说我有一个字符串：
"I say ""Hello world"" and she says ""Excuse me?"""
VBA会将此字符串解释为：
I say "Hello world" and she says "Excuse me?"
我有一个字符串：
"I say ""Did you know that she said """"Hi there!"""""""
VBA将此字符串解释为：
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!"""
我们可以继续在vba中解析字符串：
Did you know that she said "Hi there!"
最终我需要一些函数sBasicQuote（quotedStringHierarchy as string），它返回一个包含字符串层次结构中下一级别的字符串。
E.G。
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!'
我只是想不出一个可以解决这个问题的算法...你几乎需要替换所有的双引号，但是当你更换了第n个双引号时你必须跳到第n + 1个双引号？
如何在VBA中实现此功能？
答案 0 :(得分：1)
你可以做这样的事情
---load-avg--- --io/total-----io/sda------io/md0-- ---procs--- ------memory-usage----- ---paging-- -dsk/total----dsk/sda-----dsk/md0-- ---system-- ----total-cpu-usage---- -net/total-
1m 5m 15m | read writ: read writ: read writ|run blk new| used buff cach free| in out | read writ: read writ: read writ| int csw |usr sys idl wai hiq siq| recv send
6.61 6.40 3.83| 292 186 :14.6 0.40 : 283 193 |4.2 1.5 1.6|12.0G 1548k 3685M 147M| 0 0 | 47M 84M: 255k 8192B: 46M 82M| 39k 59k| 22 5 50 22 0 0| 52B 123B
6.06 6.29 3.82| 289 165 :22.8 0.70 : 273 173 |4.2 2.9 1.5|11.9G 1420k 3610M 281M| 0 0 | 45M 75M: 146k 9830B: 45M 77M| 21k 39k| 16 4 58 23 0 0| 884B 68B
5.74 6.22 3.82| 278 172 :20.2 0.70 : 265 177 |3.3 2.5 3.9|11.9G 1124k 3709M 162M| 0 0 | 44M 80M: 294k 9011B: 44M 79M| 34k 51k| 19 4 52 25 0 0| 32B 83B
5.47 6.14 3.82| 351 178 :52.6 1.50 : 305 180 |6.0 2.1 0|11.9G 1092k 3701M 175M| 0 0 | 38M 79M: 669k 26k: 37M 78M| 27k 59k| 24 4 53 18 0 0| 51B 99B
5.09 6.04 3.82| 300 176 :25.4 0.40 : 282 190 |4.2 2.4 0|12.0G 656k 3655M 182M| 0 0 | 45M 80M: 228k 3686B: 45M 82M| 32k 60k| 21 5 55 19 0 0| 228B 103B
4.84 5.95 3.81| 294 172 :24.8 0.50 : 275 175 |3.1 2.4 0|12.0G 648k 3642M 173M| 0 0 | 45M 80M: 264k 8602B: 45M 79M| 38k 48k| 17 4 57 22 0 0| 32B 59B
5.27 6.01 3.85| 295 175 :25.3 1.20 : 276 184 |2.4 2.7 0.7|12.0G 532k 3636M 180M| 0 0 | 44M 80M: 235k 19k: 44M 81M| 33k 59k| 21 4 54 21 0 0| 46B 123B
5.32 5.99 3.87| 315 169 :23.7 0.40 : 301 175 |4.6 2.2 1.4|11.9G 536k 3742M 162M| 0 0 | 50M 76M: 506k 8192B: 49M 76M| 28k 44k| 20 4 52 24 0 0| 60B 99B
另一个略微修改的版本更准确
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
答案 1 :(得分：1)
我认为以下内容将在您的嵌套引用示例中返回您要查找的内容。你的第一个例子实际上不是嵌套引号的情况。
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
答案 2 :(得分：0)
我花了更多时间思考并提出了这个解决方案。
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
该过程的第一部分是从字符串中删除第一个和最后一个单引号并在它们之间返回文本。然后我们遍历字符串，替换""的每个实例并将其替换为"。我们每次执行此操作时都会跳到下一个字符，以确定""""之类的字符串不会转到""而不是"。
还有其他人有更好/更紧凑的解决方案吗？
在本论坛的所有建议之后，我解决了这个问题。它有一些额外的错误捕获来查找验证嵌套字符串。
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