1 Correct answer
In that case you need to find all instances of the code fragments, then replace + with |+|. Something like this:
app.findGrepPreferences = null;
app.findGrepPreferences.findWhat = '\\[&.+?&\\]';
found = app.documents[0].findGrep();
for (i = found.length-1; i >= 0; i--) {
found.contents = found.contents.replace (/\+/g, '|+|');
}
Untried, but you get the picture.
P.
10
Replies
Hi Prabu,
Very simply (including the opening/closing brackets):
(\[&|\G).*?\K\+(?=.*&\])
(^/) [ advanced Grep code -- To be validated! 
]
Prabu,
Do you mean mine not?
app.findGrepPreferences = app.changeGrepPreferences = null;
app.findGrepPreferences.findWhat = "(\\[&|\\G).*?\\K\\+(?=.*&\\])";
app.changeGrepPreferences.changeTo = "|+|";
app.activeDocument.changeGrep();
(^/) 
Only some Jedis know it! [it comes from the Dark Side]
I gave it a name: the \Ghost!
(^/)
Hi Prabu,
Keep in mind there are still possible issues in using (\[&|\G).*?\K\+(?=.*&\]) as it doesn't fully deal with extra + signs.
For example in the text:
X+Y+[&A+B+C+D&]+X+Y[&A+B&]+X+&]+[&X¶
+A&]+B[&X+Y&]+[&X+Y++Z&]++C+D&]¶
we may expect GREP to alter only the well-formed red areas. But it won't be the case with uncontrolled .* in your regex.
So, in your find/change script you might need to refine the suggested syntax,
const GREP1 = "(\\[&|\\G).*?\\K\\+(?=.*&\\])";
// etc.
into something more explicit like
const GREP2 = "(\\[&|\\G)[^&+\\r]+\\K\\++(?=[^&\\r]+&\\])";
// etc.
Given the same input, the figure below shows how the + signs are captured and changed (see the red crosses ×) using GREP1 vs. GREP2. You can see that GREP1 eats a number of + signs which are not embedded in the [&…&] regions, while GREP2 is a bit more accurate:
But of course there is no definitive answer as long as we don't have more specification on the actual characters your stream may contain inside and outside the markup.
@+
Marc
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