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
<[^>]+>
Related
I have a database in this format:
username:something:UID:something:name:home_folder
Now I want to see which users have a UID ranging from 1000-5000. This is what what I tried to do:
ypcat passwd | grep '^.*:.*:[1-5][0-9]\{2\}:'
My thinking is this: I go to the third column and find numbers that start with a number from 1-5, the next number can be any number - range [0-9] and that range repeats itself 2 more times making it a 4 digit number. In other words it would be something like [1-5][0-9][0-9][0-9].
My output, however, lists even UID's that are greater than 5000. What am I doing wrong?
Also, I realize the code I wrote could potentially lists numbers up to 5999. How can I make the numbers 1000-5000?
EDIT: I'm intentionally not using awk since I want to understand what I'm doing wrong with grep.
There are several problems with your regex:
As Sundeep pointed out in a comment, ^.*:.*: will match two or more columns, because the .* parts can match field delimiters (":") as well as field contents. To fix this, use ^[^:]*:[^:]*: (or, equivalently, ^\([^:]:\)\{2\}); see the notes on bracket expressions and basic vs extended RE syntax below)
[0-9]\{2\} will match exactly two digits, not three
As you realized, it matches numbers starting with "5" followed by digits other than "0"
As a result of these problems, the pattern ^.*:.*:[1-5][0-9]\{2\}: will match any record with a UID or GID in the range 100-599.
To do it correctly with grep, use grep -E '^([^:]*:){2}([1-4][0-9]{3}|5000):' (again, see Sundeep's comments).
[Added in edit:]
Concerning bracket expressions and what ^ means in them, here's the relevant section of the re_format man page:
A bracket expression is a list of characters enclosed in '[]'. It
normally matches any single character from the list (but see below).
If the list begins with '^', it matches any single character (but see
below) not from the rest of the list. If two characters in the list
are separated by '-', this is shorthand for the full range of
characters between those two (inclusive) in the collating sequence,
e.g. '[0-9]' in ASCII matches any decimal digit.
(bracket expressions can also contain other things, like character classes and equivalence classes, and there are all sorts of special rules about things like how to include characters like "^", "-", "[", or "]" as part of a character list, rather than negating, indicating a range, class, or end of the expression, etc. It's all rather messy, actually.)
Concerning basic vs. extended RE syntax: grep -E uses the "extended" syntax, which is just different enough to mess you up. The relevant differences here are that in a basic RE, the characters "(){}" are treated as literal characters unless escaped (if escaped, they're treated as RE syntax indicating grouping and repetition); in an extended RE, this is reversed: they're treated as RE syntax unless escaped (if escaped, they're treated as literal characters).
That's why I suggest ^\([^:]:\)\{2\} in the first bullet point, but then actually use ^([^:]*:){2} in the proposed solution -- the first is basic syntax, the second is extended.
The other relevant difference -- and the reason I switched to extended for the actual solution -- is that only extended RE allows | to indicate alternatives, as in this|that|theother (which matches "this" or "that" or "theother"). I need this capability to match a 4-digit number starting with 1-4 or the specific number 5000 ([1-4][0-9]{3}|5000). There's simply no way to do this in a basic RE, so grep -E and the extended syntax are required here.
(There are also many other RE variants, such as Perl-compatible RE (PCRE). When using regular expressions, always be sure to know which variant your regex tool uses, so you don't use syntax it doesn't understand.)
ypcat passwd |awk -F: '$3>1000 && $3 <5000{print $1}'
awk here can go the task in a simple manner. Here we made ":" as the delimiter between the fields and put the condition that third field should be greater than 1000 and less then 5000. If this condition meets print first field.
I am new to ruby and I'm trying to work with regex.
I have a text which looks something like:
HEADING
Some text which is always non capitalized. Headings are always capitalized, followed by a space or nothing more.
YOU CAN HAVE MULTIPLE WORDS IN HEADING
I'm using this regular expression to choose all headings:
^[A-Z]{2,}\s?([A-Z]{2,}\s?)*$
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
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.)
I have a string seperated by \t and ,, but the number of \t is not fixed, for example :
a=["seg1\tseg2\t\tseg3,seg4"]
seg2 and seg3 is seperated by two \t.
So I try to split them by
a.split(/\t+|,/)
it print the right anwser :
["seg1", "seg2", "seg3", "seg4"]
And I also try this
a.split(/[\t+,]/)
but the answer is
["seg1", "seg2", "", "seg3", "seg4"]
Why ruby print different results?
Because \t+ inside [] does not mean "one or more tabs", it means "a tab or a plus". Since it finds two consecutive tabs, it splits twice, and the string in the middle becomes empty.
Most special characters, like . + * ? etc, when placed in an interval become "regular" characters. There are some exceptions, like ^ (which negates the interval when placed at the beginning), the \ (that escapes the next character(s), just like it does outside intervals) and the ] (that closes the interval; another [ is also disallowed there). So, [\t+,] actually means '\t' or '+' or ','.
Unfortunatly, I don't know any reference for the full set of characters that need or don't need escaping inside an interval. In doubt, I tend to escape just to be sure. In any case, an interval will always match a single character only, if you want something different you must put your quantifier outside the interval. (For example: [\t,]+, if you also admit two commas in a row; otherwise, your first regex is really the correct one)
I need to filter all lines with words starting with a letter followed by zero or more letters or numbers, but no special characters (basically names which could be used for c++ variable).
egrep '^[a-zA-Z][a-zA-Z0-9]*'
This works fine for words such as "a", "ab10", but it also includes words like "b.b". I understand that * at the end of expression is problem. If I replace * with + (one or more) it skips the words which contain one letter only, so it doesn't help.
EDIT:
I should be more precise. I want to find lines with any number of possible words as described above. Here is an example:
int = 5;
cout << "hello";
//some comments
In that case it should print all of the lines above as they all include at least one word which fits the described conditions, and line does not have to began with letter.
Your solution will look roughly like this example. In this case, the regex requires that the "word" be preceded by space or start-of-line and then followed by space or end-of-line. You will need to modify the boundary requirements (the parenthesized stuff) as needed.
'(^| )[a-zA-Z][a-zA-Z0-9]*( |$)'
Assuming the line ends after the word:
'^[a-zA-Z][a-zA-Z0-9]+|^[a-zA-Z]$'
You have to add something to it. It might be that the rest of it can be white spaces or you can just append the end of line.(AFAIR it was $ )
Your problem lies in the ^ and $ anchors that match the start and end of the line respectively. You want the line to match if it does contain a word, getting rid of the anchors does what you want:
egrep '[a-zA-Z][a-zA-Z0-9]+'
Note the + matches words of length 2 and higher, a * in that place would signel chars too.