I'm working on some code inherited from someone else and trying to understand some regular expression code in C#:
Regex.Replace(query, #"""[^""~]+""([^~]|$)",
m => string.Format(field + "_exact:{0}", m.Value))
What is the above regular expression doing? This is in relation to input from a user performing a search. It's doing a replace of the query string using the pattern provided in the second argument, with the value of the third. But what is that regular expression? For the life of me, it doesn't make sense. Thanks.
As far as I can see, xanatos' answer is correct. I tried to understand the regex, so here it comes:
"[^"~]+"([^~]|$)
You can test our regex and play with the single parts for better understanding at http://www.regexpal.com/
1.) a single character
"
The first pattern is a literal character. Since there is no statement of relative position, it can occur everywhere.
2.) a character class
[^"~]
The next expression is the []-bracket. This is a character set. It defines a quantity of characters, which maybe follow next. It is a placeholder for one single character... So lets see inside, which content is allowed:
^"~
The definition of the character class begins with an caret (^), which is a special character. Typing a caret after the opening square bracket will negate the character class. So it's "upside down": everything following, which does not match the class expression, matches and is a valid character.
In this case, every literal character is possible, except the two excluded ones: " or ~.
3.) a special character
+
The next expression, a plus, tells the engine to attempt to match the preceding token once or more.
So the defined character class should one or multiple times repeated to match the given expression.
4.) a single character
"
To match, the expression should contain furthermore one further apostrophe, which will be the corresponding apostrophe to the first one in 1.) since the character class in (2.) hence (3.) does not permit an apostrophe.
5.) a lookaround
([^~]|$)
The first structure here to examine is the ()-bracket. This is called a "Lookaround".
It is is a special kind of group. Lookaround matches a position. It does not expand the regex match.
So this means this part does not try to find any certain characters inside of an expression
rather then to localize them.
The localisation demands has two conditions, which are connected by a logical OR by the pipeline symbol: |
So the next character of the matched expression could either be
[^~] one single character out of the class everything excluding the character ~
or
$ the end of the line (or word, if multiline-mode is not used in regex engine)
I'll try to edit my answer to a better format, since this is my first post, I first have to check out how this is working.. :)
Update:
to "detect" a Asterisk/star in front/end of the line, you have to do following:
First it's a special character, so you have to escape it with an backslash: *
To define the position, you can use:
^ to look at the beginning of the line,
$ end of the line
The overall expression would be:
^* in front of the expression to search for an * at the beginning of
the line $* at the end of the regex to demand an * at the end.
.... in your case you can add the * in the last character class to detect an * in the end:
([^~]|$|$*)
and to force an * in the end, delete the other conditions:
($*)
PS:
(somehow my regex is swallowed up by formating engine, so my update is wrong...)
The # makes it necessary to escape all the " with a second ", so "". Without it to escape the " you would have used \", but I consider it better to always use # in regexes, because the \ is used quite often, and it's boring and unreadable to always have to escape it to \\.
Let's see what the regex really is:
Console.WriteLine(#"""[^""~]+""([^~]|$)");
is
"[^"~]+"([^~]|$)
So now we can look at the "real" regex.
It looks for a " followed by one or more non-" and non-~ followed by another " followed by a non-~ or the end of the string. Note that the match could start after the start of the string and it could end before the end of the string (with a non-~)
For example in
car"hello"help
it would match "hello"h
Related
I m trying to matching a string which will not allow same special character at same time
my regular expression is:
[RegularExpression(#"^+[a-zA-Z0-9]+[a-zA-Z0-9.&' '-]+[a-zA-Z0-9]$")]
this solve my all requirement except the below two issues
this is my string : bracks
acceptable :
bra-cks, b-r-a-c-ks, b.r.a.c.ks, bra cks (by the way above regular expression solved this)
not acceptable:
issue 1: b.. or bra..cks, b..racks, bra...cks (two or more any special character together),
issue 2: bra cks (two ore more white space together)
You can use a negative lookahead to invalidate strings containing two consecutive special characters:
^(?!.*[.&' -]{2})[a-zA-Z0-9.&' -]+$
Demo: https://regex101.com/r/7j14bu/1
The goal
From what i can tell by your description and pattern, you are trying to match text, which start and end with alphanumeric (due to ^+[a-zA-Z0-9] and [a-zA-Z0-9]$ inyour original pattern), and inside, you just don't want to have any two consecuive (adjacent) special characters, which, again, guessing from the regex, are . & ' -
What was wrong
^+ - i think here you wanted to assure that match starts at the beginning of the line/string, so you don't need + here
[a-zA-Z0-9.&' '-] - in this character class you doubled ' which is totally unnecessary
Solution
Please try pattern
^[a-zA-Z0-9](?:(?![.& '-]{2,})[a-zA-Z0-9.& '-])*[a-zA-Z0-9]$
Pattern explanation
^ - anchor, match the beginning of the string
[a-zA-Z0-9] - character class, match one of the characters inside []
(?:...) - non capturing group
(?!...) - negative lookahead
[.& '-]{2,} - match 2 or more of characters inside character class
[a-zA-Z0-9.& '-] - character class, match one of the characters inside []
* - match zero or more text matching preceeding pattern
$ - anchor, match the end of the string
Regex demo
Some remarks on your current regex:
It looks like you placed the + quantifiers before the pattern you wanted to quantify, instead of after. For instance, ^+ doesn't make much sense, since ^ is just the start of the input, and most regex engines would not even allow that.
The pattern [a-zA-Z0-9.&' '-]+ doesn't distinguish between alphanumerical and other characters, while you want the rules for them to be different. Especially for the other characters you don't want them to repeat, so that + is not desired for those.
In a character class it doesn't make sense to repeat the same character, like you have a repeat of a quote ('). Maybe you wanted to somehow delimit the space, but realise that those quotes are interpreted literally. So probably you should just remove them. Or if you intended to allow for a quote, only list it once.
Here is a correction (add the quote if you still need it):
^[a-zA-Z0-9]+(?:[.& -][a-zA-Z0-9]+)*$
Follow-up
Based on a comment, I suspect you would allow a non-alphanumerical character to be surrounded by single spaces, even if that gives a sequence of more than one non-alphanumerical character. In that case use this:
^[a-zA-Z0-9]+(?:(?:[ ]|[ ]?[.&-][ ]?)[a-zA-Z0-9]+)*$
So here the space gets a different role: it can optionally occur before and after a delimiter (one of ".&-"), or it can occur on its own. The brackets around the spaces are not needed, but I used them to stress that the space is intended and not a typo.
So i have the following RegEx for the purpose of finding and adding whitespace:
(\S)(\()
So for a string like "SomeText(Somemoretext)" I want to update this to "SomeText (Somemoretext)" it matches "t(" and so my replace eliminates the "t" from the string which is not good. I also do not know what the character could be, I'm merely trying to find the non-existence of whitespace.
Is there a better expression to use or is there a way to exclude the found character from the match returned so that I can safely replace without catching characters i do not want to replace?
Thanks
I find lookarounds hard to read and would prefer using substitutions in the replacement string instead:
var s = Regex.Replace("test1() test2()", #"(\S)\(", "$1 (");
Debug.Assert(s == "test1 () test2 ()");
$1 inserts the first capture group from the regex into the replacement string which is the non-space character before the opening parenthesis (.
If you need to detect the absence of space before a specific character (such as bracket) after a word, how about the following?
\b(?=[^\s])\(
This will detect words ( [a-zA-z0-9_] that are followed by a bracket, without a space).
(if I got your problem correctly) you can replace the full match with ( and get exactly what you need.
In case you need to look for absence spaces before a symbol (like a bracket) in any kind of text (as in the text may be non-word, such as punctuation) you might want to use the following instead.
^(?:\S*)(\()(?:\S*)$
When using this, your result will be in group 1, instead of just full match (which now contains the whole line, if a line is matched).
I'm trying to learn regex, but still have no clue. I have this line of code, which successfully seperates the placeholder 'FirstWord' by the '{' delimiter from all following text:
var regexp = new Regex(#"(?<FirstWord>.*?)\{(?<TextBetweenCurlyBrackets>.*?)\}");
Which reads this string with no problem:
Greetings{Hello World}
What I want to do is to replace the '{' with a character chain like for instance '/>>'
so I tried this:
var regexp = new Regex(#"(?<FirstWord>.*?)\/>>(?<OtherText>.*?)\");
I removed the last bracket and replaced the first one with '/>>' But it throws an ArgumentException. How would the correct character combination look like?
/ does not need to be escaped, unless you use it as the pattern-delimiter.:
#"(?<FirstWord>.*?)/>>(?<OtherText>.*?)\"
Also your last \ will basically escape the " which should end the String (c#-wise: remove it):
#"(?<FirstWord>.*?)/>>(?<OtherText>.*?)"
And since you want most likely fetch until the END of the String (.*? will fetch as less characters as required to satisfy the expression), you should use the $ at the end or use any other sort of delimiter (whitspace, linebreak, etc...).
#"(?<FirstWord>.*?)/>>(?<OtherText>.*?)$"
Example:
(.*?)/>>(.*?)$
Debuggex Demo
Removing the trailing $ will fetch the empty string for the second match group, because "" is the shortest string possible satisfying the expression .*?
(.*?)/>>(.*?)$ on This/>>Test One will match This and Test One
(.*?)/>>(.*?)\s on This/>>Test One will match This and Test
(.*?)/>>(.*?) on This/>>Test One will match This and ""
Note: I'm saying "" is the shortest string possible satisfying the expression .?* on purpose! A frequent Misstake is to interpret .*?a as "everything until a":
Regex is greedy by default!
Searching for the expressiong (.*?)a$ on "caba" will NOT fail to match - it will return cab!, because cab followed by a is satisfying the expression AND cab is the shortest string possible for any match.
One might also expect b to be matched - but regex is working from left to right, hence aborting once it found cab - even if b would be shorter.
I'm currently using the following line of code:
Regex Regex_Alpha = new Regex(#"[a-zA-Z]+('[a-zA-Z])?[a-zA-Z]*");
What I want to do is filter the input of text fields with the condition that input should only be letters and the apostrophe symbol (actually, I still want to add more, but I'm trying to resolve this first).
Right now, it is accepting ALL characters, even numbers.
With my understanding of Regex, I tried to formulate my own expression in the line of:
Regex Regex_Alpha = new Regex(#"^[a-zA-Z'-"+$);
It filters numbers, but doesn't accept the apostrophe symbol. Tried to remove the # sign and filter the apostrophe with the backslash escape character, but still no use.
What should be the best approach to filter the input so that it only accepts letters and apostrophe? (I'll do the rest of the symbols once I understand how this one should work)
As I've commented, your first regular expression is a pretty good shot at "letters, with a single apostrophe not at either end". However, it matchs any string with even a single letter because a regular expression looks for any match in the input, not for whether the entire input matches.
You can fix this by doing what you've done in your second regular expression - just put a ^ at the start and a $ at the end. This means the start and end of the expression have to match the start and end of the input, so it ensures the whole input is only made up of letters and a possible apostrophe.
Regarding your second regular expression, you have a few of problems.
If you want a double-quote in a #"..." string literal, you need to put two double quotes. (I think this might just be a typing mistake in your question, as what you currently have wouldn't even compile.)
You need to close your character class with a ], otherwise the [ and everything inside just get treated as a sequence of characters to match, one after the other.
If you want a hyphen in a character class, it has to go at the start or end, or it gets mistaken for a "between" hyphen (as in A-Z).
The expression #"^[a-zA-Z'""-]+$" should match "any string entirely made of letters, apostrophes, quotes or hyphens".
I have an input like the following
[a href=http://twitter.com/suddentwilight][font][b][i]#suddentwilight[/font][/a] My POV: Rakhi Sawant hits below the belt & does anything for attention... [a href=http://twitter.com/mallikaLA][b]http://www.test.com[/b][/a] has maintained the grace/decency :)
Now I need to get the string #suddentwilight and http://www.test.com that comes inside the anchor tags. there might be some [b] or [i] like tags wrapping the actual text. I need to ignore that.
Basically I need to get a string matching that starts with [a] then need to get the string/url before closing of the a tag [/a].
Please Suggest
I don't know C#, but here's a regex:
/\[a\s+[^\]]*\](?:\[[^\]]+\])*(.*?)(?:\[[^\]]+\])*\[\/a\]/
This will match [a ...][tag1][tag2][...][tagN]text[/tagN]...[tag2][tag1][/a] and capture text.
To explain:
the /.../ are common regex delimiters (like double quotes for strings). C# may just use strings to initialize regexes - in which case the forward slashes aren't necessary.
\[ and \] match a literal [ and ] character. We need to escape them with a backslash since square brackets have a special meaning in regexes.
[^\]] is an example of a character class - here meaning any character that is not a close square bracket. The square brackets delimit the character class, the caret (^) denotes negation, and the escaped close square bracket is the character being negated.
* and + are suffixes meaning match 0 or more and 1 or more of the previous pattern, respectively. So [^\]]* means match 0 or more of anything except a close square bracket.
\s is a shorthand for the character class of whitespace characters
(?:...) allows you to group the contents into an atomic pattern.
(...) groups like (?:...) does, but also saves the substring that this portion of the regex matches into a variable. This is normally called a capture, since it captures this portion of the string for you to use later. Here, we are using a capture to grab the linktext.
. matches any single character.
*? is a suffix for non-greedy matching. Normally, the * suffix is greedy, and matches as much as it can while still allowing the rest of the pattern to match something. *? is the opposite - it matches as little as it can while still allowing the rest of the pattern to match something. The reason we use *? here instead of * is so that if we have multiple [/a]s on a line, we only go as far as the next one when matching link text.
This will only remove [tag]s that come at the beginning and end of the text, to remove any that come in the middle of the text (like [a href=""]a [b]big[/b] frog[/a]), you'll need to do a second pass on the capture from the first, scrubbing out any text that matches:
/\[[^\]]+\]/