Regex with equals sign not working in Notepad++ [duplicate] - regex

This question already has answers here:
Regex plus vs star difference? [duplicate]
(9 answers)
Closed 4 years ago.
Issue
I am having trouble matching lines with nothing but repeating equals signs (=) in Notepad++. I'm searching a plain text document for any line that begins with "=" and ends with "=". For instance, all these should match my regex:
========================
==================
==
=================================================
, etc.
My code
This is my regex:
^(=*)$
Not only does the regex not find equals signs, it falsely finds blank lines instead.
Rationale
^ = line begins with an equals sign.
=* = find any sequence of one or more equals signs
$= line ends with an equals sign
But my regex doesn't work. There must be some strange exception in Notepad++ because I verified that equals signs don't have to be escaped in JavaScript and my regex works fine at this online regex tester:
https://regex101.com/
Links I've found that haven't been fruitful
regex matching expression notepad++
Difficulties with adding spaces around equal signs using regular expressions with Notepad++
http://docs.notepad-plus-plus.org/index.php/Regular_Expressions
https://www.icewarp.com/support/online_help/203030104.htm
My questions
Why is my regex only returning blank lines?
If my regex is wrong, please explain why and what the correct regex is to do my find. Eventually I will do a replace as well, but I didn't want to cloud the issue.
Feedback on proposed duplicate
It was suggested that this post might be a duplicate of this question. This post is not a duplicate, and here is why:
Even if the content of the two posts were similar:
For a user to find the suggested post with "plus vs. star" in the title would suggest that he/she already had some idea what the problem was (i.e., use "plus" instead of "star").
Anyone else having this problem and being in my same predicament wouldn't necessarily know that plus vs star was the issue.
If the suggested post had come up as a possible answer when I searched "equals sign regex not working notepad++", I wouldn't have had to take my time to write this post.

* is zero or more.
+ is one or more.
Replace * with + in your regex.
So your regex would be ^(=+)$, this would match only lines with = and skip anything else.

Related

Using just regex is there a way to skip words or chars when using a lookaround? [duplicate]

This question already has answers here:
RegExp exclusion, looking for a word not followed by another
(3 answers)
Closed 4 years ago.
I guess specifically this might be about a negative look ahead.
If i had a sentence in the form of:
this is the WORD i want and I want and this is the PHRASE I DONT WANT
is there a way to use just regex to match "WORD" but only not if "PHRASE" is present? My initial idea was a negative lookahead but that is only the immediate word following. I then tried using (?:\w+(?:\s*[\,\-\'\:\/]\s*|\s+)){0,3} and other similar tricks but this would match the words in-between and not the actual phrase. Not to mention the wonkiness of + in lookarounds. Then I thought about using a grouping like [^something] but i didnt know how to do that with full words without a lookaround. I then had the idea to nest lookarounds which i found out can happen, but that still gives me the root of the problem.
Can you skip words in the matching for a lookaround and if not how would i go about solving this issue?
Because if i nest using a lookbehind I still need to skip stuff to get to WORD in order to match it.
Assume the words are arbitrary in the sentence but the key word and the key phrase is something specific.
If I understand your question, you can try:
/(?!.*PHRASE I DONT WANT)WORD/
Demo

Checkpoint regex [duplicate]

This question already has an answer here:
Reference - What does this regex mean?
(1 answer)
Closed 5 years ago.
I found these two expressions in documentation:
To match subdomains of mydomain.com: (^|.*\.)mydomain\.com
To match domain and subdomains of mydomain.com: (^|.*\.)*mydomain\.com
I can't understand why those expressions mean what they say they mean. Can anybody explain both expressions please?
Fist it is not a good regex expressions (it except other things that it should not) but i will explain the (^|.*\.)mydomain\.com (you will figure out the second)
between the parenthesis :
^ matches the starting position of the line
| acts like a Boolean OR ,between the expression before and the expression after the operator
.matches any character except line breaks
*Matches the preceding element zero or more times
\.matches a dot . character
For more information you could read wiki doc and use a great Regex tool

Regex Meaning (Regex Golf) [duplicate]

This question already has an answer here:
Reference - What does this regex mean?
(1 answer)
Closed 7 years ago.
Regex newbie here, so I was trying this website for fun: https://regex.alf.nu
In particular, I'm concerned about the "Ranges" section here: https://regex.alf.nu/2
I was able to get as far as ^[a-f]+, and couldn't figure out the rest. By accident, I added a $ to get ^[a-f]+$ which was actually the answer.
Trying to wrap my mind around the meaning of this regex. Can someone give the plain English explanation of what's happening here?
It seems to say "a string that starts and ends with one or more of the letters a through f," but that doesn't quite make sense for me, for instance, with the word "cajac" which seems to satisfy those conditions.
For those who can't see the URL, it's asking me to match these words:
abac
accede
adead
babe
bead
bebed
bedad
bedded
bedead
bedeaf
caba
caffa
dace
dade
daff
dead
deed
deface
faded
faff
feed
But NOT match these:
beam
buoy
canjac
chymia
corah
cupula
griece
hafter
idic
lucy
martyr
matron
messrs
mucose
relose
sonly
tegua
threap
towned
widish
yite
In English it means: Match any words which contain only the letters a thru f.
Your pattern, when broken down:
^ assert position at start of the string
[a-f]+ match a single character present in the list below:
+ Between one and unlimited times, as many times as possible, giving back as needed
a-f a single character in the range between a and f (case sensitive)
$ assert position at end of the string
You can also see a quick explanation of your patterns on the Regex101 webpage.

VB.NET Regular Expressions [duplicate]

This question already has answers here:
RegEx match open tags except XHTML self-contained tags
(35 answers)
Closed 9 years ago.
I have this HTML code:
<td class="Class 1">Example</td><td class="Class2">Other Example</td>
and I am trying to use Regular Expressions in VB.NET to extract "Example" and "Other Example"
Dim parsedtext As MatchCollection = Regex.Matches(htmlcode, ">(.+)<)
(the htmlcode variable contains the html code mentioned above as a string.)
However, looking at
parsedtext(0).Groups(0)
, it is returning ">Example</td><td class="Class2">Other Example<". I do not understand why this is happening, and I have tried many other pattern strings and cannot figure this problem out. How would one extract all text between two specific characters such as > and < in the example above?
I agree with #ColeJohnson (no one on SO is allowed to believe otherwise, at this point), but it's a good example for teaching the concept of greedy versus non-greedy matching.
By default, regular expressions quantifiers (+, *, ?) "eat up" as much as possible, and only eat less when some part of the match fails. That's called greedy matching. To make it non-greedy, you use non-greedy quantifiers: +?, *?, ??.
That is,
">(.+?)<"
In other words, your .+ continued to match as many character as possible, before finding a <; so you see, your output was to be expected. If, however, hypothetically, it had not found that last <, it would have backtracked to the last time it "saw" a <.

Regex - How to search for singular or plural version of word [duplicate]

This question already has answers here:
Regex search and replace with optional plural
(4 answers)
Closed 6 years ago.
I'm trying to do what should be a simple Regular Expression, where all I want to do is match the singular portion of a word whether or not it has an s on the end. So if I have the following words
test
tests
EDIT: Further examples, I need to this to be possible for many words not just those two
movie
movies
page
pages
time
times
For all of them I need to get the word without the s on the end but I can't find a regular expression that will always grab the first bit without the s on the end and work for both cases.
I've tried the following:
([a-zA-Z]+)([s\b]{0,}) - This returns the full word as the first match in both cases
([a-zA-Z]+?)([s\b]{0,}) - This returns 3 different matching groups for both words
([a-zA-Z]+)([s]?) - This returns the full word as the first match in both cases
([a-zA-Z]+)(s\b) - This works for tests but doesn't match test at all
([a-zA-Z]+)(s\b)? - This returns the full word as the first match in both cases
I've been using http://gskinner.com/RegExr/ for trying out the different regex's.
EDIT: This is for a sublime text snippet, which for those that don't know a snippet in sublime text is a shortcut so that I can type say the name of my database and hit "run snippet" and it will turn it into something like:
$movies= $this->ci->db->get_where("movies", "");
if ($movies->num_rows()) {
foreach ($movies->result() AS $movie) {
}
}
All I need is to turn "movies" into "movie" and auto inserts it into the foreach loop.
Which means I can't just do a find and replace on the text and I only need to take 60 - 70 words into account (it's only running against my own tables, not every word in the english language).
Thanks!
- Tim
Ok I've found a solution:
([a-zA-Z]+?)(s\b|\b)
Works as desired, then you can simply use the first match as the unpluralized version of the word.
Thanks #Jahroy for helping me find it. I added this as answer for future surfers who just want a solution but please check out Jahroy's comment for more in depth information.
For simple plurals, use this:
test(?=s| |$)
For more complex plurals, you're in trouble using regex. For example, this regex
part(y|i)(?=es | )
will return "party" or "parti", but what you do with that I'm not sure
Here's how you can do it with vi or sed:
s/\([A-Za-z]\)[sS]$/\1
That replaces a bunch of letters that end with S with everything but the last letter.
NOTE:
The escape chars (backslashes before the parens) might be different in different contexts.
ALSO:
The \1 (which means the first pattern) may also vary depending on context.
ALSO:
This will only work if your word is the only word on the line.
If your table name is one of many words on the line, you could probably replace the $ (which stands for the end of the line) with a wildcard that represents whitespace or a word boundary (these differ based on context).