我怎样才能find一个string中的引号文本？
例
说我有一个string：
"I say ""Hello world"" and she says ""Excuse me?"""
VBA会将这个string解释为：
I say "Hello world" and she says "Excuse me?"
一个更复杂的例子：
我有一个string：
"I say ""Did you know that she said """"Hi there!"""""""
VBA将此string解释为：
I say "Did you know that she said ""Hi there!"""
如果我们删除“我说”
"Did you know that she said ""Hi there!"""
我们可以继续parsingvba中的string：
Did you know that she said "Hi there!"
问题
最终，我想要一些函数，sBasicQuote（quotedStringHierarchy as string），它返回一个string层次结构中包含下一级的string。
例如
dim s as string s = "I say ""Did you know that she said """"Hi there!""""""" s = sBasicQuote(s) ' returns 'I say "Did you know that she said ""Hi there!"""' s = sBasicQuote(s) ' returns 'Did you know that she said "Hi there!"' s = sBasicQuote(s) ' returns 'Hi there!'
我只是不能找出一个algorithm，可以用这个…你几乎需要replace所有的双引号，但是当你已经取代了第n个双引号，你必须跳到第n + 1个双引号？
在VBA中如何实现这一点？
你可以做这样的事情
Public Sub test() Dim s As String s = "I say ""Did you know that she said """"Hi there!""""""" Debug.Print DoubleQuote(s, 0) Debug.Print DoubleQuote(s, 1) Debug.Print DoubleQuote(s, 2) End Sub Public Function DoubleQuote(strInput As String, intElement As Integer) As String Dim a() As String strInput = Replace(strInput, String(2, Chr(34)), String(1, Chr(34))) a = Split(strInput, chr(34)) DoubleQuote = a(intElement) End Function
另一个稍微修改的版本是更精确一点
`Public Function DoubleQuote(strInput As String, intElement As Integer) As String Dim a() As String Dim b() As String Dim i As Integer ReDim b(0) a = Split(strInput, Chr(34)) ' ***** See comments re using -1 ******* For i = 0 To UBound(a) - 1 If Len(a(i)) = 0 Then b(UBound(b)) = Chr(34) & a(i + 1) & Chr(34) i = i + 1 Else b(UBound(b)) = a(i) End If ReDim Preserve b(UBound(b) + 1) Next i DoubleQuote = b(intElement) End Function`
我认为以下内容将返回您在嵌套报价示例中查找的内容。 你的第一个例子并不是真正的嵌套引号的情况。
Option Explicit Sub NestedQuotes() Const s As String = "I say ""Did you know that she said """"Hi there!""""""" Dim COL As Collection Dim Start As Long, Length As Long, sTemp As String, V As Variant Set COL = New Collection sTemp = s COL.Add sTemp Do Until InStr(sTemp, Chr(34)) = 0 sTemp = COL(COL.Count) sTemp = Replace(sTemp, String(2, Chr(34)), String(1, Chr(34))) Start = InStr(sTemp, Chr(34)) + 1 Length = InStrRev(sTemp, Chr(34)) - Start sTemp = Mid(sTemp, Start, Length) COL.Add sTemp Loop For Each V In COL Debug.Print V Next V End Sub
我的解决scheme
我花了更多的时间思考，并提出了这个解决scheme。
Function sMineDoubleQuoteHierarchy(s As String) As String 'Check the number of quotes in the string are even - sanity check If (Len(s) - Len(Replace(s, """", ""))) Mod 2 <> 0 Then sMineDoubleQuoteHierarchy = "Error - Odd number of quotes found in sMineDoubleQuoteHierarchy() function": Exit Function 'First thing to do is find the first and last *single* quote in the string Dim lStart, lEnd, i As Long, fs As String lStart = InStr(1, s, """") lEnd = InStrRev(s, """") 'After these have been found we need to remove them. s = Mid(s, lStart + 1, lEnd - lStart - 1) 'Start at the first character i = 1 Do While True 'Find where the next double quote is i = InStr(1, s, """""") 'if no double quote is found then concatenate with fs with the remainder of s If i = 0 Then Exit Do 'Else add on the string up to the char before the ith quote fs = fs & Left(s, i - 1) 'Replace the ith double quote with a single quote s = Left(s, i - 1) & Replace(s, """""", """", i, 1) 'Increment by 1 (ensuring the recently converted double quote is no longer a single quote i = i + 1 Loop 'Return fs sMineDoubleQuoteHierarchy = s End Function
这个解决scheme是怎么回事？
该过程的第一部分是从string中删除第一个和最后一个单引号，并返回它们之间的文本。 然后我们通过stringreplace每个""实例并replace为" 。每次我们这样做，我们跳到下一个字符不确定string，如""""去""而不是" 。
其他人是否有更好/更紧凑的解决scheme？
编辑
在这个论坛的所有build议之后，我解决了这个问题。 它有一些额外的错误陷阱findvalidation嵌套的string。
Public Function DoubleQuoteExtract(ByVal s As String, Optional ByRef ErrorLevel As Boolean) As String 'This effectively parses the string like BASIC does by removing incidents of "" and replacing them with " 'SANITY CHECK - Check even number of quotes Dim countQuote As Double countQuote = Len(s) - Len(Replace(s, """", "")) 'Calculate whether or not quote hierarchy is correct: '"..." - Is okay - Count Quotes = 2 - Count Quotes / 2 = 1 '""..."" - Is not okay - Count Quotes = 4 - Count Quotes / 2 = 2 '"""...""" - Is okay - Count Quotes = 6 - Count Quotes / 2 = 3 '""""..."""" - Is not okay - Count Quotes = 8 - Count Quotes / 2 = 4 'etc. 'Ultimately: IF CountQuotes/2 = Odd The string hierarchy is setup fine ' IF CountQuotes/2 = Even, The string Hierarchy is setup incorrectly. Dim X As Double: X = countQuote / 2 Dim ceil As Long: ceil = Int(X) - (X - Int(X) > 0) If ceil Mod 2 <> 0 Then sDoubleQuoteExtract = "#Error - Incorrect number of double quotes forming an incomplete hierarchy.": GoTo ErrorOccurred 'If an odd number of quotes are found then they cannot be paired correctly, thus throw error If countQuote Mod 2 <> 0 Then sDoubleQuoteExtract = "#Error - Odd number of quotes found in sMineDoubleQuoteHierarchy() function": GoTo ErrorOccurred 'Find the next incident of single quote. Trim the string to this s = Mid(s, InStr(1, s, String(1, Chr(34)))) 'replace all instances of "" with " s = Replace(s, String(2, Chr(34)), String(1, Chr(34))) 'Finally trim off the first and last quotes DoubleQuoteExtract = Mid(s, 2, Len(s) - 2) ErrorLevel = False Exit Function ErrorOccurred: ErrorLevel = True End Function