VBstring "People's Verdict" rejected with Syntex err message

I am sending an string like "People's Verdict" from a VB application to MSSQLServer for an update using DAO. It gives syntex error because of apostrophe i.e. '. (If the input is like "Peoples Verdict" there is no error.)
Can any body suggest how to handle this.
Thanks
devesh032698Asked:
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devesh032698Author Commented:
Edited text of question
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percosolatorCommented:
Here is a function that I wrote to handle such situations for MS-SQL Server.

It handles both single and double quotes.

Ex:

Brian O'neil             becomes    Brian O' + CHAR(39) + 'neil

6" Screwdriver           becomes    6' + CHAR(34) + ' Screwdriver

-----------------------
this is a very simple piece of code that is designed for fixing quotation problems within variables.

The optional boolean variable defaults to true, and signifies that the string is to be parsed and contained within single quotation marks.
-----------------------

Option Explicit

Public Function FixQuotes(ByVal pstrSource As String, Optional pblnEnquote As Boolean = True) As String

    If (Len(pstrSource) = 0) Then Exit Function
   
    'replace all single quotation marks with a CHR(255)
    'because we need to have apostrophes inside to delineate
    'where the original string stops and starts on either
    'side of the " + CHAR(39) + ". Otherwise you will get
    'an infinite loop
    While (InStr(pstrSource, "'") > 0)

        pstrSource = Left(pstrSource, InStr(pstrSource, "'") - 1) + Chr(255) + " + CHAR(39) + " + Chr(255) + Mid(pstrSource, InStr(pstrSource, "'") + 1, Len(pstrSource))

    Wend

    'now we have only valid apostrophes inside, replace all the
    'Chr(255)'s with apostrophes
    While (InStr(pstrSource, Chr(255)) > 0)

        pstrSource = Left(pstrSource, InStr(pstrSource, Chr(255)) - 1) + "'" + Mid(pstrSource, InStr(pstrSource, Chr(255)) + 1, Len(pstrSource))

    Wend

    'look through the string and replace all quotation marks
    'with their SQL equivalent
    While (InStr(pstrSource, """") > 0)
   
        pstrSource = Left(pstrSource, InStr(pstrSource, """") - 1) + "' + CHAR(34) + '" + Mid(pstrSource, InStr(pstrSource, """") + 1, Len(pstrSource))
   
    Wend
   
    If (pblnEnquote) Then pstrSource = Enquote(pstrSource)
   
    FixQuotes = pstrSource
   
End Function

Public Function Enquote(pstrSource As String) As String

    Enquote = "'" & pstrSource & "'"
   
End Function

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trkcorpCommented:
devesh,
  If you double the single quotes within your string then they will be interpreted as single.  You can drop the following subroutine into a module and execute it from anywhere within your program by passing the string where this condition can occur to this routine.


Sub Fix_Quotes(ByRef TextIn As String)
' replaces single quotes (') with double ('') so RDO engine can interpret them as single
Dim pos As Long
Dim pos2 As Long
Dim length As Long
pos2 = 1
Do
  pos = InStr(pos2, TextIn, "'", vbBinaryCompare)
  If pos > 0 Then
    length = Len(TextIn)
    TextIn = Mid$(TextIn, 1, pos - 1) & "''" & _
      Mid$(TextIn, pos + 1, length - (pos))
    pos2 = pos + 2 ' reset starting point for instr search after apostrophes
  End If
Loop Until pos = 0
End Sub

I use this as a generic routine over TEXT, CHAR & VARCHAR data. It works very well.  Percosolator answered the question while I was typing. Let me know if you decide you want my answere so I can answer after possible rejection of p's answer.  trkcorp@ixlmemphis.com

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RazdanCommented:
Define a variable
Var = "People" & "'" & "s verdict"
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trkcorpCommented:
dim Var as String
Var = "People" & "'" & "s verdict"
Fix_Quotes Var

The result is "People''s verdict".  When this string is sent to the SQL Server it is stored as "People's verdict".
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devesh032698Author Commented:
I am using VB50

For percosolator :
-----------------
I pass value to pstrsource as "people's verdict"
when I reach the following part of the code, it get's into infinite loop i.e. it does not change value of pstrsource , however the right part of the statement i.e Left(pstrSource... does show the desired value

While (InStr(pstrSource, "'") > 0)
       
        pstrSource = Left(pstrSource, InStr(......

Wend


For trkcorp :
-----------

The behavior is same as above.

Further suggestions please!

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trkcorpCommented:
I do not understand.  The behavior is the same as above? After returning from Fix_Quotes the value of Var will be "People''s verdict".  There is no way to get caught in my loop...
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percosolatorCommented:
hmmm.

just copied the code off of the web page and ran some tests, works.  maybe I am doing something different.

here's what I tested in the immediate pane:
-----------------------------------------
?fixquotes("people" & chr(39) & "s court")
'people' + CHAR(39) + 's court'   <- result


A = "people" & chr(39) & "s court"
?a
people's court                    <- result

?fixquotes(A)
'people' + CHAR(39) + 's court'   <- result



b = "people's Court
?b
people's Court

?fixquotes(b)
'people' + CHAR(39) + 's Court'

------------------------------------------

what am I doing differently to test?  It seems to work with no infinate loop.
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devesh032698Author Commented:
Both the solutions work, however just cut and paste does not work
Anyway thanks to you all.

I am sorry trkcorp, The points can not be shared as per the rules of this site, however I am sure you deserve 50% of the points for equally good answer. My sincere regards for your help.
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trkcorpCommented:
That's OK devesh. Thanks.  I hadn't taken double quotes into account and percosolator made me aware that I should.  That is reward enough.
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