Hi Alex,
You're right about the additional assumptions, although one might
call the function within another repeat loop for each line, if each
pair of quotes is in one single line. As other solutions have shown,
one may also replace the return with a unique string before calling
this function.
As to the spaces, one will not want to apply any of the available
solutions, if the number of characters of the new string needs to be
equal to the number of characters of the old string and normally you
don't want to have 2 spaces subsequently. It is possible, though, to
apply another correction, by adding spaces as necessary. Also,
starting and trailing spaces of the entire text would have to be
added. All other spaces are included in the new string.
Best,
Mark
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Op 17-mei-2006, om 17:47 heeft Alex Tweedly het volgende geschreven:
Mark Schonewille wrote:
Again, no regex. I looked into regex, but I can't find how to
replace quoted strings without knowing the whole string in
advance. It would be great if someone came up with a regex
solution for this. In the mean time, this function seems to do
the job quite nicely:
function decentQuotes theStr
set the itemDel to space
repeat for each word myWord in theStr
if number of items of myWord > 1
then put quote & word 1 to -1 of (char 2 to -2 of myWord) & ¬
quote & space after myNewStr else put myWord & space after
myNewStr
end repeat
return char 1 to -2 of myNewStr
end decentQuotes
I think it also assumes the input string is on a single line, since
cr is a word delimiter, so input like
first line
second line
becomes
first line second line
And input like ("."s to indicate where the spaces are ....)
first ".line..
..second.." line
get badly mangled into
first "line" second "lin"
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