格

    Dim sFirstWord As String = "qwerty"
    Dim sSecndWord As String = "qseftg"
    Dim sResult As String = ""

    For i As Integer = 0 To Math.Min(sFirstWord.Length, sSecndWord.Length) - 1
        If sFirstWord(i) = sSecndWord(i) Then
            sResult &= sFirstWord(i)
        Else
            sResult &= " "
        End If
    Next

sResult将保持:“q e t”

您必须循环字符串的每个索引,将该索引处的字符与其他字符串进行比较,并确定它们是否相等。你能展示一下你的努力,让我们看看你做错了什么吗?一个问题。在第一个示例中,您说您只想要“a”。那么准确,或者你真的想要“a-”?使用某人比使用&=?我不是在评判,只是好奇。还是某人在后台这样做?@Keith Stringbuilder将在后台创建一个单字符缓冲区一次。&=将分配和复制一个全新的字符串和每个附加。但忘了这一点:我找到了一种更有效的方法。我知道我想要角色阵列是有原因的:)@Jens我找到了一种比Stringbuilder@JoelCoehoorn啊,好的。谢谢你提供的信息。