I need to find lines that are 3 digits and 3 other characters: I thought I use the following RegEx:
^\d{3}\D{3}$
But take the following sample text file and run the RegEx above (the text must have the empty lines in it):
1
12
123xxx
123y
aaabb
The problem is that there are two matches: 123xxx (which is fine), but also 123y is matched!
I suspect the reason is that "y" + the end-of-line + the beginning-of-next-line are also matched.
How can I tell the regex engine to ignore line beginnings and endings with \D and match characters only, not positions?
The behavior of $ in UltraEdit changes depending on whether you have "Match Whole Word Only" checked or not. To get the behavior you want you need to make sure that that option is checked. Your regular expression doesn't need to change.
Maybe:
/^\d{3}\D{3}$/m
The m means
Treat string as multiple lines. That is, change "^" and "$" from matching the start or end of the string to matching the start or end of any line anywhere within the string.
http://perldoc.perl.org/perlre.html
I don't know about UltraEdit exactly but I expect it will have something similar.
Try this :
^\d{3}[\S]{3}$
Match lines with 3 digits followed by three characters that are not blank characters.
Related
I'm using SubtitleEdit and I'd like to locate all the lines that do not contain a line break.
Because lines containing a line break indicates they are bilingual, which I want.
But those that do not have line breaks are mono-lingual, and I'd like to quickly locate them all and delete them. TIA!
Alternatively, if there is a regex expression that can find lines which do not contain any English characters, that would also work.
The confusion here was caused by 2 facts:
What SubtitleEdit calls a line is actually a multiline, containing
newlines.
The newline displayed is not the one used internally (so it would never match <br>).
Solution 1:
Now that we have found out it uses either \r\n or just \n, we can write a regex:
(?-m)^(?!.*\r?\n)[\s\S]*$
Explanation:
(?-m) - turn off the multiline option (which is otherwise enabled).
^ - match from start of text
(?!.*\r?\n) - negative look ahead for zero or more of any characters followed by newline character(s) - (=Contains)
[\s\S]*$ - match zero or more of ANY character (including newline) - will match the rest of text.
In short: If we don't find newline characters, match everything.
Now replace with an empty string.
Solution 2:
If you want to match lines that doesn't have any English characters, you can use this:
(?-m)^(?![\s\S]*[a-zA-Z])[\s\S]*$
Explanation:
(?-m) - turn off the multiline option (which is otherwise enabled).
^ - match from start of text
(?![\s\S]*[a-zA-Z]) - negative look ahead for ANY characters followed by an English character.
[\s\S]*$ - match zero or more of ANY character (including newline) - will match the rest of text.
In short: If we don't find an English character, match everything.
Now replace with an empty string.
You should use regex assert. Given test lines:
something_1
some<br>thing_2
something_3<br>
<br>something_4
something_5
This is an expression that will match lines 1 and 5
^(?!.*<br>).*$
In this regular expression we have the negative lookahead assertion (?!.*<br>) that allows us to define what line is suitable for us
I cannot figure a way to make regular expression match stop not on end of line, but on end of file in VS Code? Is it a tool limitation or there is some kind of pattern that I am not aware of?
It seems the CR is not matched with [\s\S]. Add \r to this character class:
[\s\S\r]+
will match any 1+ chars.
Other alternatives that proved working are [^\r]+ and [\w\W]+.
If you want to make any character class match line breaks, be it a positive or negative character class, you need to add \r in it.
Examples:
Any text between the two closest a and b chars: a[^ab\r]*b
Any text between START and the closest STOP words:
START[\s\S\r]*?STOP
START[^\r]*?STOP
START[\w\W]*?STOP
Any text between the closest START and STOP words:
START(?:(?!START)[\s\S\r])*?STOP
See a demo screenshot below:
To matcha multi-line text block starting from aaa and ending with the first bbb (lazy qualifier)
aaa(.|\n)+?bbb
To find a multi-line text block starting from aaa and ending with the last bbb. (greedy qualifier)
aaa(.|\n)+bbb
If you want to exclude certain characters from the "in between" text, you can do that too. This only finds blocks where the character "c" doesn't occur between "aaa" and "bbb":
aaa([^c]|\n)+?bbb
I'm trying to improve with regex as I'm tired of constantly having to look up existing solutions instead of creating my own. Having a bit of difficulty understanding why this isn't working though:
Trying to extract both phone numbers from the following string (numbers and address are random):
+1-541-754-3010 156 Alphand_St. <J Steeve>\n 133, Green, Rd. <E Kustur> NY-56423 ;+1-541-914-3010\n"
So I'm using the following expression:
/\+(.+)(?:\s|\b)/
These are the matches I'm getting back:
1-541-754-3010 156 Alphand_St.
1-541-914-3010
So I'm getting the last one correctly, but not the first one. Based on the expression, it should match anything from between a + and a space/boundary. But for some reason it's not stopping at the space after the first number. Am I going about this the wrong way?
In the format you provided for the search string, and since you are starting with a literal "+", I would just include the next following string of decimals and separators, like the hyphen:
/\+([0-9\-]+)/
Your ".+" says to match everything until there's a \s. However that also includes \s on the way to the \s.
Remember that dashes - are not word characters, so \b will match between, for example, 1- and -5 and so on. Also, your current regex is greedy - it'll try to match as many characters as it can with the repeated ., which is why it goes all the way to the end of the first line (because after the last character in the line matches \b). Making it lazy (with .+?) wouldn't fix it, though, because then it would terminate right after the 1 in 1-541 (because between 1- is a word boundary)
Try using a character set of digits and - instead:
\+([\d-]+)
https://regex101.com/r/ktbcHJ/1
I have tried (^[.*]{1,50}$)/gm but it simply does not work.
I'd like a line made up of any characters to match this regex.
Qwertyuiop
$$%%^^89e7hbequdwanjk
etc should all match, including this line
However, lines over 50 characters long should not match.
You are specifying a string of 1-50 occurrences of either . or *. If you want a string of any characters, the [...] character class is wrong (it enumerates literal characters you want to match); you are looking for . without square brackets, which matches any one character.
The regular expression for that is
^.{1,50}$
Some languages require you to specify a separator such as /.../ around your regex, but it's hard to tell from your example whether yours is one of them; in this case, you are missing the beginning separator.
The /g flag only makes sense if you need to find multiple occurrences on the same line. The /m flag makes sense if the ^ and $ anchors should match newlines in multi-line text.
If the title of your question is correct, and you want properly under 50 characters, change the 50 to 49 (and maybe the 1 to 0).
Your regex, [.*] matches only dots . and *, since inside [] both are treated literally. Try
/^.{1,50}$/gm
It'll match between 1 and 50 of anything. If you also want to capture it add back the parenthesis
/(^.{1,50}$)/gm
I have a logfile that contains lines like:
....
unit:
...
unit.integration:
....
I would like to run a RegEx search on the file using notepad++ that returns all lines that: start with a word with no blanks and ends with :\n. I have tried:
(.:\n)
but that gives 0 results. I have looked at:
http://www.aivosto.com/vbtips/regex.html
EDIT: Updated with more specific requirements to the starting word.
I would like to run a RegEx search ... that returns all lines that start with anything but ends with :\n.
The first part of the match is :, and because you're looking for line breaks, you'll want to match the end of the line, which is $.
You said you don't care about what comes before it, so don't even include it in the regex:
:$
Update to address edit:
I would like to run a RegEx search ... that returns all lines that: start with a word with no blanks and ends with :\n.
This differs from your original post, now you want to match a "word with no blanks", however that implies that it's a single word, and doesn't contain special characters.
It seems to me that you'd like to match unit.integration which is two words and a separating . character.
If you want what you asked for ("a word with no blanks") then just prepend ^\w+ to the regex:
^\w+:$
(matches unit:, but not unit.integration:)
If, instead you want to match lines that don't contain spaces and end in :, then you should use ^\S+ instead:
^\S+$
(matches unit: and unit.integration:, but also matches ##*()$&*(&:)
The details matter, so avoid assumptions and be as explicit as possible in what you want matched.
To get all lines that have no spaces and end in :, use
^\S+:\R
\S matches non-whitespace symbols only, and \R means any line break.
See screenshot:
If you plan to match the last line, too, replace \R with $ (end of line or whole file metacharacter).