I want to make a negated character class to match a square bracket tag like this [square bracket tag]. The problem is, the ] character ends the character class!
I tried
\[[^\]]+]
but I get a syntax error when I run it. (This is in the find and replace regex engine which is slightly different than the standard .NET engine fyi).
You forgot to escape the final end bracket:
\[[^\]]+\]
The first example in msdn uses \\ for escaping the \ which then escapes the .. So you should do something like \\[[^\\]]+\\] and also as Damien_The_Unbeliever said you haven't closed the final bracket.
I definitely expected escaping with "\" but it didn't work for me (grep#MacOS) but this:
[^]]
did the job.
Just place ] as the first character in class.
I actually used something like:
[^]?[]
Related
This regex:
"REGION\\((.*?)\\)(.*?)END_REGION\\((.*?)\\)"
currently finds this info:
REGION(Test) my user typed this
END_REGION(Test)
I need it to instead find this info:
#region REGION my user typed this
#endregion END_REGION
I have tried:
"#region\\ (.*?)\\\n(.*?)#endregion\\ (.*?)\\\n"
It tells me that the pattern assignment has failed. Can someone please explain what I am doing wrong? I am new to Regex.
It seems the issue lies in the multiline \n. My recommendation is to use the modifier s to avoid multiline complexities like:
/#region\ \(.*?\)(.*?)\s#endregion\s\(.*?\)/s
Online Demo
s modifier "single line" makes the . to match all characters, including line breaks.
Try this:
#region(.*)?\n(.*)?#endregion(.*)?
This works for me when testing here: http://regexpal.com/
When using your original text and regex, the only thing that threw it off is that I did not have a new line at the end because your sample text didn't have one.
Constructing this regex doesn't fail using boost, even if you use the expanded modifier.
Your string to the compiler:
"#region\\ (.*?)\\\n(.*?)#endregion\\ (.*?)\\\n"
After parsed by compiler:
#region\ (.*?)\\n(.*?)#endregion\ (.*?)\\n
It looks like you have one too many escapes on the newline.
if you present the regex as expanded to boost, an un-escaped pound sign # is interpreted as a comment.
In that case, you need to escape the pound sign.
\#region\ (.*?)\\n(.*?)\#endregion\ (.*?)\\n
If you don't use the expanded modifier, then you don't need to escape the space characters.
Taking that tack, you can remove the escape on the space's, and fixing up the newline escapes, it looks like this raw (what gets passed to regex engine):
#region (.*?)\n(.*?)#endregion (.*?)\n
And like this as a source code string:
"#region (.*?)\\n(.*?)#endregion (.*?)\\n"
Your regular expression has an extra backslash when escaping the newline sequence \\\n, use \\s* instead. Also for the last capturing group you can use a greedy quantifier instead and remove the newline sequence.
#region\\ (.*?)\\s*(.*?)#endregion\\ (.*)
Compiled Demo
Regular expressions are not strong point.
I can do simple stuff, but this one has just got my goat !!
So could someone give me a hand with this one.
Here's the comment in the code :
// If utf8 detection didnt work before, strip those weird characters for an underscore, as a last resort.
eregi_replace("[^a-z0-9 \-\.\(\)\/\\]","_",$str);
to (here's what I tried)
preg_replace("{[^a-z0-9 \-\.\(\)\/\\]}i","_",$str);
Any regex pros out there who give me a hand?
You need to specify regexp identifier such as # or /
preg_replace("#[^a-z0-9 \-\.\(\)\/\\]#i","_",$str);
So you should enclose your regular expression in those identifier characters.
First, I believe the { and } are fine as delimiters for the expression from the flags, but I know there are some regex flavors that don't support it, so it might be a good idea to just use something like ! or #
Second, I am not sure how the expression before worked, because AFAIK escaping with a \ character does not work with ERE expressions. You have to represent special characters like ^, -, and ] by their position within the class (^ cannot be the first character, ] must be the first character, and - must be either the first or the last character). The - character in the first expression would be interpreted as a range specifier (in this case a character in the range between \ and \). Additionally, the \ characters are treated literally, so you've got a confusing looking and largely redundant regex.
The replacement expression, however, needs to be in preg notation/flavor, so there are rule changes:
Very few things need to be escaped in a character class, even with the new rules
The \ character needs to be escaped twice - once for the string, and then one more time for the regex - otherwise, it will escape the closing bracket ]
Assuming you want to match a dash (or rather match something OTHER than a dash, it needs to be moved to the end of the class
So, here is some code (link) that I believe does what you need it to do:
$source = 'hello! ##$%^&* wazzup-dawg?.()/\\[]{}<>:"';
$blah = preg_replace('![^a-z0-9 .()/\\\\-]!i','_',$source);
print($blah);
preg_replace("{[^a-z0-9]-.()/\/}i","_",$str)
works just fine.
I tried it with all # and / and { and they all worked.
I have data that looks like this:
[Shift]);[Ctrl][Ctrl+S][Left mouse-click][Backspace][Ctrl]
I want to find all [.*] tags that have the word mouse in them. Keeping in mind non-greedy specifiers, I tried this in Vim: \[.\{-}mouse.\{-}\], but this yielded this result,
[Shift]);[Ctrl][Ctrl+S][Left mouse-click]
Rather than just the desired,
[Left mouse-click]
Any ideas? Ultimately I need this pattern in Perl syntax as well, so if anyone has a solution in Perl that would also be appreciated.
\[[^]]*mouse[^[]*\]
That is, match a literal opening bracket, then any number of characters that aren't closing brackets, then "mouse," then any number of non-opening-brackets, and finally a literal closing bracket. Should be the same in Perl.
You can use the following regex:
\[[^\]]*mouse.*?\]
I want a regex that matches a square bracket [. I haven't found one yet. I think I tried all possibilities, but haven't found the right one. What is a valid regex for this?
How about using backslash \ in front of the square bracket. Normally square brackets match a character class.
Try using \\[, or simply \[.
If you want to match an expression starting with [ and ending with ], use \[[^\]]*\].
Here is the meaning of each part (as explained at www.regexr.com):
Are you escaping it with \?
/\[/
Here's a helpful resource to get started with Regular Expressions:
Regular-Expressions.info
If you're looking to find both variations of the square brackets at the same time, you can use the following pattern which defines a range of either the [ sign or the ] sign: /[\[\]]/
In general, when you need a character that is "special" in regexes, just prefix it with a \. So a literal [ would be \[.
does it work with an antislash before the [ ?
\[ or \\[ ?
If you want to remove the [ or the ], use the expression: "\\[|\\]".
The two backslashes escape the square bracket and the pipe is an "or".
For a pure exgex, it's simple:
1 /[]abcde]/ - it's the way to include the ']' in the class.
2 /[abc[de]/ - freely put anything else inside the brackets, including the '['. (Most of the meta-characters lose their special meaning inside '[]').
3 Test(verify) your regex w/ 'grep' or 'vim' etc. first.(They are easy-going guys.)
4 It's not too late to try inserting '\' at this moment if your scripting environment doesn't agree.
The below expression is able to detect the timing 65
\[(?<timeElpsed>\d+)ms\]
from the below log:
[2022-12-16T04:51:55.993+0000] [ INFO] [scala-execution-context-global-75] [200 OK]: [100.107.99.132] [65ms] "content-length:
I think I'm burnt out, and that's why I can't see an obvious mistake. Anyway, I want the following regex:
#BIZ[.\s]*#ENDBIZ
to grab me the #BIZ tag, #ENDBIZ tag and all the text in between the tags. For example, if given some text, I want the expression to match:
#BIZ
some text some test
more text
maybe some code
#ENDBIZ
At the moment, the regex matches nothing. What did I do wrong?
ADDITIONAL DETAILS
I'm doing the following in PHP
preg_replace('/#BIZ[.\s]*#ENDBIZ/', 'my new text', $strMultiplelines);
The dot loses its special meaning inside a character class — in other words, [.\s] means "match period or whitespace". I believe what you want is [\s\S], "match whitespace or non-whitespace".
preg_replace('/#BIZ[\s\S]*#ENDBIZ/', 'my new text', $strMultiplelines);
Edit: A bit about the dot and character classes:
By default, the dot does not match newlines. Most (all?) regex implementations have a way to specify that it match newlines as well, but it differs by implementation. The only way to match (really) any character in a compatible way is to pair a shorthand class with its negation — [\s\S], [\w\W], or [\d\D]. In my personal experience, the first seems to be most common, probably because this is used when you need to match newlines, and including \s makes it clear that you're doing so.
Also, the dot isn't the only special character which loses its meaning in character classes. In fact, the only characters which are special in character classes are ^, -, \, and ]. Check out the "Metacharacters Inside Character Classes" section of the character classes page on Regular-Expressions.info.
// Replaces all of your code with "my new text", but I do not think
// this is actually what you want based on your description.
preg_replace('/#BIZ(.+?)#ENDBIZ/s', 'my new text', $contents);
// Actually "gets" the text, which is what I think you might be looking for.
preg_match('/(#BIZ)(.+?)(#ENDBIZ)/s', $contents, $matches);
list($dummy, $startTag, $data, $endTag) = $matches;
This should work
#BIZ[\s\S]*#ENDBIZ
You can try this online Regular Expression Testing Tool
The mistake is the character group [.\s] that will match a dot (not any character) or white space. You probably tried to get .* with . matching newline characters, too. You achieve this by enabling the single line option ((?s:) does this in .NET regex).
(?s:#BIZ.*?#ENDBIZ)
Depending on the environment you're using your regex in, it may need special care to properly parse multiline text, eg re.DOTALL in Python. So what environment is that?
you can use
preg_replace('/#BIZ.*?#ENDBIZ/s', 'my new text', $strMultiplelines);
the 's' modifier says "match the dot with anything, even the newline character". the '?' says don't be greedy, such as for the case of:
foo
#BIZ
some text some test
more text
maybe some code
#ENDBIZ
bar
#BIZ
some text some test
more text
maybe some code
#ENDBIZ
hello world
the non-greediness won't get rid of the "bar" in the middle.
Unless I am missing something, you handle this the same way that you would in Perl, with either the /m or /s modifier at the end? Oddly enough the other answers that rather correctly pointed this out got down voted?!
It looks like you're doing a javascript regex, you'll need to enable multiline by specifying the m flag at the end of the expression:
var re = /^deal$/mg