On Feb 12, 2007, at 11:16 AM, Chuck Pelto wrote:
Morning Charles,
On Feb 11, 2007, at 8:56 AM, Charles Yeomans wrote:
On Feb 10, 2007, at 12:54 PM, Chuck Pelto wrote:
Greetings,
Where can I find a good site to go to for advice on how to code a
RegEx request that will allow for double-quotation marks while
excluding any that are preceded by a symbol indicating an ESCAPE
character?
I'd suggest the REALbasic mailing lists. We'd need a little more
information about the regular expression you're trying to write.
I'm trying to write a RegEx call that will identify text between
pairs of doube-quotation marks, excluding any such mark that is
preceded by an ESCAPE character. The ESCAPE character, for the sake
of discussion, is the backwards slash (\) character.
Therefore, in a string of text from and EditField that reads as
follows:
This is a test by "Chuck Pelto" for "Susan Pelto" done in "\"Olde\"
English"
It would find the following items:
[1] "Chuck Pelto"
[2] "Susan Pelto"
[3] ""Olde" English"
I've got the part of getting the stuff between double-quotation
marks working. It reads as follows:
// method to test aspects of Regular Expressions
dim rg as New RegEx
dim theMatch as RegExMatch
dim strInput as string
dim strQuots as string
dim srchPatt as string
dim iCount as integer
dim i as integer
iCount = 0
strInput = EditField1.text
srchPatt = chr(34) + ".*?" + chr(34) // HERE is the pattern that
looks for text between double-quotation marks in a non-greedy manner
rg.SearchPattern = srchPatt
theMatch = rg.Search(strInput)
while theMatch<> nil
iCount = iCount + 1
strQuots = theMatch.subExpressionString(0)
theMatch = rg.search()
wend
What I need now is to figure out how to get it to exclude double-
quotation marks that are preceded by the ESCAPE character (\).
I'm learning how to do this using TextWrangler.
I'm toying around with the following variation on the basic search
pattern.....
srchPatt = chr(34) + ".*?" + chr(34) + "[^(\")]"
Do you think THAT would work?
No. But the following does appear to work.
rg.SearchPattern = "\x22.*?(?<!\\)\x22"
In other words -- match ", then keep matching until you hit a " not
preceded by \.
See the section entitled "Extension Mechanism" in the RegEx
documentation for explanation of this incantation.
Charles Yeomans
P.S. The Getting Started list is being killed off, so I've moved
this to the NUG.
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