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I tried to find all comments beginning with // that don't have a space after the slashes.
I want to select only the slashes. No whitespace or text before that, no whitespace or text after that.
So far I've reached to [\s].(\/\/(?! )) but it catches the space before the slashes as well.
Basically I wanna make sure my line comments have a space after the slashes.
I'm trying to do this either in JavaScript or in any text editor.
Since javascript doesn't have the lookbehind feature, you can't.
The workaround (for instance, in a replacement context) is to use a capture group for the character before the two slashes and to start the replacement string with a reference to this group ('$1replacement'):
([^/\s]|^)//(?! )
You can use the following regex:
.*(\/\/(?= )) demo
The idea is to use positive lookahead and capture the // iff it is followed by a space.
EDIT: Just noticed that your question is contradictory. So if you want to capture if the // is not followed by a space, use this: .*(\/\/(?=\S)). Otherwise use the one above.
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The problem in Akeneo seems to be that simple regex combinations not working. I think the functionality (a single or group regex combination) is not integrated/implemented proper in Akeneo. If there is anybody out there who knows a trick to do a regex combination please let me know.
Tried to figure out how to make regex with | OR working in Akeneo "attributes".
the simple Example not working either a syntax error or no matching in Akeneo:
find this "323"
or find "123456"
\d{3}|\d{6}
Can anybody help?
According to the documentation, you need to use regex literal notation, and anchor the match both at the start and end of the string (so, add a grouping):
/^(\d{6}|\d{3})$/
Here, / are regex delimiters, ^ matches the start of string, (...) is a capturing group that contains two alternatives, six digits or three digits, and then end of string anchor, $, follows.
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My User data can come in any of the following 3 ways -
user="dc\AAA", user="BBB", user=CCCC,
Now, the bottom two I am able to extract it easily but issue comes when user data has an additional prefix of "dc" to it
I am trying to remove that prefix using regex and format all user data in single regex as below, but the unable to do so
user=AAA user=BBB user=CCC
Can someone please help.
This regex should do the work: (?:.*\\)?(.*).
Let's split this regex into parts:
(?: ) - A non-capturing group
.*\\ - Any characters many times, trailing by backslash
? (after the brackets) indicates the data in the brackets may occur once or not at all
(.*) Any characters
Overall - Capturing the data after the backslash if exists
I suggest using this amazing website for trying regex
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Is there a RegEx to remove everything but digits and spaces in Notepad++?
I know there is one that removes only digits but i dont need that.
PS: I do not want the lines to be removed
Example:
11234123 alex-james
1412412 mmafsmasdas
After regex:
11234123
1412412
As the pattern use [^\d ]+. Almost what Poul Bak proposed, but change * into +, i.e. the sequence of chars to match should be non-empty.
There is no point in searching for an empty string and replace it with another empty string.
correction:
try this:
([^0-9| ]+)
this will surely work!!!
open the file in notepad++ press CTRL+H check the box with search mode regular expression and put in the above regex and click replace all
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I want any regex to remove everything after underscore in all anchor Tag e.g
input: Text
Output Text
Although you should avoid parsing HTML with regex, but since this is a case of anchor tag which won't be nested, hence you can do a quick work using regex. Use this regex to match the data in group1 and group2,
(<a\s+[^>]*?href=["'][^']*?)_.*?(["'])
and replace it with \1\2 (or $1$2 as per the language)
Check the demo
You haven't mentioned how should the data be replaced in case there are multiple underscores in the href attribute, so for now I have done it in a way where it replaces everything from first occurrence of underscore but you can easily do it for last occurrence of underscore by making the regex as greedy.
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I hate to ask specific question, but i need the regex code for matching strings like :
{block any_single_word_here}
Anything Here
{/block}
Your original query is very close, albeit a little verbose. It can be shortened to:
/{block (.+?)}(.+?){\/block}/
(The ? modifier stops the + from being "greedy", so you don't have to explicitly stop the match at the next } or {.)
Next you have to consider that . won't match newlines by default. You can change this with the /s flag:
/{block (.+?)}(.+?){\/block}/s
Here's a demo.
And here's the documentation from man perlre:
s
Treat string as single line. That is, change "." to match any character whatsoever, even a newline, which normally it would not match.
In javascript (added escape to end of lines in original string) I added a \s* (zero or more spaces) to your regex and it outputs a match for any_single_word_here and Anything Here fine...
alert( "{block any_single_word_here}\
Anything Here\
{/block}".match(/\{block ([^\}]+)\}\s*([^\{]+)?\{\/block}/) )
For simpler regex, remove the unneccessary escaping, and just capture with . rather than complex [^\}]
/{block (.+)}\s*(.+){\/block}/
Does this do the job?
/\{block \w+\}\r?\n.*?\r?\n\{\/block\}/i