I need my C# regex to only match full words, and I need to make sure that +-/*() delimit words as well (I'm not sure if the last part is already set that way.) I find regexes very confusing and would like some help on the matter.
Currently, my regex is:
public Regex codeFunctions = new Regex("draw_line|draw_rectangle|draw_circle");
Thank you! :)
Try
public Regex codeFunctions = new Regex(#"\b(draw_line|draw_rectangle|draw_circle)\b");
The \b means match a word boundary, i.e. a transition from a non-word character to a word character (or vice versa).
Word characters include alphabet characters, digits, and the underscore symbol. Non-word characters include everything else, including +-/*(), so it should work fine for you.
See the Regex Class documentation for more details.
The # at the start of the string makes the string a verbatim string, otherwise you have to type two backslashes to make one backslash.
Do you want to match any words, or just the words listed above? To match an arbitrary word, substitute this for the bit that creates the Regex object:
new Regex (#"\b(\w+)\b");
In the future, if you want more characters to be treated as whitespace (for example, underscores), I would recommend String.Replace-ing them to a space character. There may be a clever way to get the same effect with regular expressions, but personally I think it would be too clever. The String.Replace version is obvious.
Also, I can't help but recommend that you read up on regular expressions. Yes, they look like line noise until you get used to them, but once you do they're convenient and there are plenty of good resources out there to help you.
Related
I have searched through some questions but couldn't find the exact answer i am looking for.
I have a requirement to search through large strings of text looking for keywords matches. I was using IndexOf, however, i require to find whole word matches e.g. if i search for Java, but the text contains JavaScript, it shouldn't match. This works fine using \b{pattern}\b, but if i search for something like C#, then it doesn't work.
Below is a few examples of text strings that i am searching through:
languages include Java,JavaScript,MySql,C#
languages include Java/JavaScript/MySql/C#
languages include Java, JavaScript, MySql, C#
Obviously the issue is with the special character '#'; so this also doesn't work when searching for C++.
Escape the pattern using Regex.Escape and replace the context-dependent \b word boundaries with (?<!\w) / (?!\w) lookarounds:
var rx = $#"(?<!\w){Regex.Escape(pattern)}(?!\w)";
The (?<!\w) is a negative lookbehind that fails the match if there is a start of string or a non-word char immediately before the current location, and (?!\w) is a negative looahead that fails the match if there is an end of string or a non-word char immediately after the current location.
Yeah, this is because there isn't a word boundary (a \b) after the #, because # isn't a "word" character. You could use a regular expression like the following, which searches for a character that isn't a part of a language name [^a-zA-Z+#] after the language:
\b{pattern}[^a-zA-Z+#]
Or, if you believe you can list all of the possible characters that aren't part of a language name (for example, whitespace, ,, ., and ;):
[\s,.;]{pattern}[\s,.;]
Alternately, if it is possible that a language name is at the very end of a string (depending on what you're getting the data from), you might need to also match the end of the string $ in addition to the separators, or similarly, the beginning of the string ^.
[\s,.;]{pattern}(?:[\s,.;]|$)
I know the regex for excluding words, roughly anyway, It would be (!?wordToIgnore|wordToIgnore2|wordToIgnore3)
But I have an existing, complicated regex that I need to add this to, and I am a bit confused about how to go about that. I'm still pretty new to regex, and it took me a very long time to make this particular one, but I'm not sure where to insert it or how ...
The regex I have is ...
^(?!.*[ ]{2})(?!.*[']{2})(?!.*[-]{2})(?:[a-zA-Z0-9 \:/\p{L}'-]{1,64}$)$
This should only allow the person typing to insert between 1 and 64 letters that match that pattern, cannot start with a space, quote, double quote, special character, a dash, an escape character, etc, and only allows a-z both upper and lowercase, can include a space, ":", a dash, and a quote anywhere but the beginning.
But I want to forbid them from using certain words, so I have this list of words that I want to be forbidden, I just cannot figure out how to get that to fit into here.. I tried just pasting the whole .. "block" in, and that didn't work.
?!the|and|or|a|given|some|that|this|then|than
Has anyone encountered this before?
ciel, first off, congratulations for getting this far trying to build your regex rule. If you want to read something detailed about all kinds of exclusions, I suggest you have a look at Match (or replace) a pattern except in situations s1, s2, s3 etc
Next, in your particular situation, here is how we could approach your regex.
For consision, let's make all the negative lookarounds more compact, replacing them with a single (?!.*(?: |-|'){2})
In your character class, the \: just escapes the colon, needlessly so as : is enough. I assume you wanted to add a backslash character, and if so we need to use \\
\p{L} includes [a-zA-Z], so you can drop [a-zA-Z]. But are you sure you want to match all letters in any script? (Thai etc). If so, remember to set the u flag after the regex string.
For your "bad word exclusion" applying to the whole string, place it at the same position as the other lookarounds, i.e., at the head of the string, but using the .* as in your other exclusions: (?!.*(?:wordToIgnore|wordToIgnore2|wordToIgnore3)) It does not matter which lookahead comes first because lookarounds do not change your position in the string. For more on this, see Mastering Lookahead and Lookbehind
This gives us this glorious regex (I added the case-insensitive flag):
^(?i)(?!.*(?:wordToIgnore|wordToIgnore2|wordToIgnore3))(?!.*(?: |-|'){2})(?:[\\0-9 :/\p{L}'-]{1,64}$)$
Of course if you don't want unicode letters, replace \p{L} with a-z
Also, if you want to make sure that the wordToIgnore is a real word, as opposed to an embedded string (for instance you don't want cat but you are okay with catalog), add boundaries to the lookahead rule: (?!.*\b(?:wordToIgnore|wordToIgnore2|wordToIgnore3)\b)
use this:
^(?!.*(the|and|or|a|given|some|that|this|then|than))(?!.*[ ]{2})(?!.*[']{2})(?!.*[-]{2})(?:[a-zA-Z0-9 \:\p{L}'-]{1,64}$)$
see demo
I'm having some difficulties with the regex boundary \b character. I need to search for an exact keyword inside some loaded text(either plain textual data or Xml). Because the need for exact matches I use the \bkeyword\b pattern but I get a different behaviour than what I expect when the keyword starts with a special character. For example the pattern \b€ 3,5\b doesn't match in I have € 3,5 to spend!. This is the case with any special characters.
I've search around but came up with no solution. Is there some mechanism that acts like a \b but for special characters ? Also note that i cannot alter the keyword.
Any help would be appreciated.
You can perhaps make use of a positive lookbehind:
(?<=^|\s)€ 3,5\b
The positive lookbehind will match either the beginning of the string or a \s, without including them in the match itself.
I'm trying to come up with a regular expression that will stop at the first occurence of </ol>. My current RegEx sort of works, but only if </ol> has spaces on either end. For instance, instead of stopping at the first instance in the line below, it'd stop at the second
some random text and HTML</ol></b> bla </ol>
Here's the pattern I'm currently using: string pattern = #"some random text(.|\r|\n)*</ol>";
What am I doing wrong?
string pattern = #"some random text(.|\r|\n)*?</ol>";
Note the question mark after the star -- that tells it to be non greedy, which basically means that it will capture as little as possible, rather than the greedy as much as possible.
Make your wild-card "ungreedy" by adding a ?. e.g.
some random text(.|\r|\n)*?</ol>
^- Addition
This will make regex match as few characters as possible, instead of matching as many (standard behavior).
Oh, and regex shouldn't parse [X]HTML
While not a Regex, why not simply use the Substring functions, like:
string returnString = someRandomText.Substring(0, someRandomText.IndexOf("</ol>") - 1);
That would seem to be a lot easier than coming up with a Regex to cover all the possible varieties of characters, spaces, etc.
This regex matches everything from the beginning of the string up to the first </ol>. It uses Friedl's "unrolling-the-loop" technique, so is quite efficient:
Regex pattern = new Regex(
#"^[^<]*(?:(?!</ol\b)<[^<]*)*(?=</ol\b)",
RegexOptions.IgnoreCase);
resultString = pattern.Match(text).Value;
Others had already explained the missing ? to make the quantifier non greedy. I want to suggest also another change.
I don't like your (.|\r|\n) part. If you have only single characters in your alternation, its simpler to make a character class [.\r\n]. This is doing the same thing and its better to read (I don't know compiler wise, maybe its also more efficient).
BUT in your special case when the alternatives to the . are only newline characters, this is also not the correct way. Here you should do this:
Regex A = new Regex(#"some random text.*?</ol>", RegexOptions.Singleline);
Use the Singleline modifier. It just makes the . match also newline characters.
example strings
785*()&!~`a
##$%$~2343
455frt&*&*
i want to capture the first and the third but not the second since it doesnt contain any alphabet character plz help
In fact, I think [a-zA-Z] might suffice to match your strings.
To capture the whole thing, try: ^.*[a-zA-Z].*$
Here is one possible way:
.*[a-zA-Z]+
You should maybe clarify a bit what you mean by 'catpuring': do you want the whole string of just the ascii bits?
Also, you don't say if it should match just plain Roman alphabet (A to Z) or if it should also match Unicode chars to match strings in other languages.
If you just need to test your string, in C# you would do:
bool matching = Regex.IsMatch(myString, "[a-zA-Z]");
You wouldn't need anything else, since just one letter anywhere in the myString string will match (according to your definition).
This is my favorite RegEx testing site: Javascript Regexp Tester and Cheat Sheet
If you want to match all letters (including non-ascii ones), use p{L} instead of [a-zA-Z]. See Unicode categories.