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:
Related
I have this text:
2015-10-01 15:15:30 subject: Announcement: [Word To Find] Some other thext
My Goal is to match the date with the time:
(?s)(?<=^)(.+?)(?= subject\: Announcement\: )
And also the text within [ ]
(?s)(?<=\[)(.+?)(?=\])
How to get those two results in a single regex?
I'm going to chime in with a working regex, which although similar to other answers, has all redundancies removed:
^(?s)(.*?) subject: Announcement: \[(.*?)]
Which yields groups:
1. "2015-10-01 15:15:30"
2. "Word To Find"
See live demo.
Redundancies:
It is not necessary to escape ] except within a character class
It is never necessary to escape a colon :
The look behind (?<=^) is identical to simply ^, since both are zero-width assertions
Use regex alternation operator.
^(?s).*?(?= subject\: Announcement\: )|(?<=\[)[^\]]*(?=\])
DEMO
You can use a simple regex for that:
(.*)\s+subject.*\[(.*?)\]
Or
(.*)\s+subject.*\[([^]]+)\]
The first group contains the date, the second contains the text within the [ ].
You can use following regex to get both match :
(?<=^|\[)(.*?)(?=subject|\])
see demo https://regex101.com/r/hU2iZ3/2
Note that all you need is use a logical OR (|) between your precede tokens and next tokens.
Also note that if your have another brackets within your text you should use a negated character class instead .*:
(?<=^|\[)([^[\]]*?)(?=subject|\])
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:
[^]?[]
The code uses the following regular expression
img[src~=(?i)\\.(png|jpe?g)]
I'm not sure if the . is escaped or the \
the \ is escaped, which appears to be an error given what it's trying to do....
actually, you've taken that out of context. that's probably in a string. if it's in a string, then it's escaping the slash, and then that slash is escaping the dot.
the ~= means "ends with" and the (?i) switches it into case-insensitive mode.
errr... now that i think about it, that actually looks like a hybrid between a CSS selector (probably used in jquery) and a regex (being familiar with both syntaxes, I thought nothing of it!). The ~= doesn't do anything in a regex (they're literal chars) the [ and ] represent a character set though.
So...I don't know what the result of this is. I suspect someone got confused and tried mixing the two.
It means match case insensitively, any string that ends in:
\.png
\.jpeg
\.jpg
But this is dependant on context. If used in a context, were \ need to be escaped out at a higher level, then it means match case insensitively:
.png
.jpeg
.jpg
In this expression , '/' is escaped ,which in turn escapes the '.'
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 can't figure out how to match this:
[songstart]string with !##$%^&*( and stuff[songend]
I figured out how to match the songstart and songend, but not [songstart] and [songend].
Any tips? =/.
\[songstart\].*\[songend\]
That regex should work. See # http://www.rubular.com/r/TtuS1CZeJf
Are you "escaping" the brackets in your regex?
\[songstart\]
[ and ] are special characters so they need to be treated special.