I am relatively new to scripting and within an InDesign Script I am trying to change all the first letters of all sentences to uppercase (many of the are lowercase, since I randomly generated the setences from different text sources).
I am so far able to find the text parts with this Grep expression:
\.(\s)+\l
I also found this script by Peter Kahrel, that he shares on InDesign Secrets:
app.findGrepPreferences.findWhat = "^.";
found = app.activeDocument.findGrep();
for (i = 0; i < found.length; i++)
found[i].characters[0].changecase (ChangecaseMode.lowercase);
However, when I now replace the ^. with my own expression, and change lowercase to uppercase, the script does not work, which makes sense, since I do not want to change the first character of my findGrep results, but the last one. But how can I find the last character? The breaks between the sentences have different lengths, so I cannot simply type 2 instead of 0.
Any help would be very appreciated! Thank you!
Edit: I'm working on CS6.
Your GREP returns matches that start with a period, then have any number of spaces (including hard returns, probably), and always end with one lowercase character. So far, so good. You can access the last character (and in fact any last item in any InDesign object collection) in this way:
found[i].characters[-1].changecase (ChangecaseMode.lowercase);
which 'indexes' from the end, rather than from the start.
However! The only character in your matches, other than the period and spaces, is always going to be a lowercase letter. So you can skip the entire "how to find the correct index" thing, and probably slightly speed up the script as well, by simply applying lowercase (or, as you are using it, uppercase) to the entire match:
found[i].changecase (ChangecaseMode.lowercase);
because nothing will happen to not-lowercaseable characters (a word I declare to signify "having the Unicode-defined property of being lowercase and having an uppercase equivalent). (Or the other way around, if I understand your purpose correct.)
Related
I am trying to check if a string contains at least one lowercase letter, uppercase letter, and a number, but not punctuation (including spaces).
For example
4aBc8Fk3 should match
4aBc 8.;3 should not match
I tried the following, but it matches spaces:
^(?=.*[a-z])(?=.*[A-Z])(?=.*[0-9]).{6,}[^[:punct:]]$
Any ideas how to not match strings containing punctuation including spaces?
The regular expression you have got there does the following for as far as I understand (I'm not familiar with the ruby variety, and still quite new to regex myself; this will give you an idea, but may not be 100% correct):
Go to the beginning of the string
Ensure the string matches any number of any characters followed by a lowercase letter, e.g. --a
Ensure the string matches any number of any characters followed by an uppercase letter, e.g.--aA
Ensure the string matches any number of any characters followed by a number, e.g. --aA0
If that is all true, make sure the beginning of the string is followed by at least 6 random characters, e.g.--aA0-
Ensure that is followed by a single non-punctuation character (although this is the part I'm not sure about, as I haven't used character classes before, and don't know if it's [^[:punct:]] or [^:punct:]), e.g. --aA0-c
Ensure that is followed directly by the end of the string
Now, the lookaheads would also allow a different order of occurrences, e.g. 0---Aa, as long as the string contains any characters followed by what they are looking for.
What you probably want is ^[a-zA-Z0-9]{6,}$, i.e. at least six characters, with the characters being letters and numbers (though that would also allow aaaaaa, for example).
Maybe try ^(?=.*[a-z])(?=.*[A-Z])(?=.*[0-9])[a-zA-Z0-9]{6,}$ to make sure each group is present, and to get alpha-numerical characters (at least six of them) only.
I always use a tool such as http://www.regexpal.com/ to slowly build up my regex and to see where I go wrong, deconstructing a "bad" regex until I get to a "good" one, then slowly adding to it again.
Hope that helps. :)
P.S.: I'm still a bit unclear how many characters you want to match in total, i.e. if the string is fixed length or not...?
I need some GREP help. I'm trying to search for text in an InDesign file that has lesser-than "<" and greater-than ">" characters on either end. The text could be one word or more and could include spaces and even numbers. But, here's here's the catch, there may be multiple words on a line, such as , , <12 peaches> and <3 plums>.
I tried using <(.+)> but this picks up the whole paragraph and removes the beginning and ending brackets < > but leaves the ones in the middle.
Anyone know the proper wildcard structure that would find any text or number between these brackets and even if more than one set of "<>" appear in a single paragraph?
FYI, GREP in InDesign is not exactly the same as it's used on the web.
The + metacharacter is greedy meaning it tends to catch more than needed. You can restrain it with a question mark like this :
<.+?>
or change your pattern with
<[^>]+>
I am new to ruby and I'm trying to work with regex.
I have a text which looks something like:
HEADING
Some text which is always non capitalized. Headings are always capitalized, followed by a space or nothing more.
YOU CAN HAVE MULTIPLE WORDS IN HEADING
I'm using this regular expression to choose all headings:
^[A-Z]{2,}\s?([A-Z]{2,}\s?)*$
However, it matches all headings which does not contain chars as Č, Š, Ž(slovenian characters).
So I'm guessing [A-Z] only matches ASCII characters? How could I get utf8?
You are right in that when you define the ASCII range A-Z, the match is made literally only for those characters. This is to do with the history of characters on computers, more and more characters have been added over time, and they are not always structured in an encoding in ways that are easy to use.
You could make a larger character class that matches the slovenian characters you need, by listing them.
But there is a shortcut. Someone else has already added necessary data to the Unicode data so that you can write shorter matches for "all uppercase characters": /[[:upper:]]/. See http://ruby-doc.org//core-2.1.4/Regexp.html for more.
Altering your regular expression with just this adjustment:
^[[:upper:]]{2,}\s?([[:upper:]]{2,}\s?)*$
You may need to adjust it further, for instance it would not match the heading "I AM A HEADING" due to the match insisting each word is at least two letters long.
Without seeing all your examples, I would probably simplify the group matching and just allow spaces anywhere:
^[[:upper:]\s]+$
You can use unicode upper case letter:
\p{Lu}
Your regex:
\b\p{Lu}{2,}(?:\s*\p{Lu}{2,})\b
RegEx Demo
This seems like a simple one, but I am missing something.
I have a number of inputs coming in from a variety of sources and in different formats.
Number inputs
123
123.45
123,45 (note the comma used here to denote decimals)
1,234
1,234.56
12,345.67
12,345,67 (note the comma used here to denote decimals)
Additional info on the inputs
Numbers will always be less than 1 million
EDIT: These are prices, so will either be whole integers or go to the hundredths place
I am trying to write a regex and use gsub to strip out the thousands comma. How do I do this?
I wrote a regex: myregex = /\d+(,)\d{3}/
When I test it in Rubular, it shows that it captures the comma only in the test cases that I want.
But when I run gsub, I get an empty string: inputstr.gsub(myregex,"")
It looks like gsub is capturing everything, not just the comma in (). Where am I going wrong?
result = inputstr.gsub(/,(?=\d{3}\b)/, '')
removes commas only if exactly three digits follow.
(?=...) is a lookahead assertion: It needs to be possible to be matched at the current position, but it's not becoming part of the text that is actually matched (and subsequently replaced).
You are confusing "match" with "capture": to "capture" means to save something so you can refer to it later. You want to capture not the comma, but everything else, and then use the captured portions to build your substitution string.
Try
myregex = /(\d+),(\d{3})/
inputstr.gsub(myregex,'\1\2')
In your example, it is possible to tell from the number of digits after the last separator (either , or .) that it is a decimal point, since there are 2 lone digits. For most cases, if the last group of digits does not have 3 digits then you can assume that the separator in front is decimal point. Another sign is the multiple appearance of a separator in big numbers allows us to differentiate between decimal point and separators.
However, I can give a string 123,456 or 123.456 without any sort of context. It is impossible to tell whether they are "123 thousand 456" or "123 point 456".
You need to scan the document to look for clue whether , is used for thousand separator or decimal point, and vice versa for .. With the context provided, then you can safely apply the same method to remove the thousand separators.
You may also want to check out this article on Wikipedia on the less common ways to specify separators or decimal points. Knowing and deciding not to support is better than assuming things will work.
I have the following
address.gsub(/^\d*/, "").gsub(/\d*-?\d*$/, "").gsub(/\# ?\d*/,"")
Can this be done in one gsub? I would like to pass a list of patterns rather then just one pattern - they are all being replaced by the same thing.
You could combine them with an alternation operator (|):
address = '6 66-666 #99 11-23'
address.gsub(/^\d*|\d*-?\d*$|\# ?\d*/, "")
# " 66-666 "
address = 'pancakes 6 66-666 # pancakes #99 11-23'
address.gsub(/^\d*|\d*-?\d*$|\# ?\d*/,"")
# "pancakes 6 66-666 pancakes "
You might want to add little more whitespace cleanup. And you might want to switch to one of:
/\A\d*|\d*-?\d*\z|\# ?\d*/
/\A\d*|\d*-?\d*\Z|\# ?\d*/
depending on what your data really looks like and how you need to handle newlines.
Combining the regexes is a good idea--and relatively simple--but I'd like to recommend some additional changes. To wit:
address.gsub(/^\d+|\d+(?:-\d+)?$|\# *\d+/, "")
Of your original regexes, ^\d* and \d*-?\d*$ will always match, because they don't have to consume any characters. So you're guaranteed to perform two replacements on every line, even if that's just replacing empty strings with empty strings. Of my regexes, ^\d+ doesn't bother to match unless there's at least one digit at the beginning of the line, and \d+(?:-\d+)?$ matches what looks like an integer-or-range expression at the end of the line.
Your third regex, \# ?\d*, will match any # character, and if the # is followed by a space and some digits, it'll take those as well. Judging by your other regexes and my experience with other questions, I suspect you meant to match a # only if it's followed by one or more digits, with optional spaces intervening. That's what my third regex does.
If any of my guesses are wrong, please describe what you were trying to do, and I'll do my best to come up with the right regex. But I really don't think those first two regexes, at least, are what you want.
EDIT (in answer to the comment): When working with regexes, you should always be aware of the distinction between a regex the matches nothing and a regex that doesn't match. You say you're applying the regexes to street addresses. If an address doesn't happen to start with a house number, ^\d* will match nothing--that is, it will report a successful match, said match consisting of the empty string preceding the first character in the address.
That doesn't matter to you, you're just replacing it with another empty string anyway. But why bother doing the replacement at all? If you change the regex to ^\d+, it will report a failed match and no replacement will be performed. The result is the same either way, but the "matches noting" scenario (^\d*) results in a lot of extra work that the "doesn't match" scenario avoids. In a high-throughput situation, that could be a life-saver.
The other two regexes bring additional complications: \d*-?\d*$ could match a hyphen at the end of the string (e.g. "123-", or even "-"); and \# ?\d* could match a hash symbol anywhere in string, not just as part of an apartment/office number. You know your data, so you probably know neither of those problems will ever arise; I'm just making sure you're aware of them. My regex \d+(?:-\d+)?$ deals with the trailing-hyphen issue, and \# *\d+ at least makes sure there are digits after the hash symbol.
I think that if you combine them together in a single gsub() regex, as an alternation,
it changes the context of the starting search position.
Example, each of these lines start at the beginning of the result of the previous
regex substitution.
s/^\d*//g
s/\d*-?\d*$//g
s/\# ?\d*//g
and this
s/^\d*|\d*-?\d*$|\# ?\d*//g
resumes search/replace where the last match left off and could potentially produce a different overall output, especially since a lot of the subexpressions search for similar
if not the same characters, distinguished only by line anchors.
I think your regex's are unique enough in this case, and of course changing the order
changes the result.