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Embedded Quotes

Posted on 2001-08-07
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Last Modified: 2010-05-02
What is the best way to get rid of embedded quotes as in
Chain 3.75" Pitch 3/8" Diameter.
I just want to eliminate the " or CHR(34) and leave nothing.
I import several thousand rows and 15-20 have embedded quotes in them. I want a function to clean these out.

Thanks
Paul
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Question by:PaulCr125
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6 Comments
 
LVL 8

Expert Comment

by:Dave_Greene
ID: 6361139
StrResult = Replace(StringIN, Char(34), "")
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Author Comment

by:PaulCr125
ID: 6361285
Dave
I'm Access 97 no replace function.
But I do have VB6 and when I run
Replace("Chain 3.75" Pitch 3/8" Diameter",Chr(34),"")
... I get an expected list seperator or ) error

I appreciate your help
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LVL 8

Accepted Solution

by:
Dave_Greene earned 50 total points
ID: 6361335
You need to put the text in a variable...  the function doesn't know how to react to the text you have entered

Try this

StringX = Chr(34) & "Chain 3.75" & Chr(34) & " Pitch 3/8" & Chr(34) & " Diameter" & Chr(34)

Then

Replace(StringX, Chr(34), "")


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LVL 22

Expert Comment

by:rspahitz
ID: 6361438
Dave's right--you can't embed a quoted string into VB as though it's coming from a database and assume the syntax is correct!

My only recommendation, for clarity, is to replace the "constants":

Const DoubleQuote as string = """" ' chr$(34)

OriginalString = "Chain 3.75" & DoubleQuote & " Pitch 3/8" & DoubleQuote & " Diameter '---> Chain 3.75" Pitch 3/8" Diameter

NewString = Replace(OriginalString, DoubleQuote, vbNullString)

Also, if you want to do this in Access97, use the same code, but create your own replace function:

Function Replace(OriginalString, SearchString, ReplaceString) as String
  ' Note that this will not work if ReplaceString is a substring of SearchString

  dim strTemp as string
  dim iFoundPosit as integer

  strTemp = OriginalString

  do
    iFoundPosit = instr(strTemp, SearchString)
    if iFoundPosit = 0 then exit do
    strTemp = left$(strTemp, iFoundPosit-1) _
            & ReplaceString _
            & mid$(strTemp, iFoundPosit+len(SearchString)+1)
  loop

  Replace = strTemp
end Function
0
 

Author Comment

by:PaulCr125
ID: 6361482
Thanks Dave,
   I guess this errant embedded quote is a bigger problem than I thought.
I am parsing a database so I'm out of luck to do it in a function?

Paul
0
 

Author Comment

by:PaulCr125
ID: 6361577
Thanks Dave,
   I guess this errant embedded quote is a bigger problem than I thought.
I am parsing a database so I'm out of luck to do it in a function?

Paul
0

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