I want to validate Winforms text box with regex.
The input sting example:
ZX1 OR N?V OR 2L? OR ?55 (any sequence of three symbols length strings with OR between them)
What is the regex that you would advise?
UPDATE:
Trying this one but seams to be it is not 100% correct
string text = "ZX1 OR N?V OR 2L? OR ?55";
Regex r = new Regex("([0-9A-Z?]{3} OR )*[0-9A-Z?]{3}");
"^\\s*\\S{3}(?:\\s+OR\\s+\\S{3})*\\s*$"
should work in a variety of languages.
\\S
matches any non-space character, and
\\s
matches any space character, so the regex above matches any number of triplets of non-space characters separated by the string "OR" surrounded by space characters.
The ^ and $ serve to ensure that it matches the whole string so you can take those out if you want to find this pattern inside a larger string.
What is the list of possible symbols you can have? can you have at most one question mark?
This will match what you've given, but it will also match multiple question marks.
([A-Z?]{3} OR )*[A-Z?]{3}
try...
(([\w\S]{3}\s+)or\s+)+[\w\S]{3}
Related
I'm trying to find a regular expression for a Tokenizer operator in Rapidminer.
Now, what I'm trying to do is to split text in parts of, let's say, two words.
For example, That was a good movie. should result to That was, was a, a good, good movie.
What's special about a regex in a tokenizer is that it plays the role of a delimiter, so you match the splitting point and not what you're trying to keep.
Thus the first thought is to use \s in order to split on white spaces, but that would result in getting each word separately.
So, my question is how could I force the expression to somehow skip one in two whitespaces?
First of all, we can use the \W for identifying the characters that separate the words. And for removing multiple consecutive instances of them, we will use:
\W+
Having that in mind, you want to split every 2 instances of characters that are included in the "\W+" expression. Thus, the result must be strings that have the following form:
<a "word"> <separators that are matched by the pattern "\W+"> <another "word">
This means that each token you get from the split you are asking for will have to be further split using the pattern "\W+", in order to obtain the 2 "words" that form it.
For doing the first split you can try this formula:
\w+\W+\w+\K\W+
Then, for each token you have to tokenize it again using:
\W+
For getting tokens of 3 "words", you can use the following pattern for the initial split:
\w+\W+\w+\W+\w+\K\W+
This approach makes use of the \K feature that removes from the match everything that has been captured from the regex up to that point, and starts a new match that will be returned. So essentially, we do: match a word, match separators, match another word, forget everything, match separators and return only those.
In RapidMiner, this can be implemented with 2 consecutive regex tokenizers, the first with the above formula and the second with only the separators to be used within each token (\W+).
Also note that, the pattern \w selects only Latin characters, so if your documents contain text in a different character set, these characters will be consumed by the \W which is supposed to match the separators. If you want to capture text with non-Latin character sets, like Greek for example, you need to change the formula like this:
\p{L}+\P{L}+\p{L}+\K\P{L}+
Furthermore, if you want the formula to capture text on one language and not on another language, you can modify it accordingly, by specifying {Language_Identifier} in place of {L}. For example, if you only want to capture text in Greek, you will use "{Greek}", or "{InGreek}" which is what RapidMiner supports.
What you can do is use a zero width group (like a positive look-ahead, as shown in example). Regex usually "consumes" characters it checks, but with a positive lookahead/lookbehind, you assert that characters exist without preventing further checks from checking those letters too.
This should work for your purposes:
(\w+)(?=(\W+\w+))
The following pattern matches for each pair of two words (note that it won't match the last word since it does not have a pair). The first word is in the first capture group, (\w+). Then a positive lookahead includes a match for a sequence of non word characters \W+ and then another string of word characters \w+. The lookahead (?=...) the second word is not "consumed".
Here is a link to a demo on Regex101
Note that for each match, each word is in its own capture group (group 1, group 2)
Here is an example solution, (?=(\b[A-Za-z]+\s[A-Za-z]+)) inspired from this SO question.
My question sounds wrong once you understand that is a problem of an overlapping regex pattern.
I'm trying to match the last four characters (alphanumeric) of all words beginning with the sequence &c.
For instance, in the string below, I'd like to match the pieces in bold:
Colour one is &cFF2AC3 and colour two is &c22DE4A.
Can anybody help me with the correct regex expression? I've spent hours on this great resource to no avail.
it looks like hexadecimal numbers, so use this pattern
&c[0-9A-F]{2}\K([0-9A-F]{4})
DEMO
This:
/(?i)\s*&c(?:[a-z0-9]{2})([a-z0-9]{4})\b/
append a g to the end of it if you want it to find all matches in a given text
Try this
/(?:^| )&c\w*(\w{4})\b/
If you want to try it in the regex tester you linked to, make sure to use the g modifier to see all matches.
Explanation: (?:^| ) matches either a space or the start of the string, &c\w* matches the ampersand and the the first however many characters of the word, and then \w{4} captures the last 4 characters. \b on the end asserts a word break (a "non-word" character or the end of the string).
Hello I have strings like
tda2030 100.200.300 circuit
I want to check if this string contains any keyword (separated by whitespace but can be on start or end) that contain more than 1 dot and then remove the dots.
The result should be
tda2030 100200300 circuit
in the example.
I tried a lot but I think I need a regex-pert :) Thanks in advance.
That's an interesting question because of your requirement to have multiple dots and the fact that PCRE does not allow infinite-width lookbehinds to see if we might have a dot behind us. We'll get over that limitation by using \K and \G.
Here is a regex that will find the right dots (see online demo)
(?<=\w)\.(?=\w+\.)|\G\w+\K\.
Use preg_replace to replace with an empty string:
$replaced = preg_replace("~(?<=\w)\.(?=\w+\.)|\G\w+\K\.~","",$string);
How does it work?
We have two cases separated by an | (OR)
Match a dot that is preceded by at least one word character and followed by some word characters and a dot
Match a dot that follows the previous match (which had to be a dot) and some word characters
This seems like it should be trivial, but I'm not so good with regular expressions, and this doesn't seem to be easy to Google.
I need a regex that starts with the string 'dbo.' and ends with the string '_fn'
So far as I am concerned, I don't care what characters are in between these two strings, so long as the beginning and end are correct.
This is to match functions in a SQL server database.
For example:
dbo.functionName_fn - Match
dbo._fn_functionName - No Match
dbo.functionName_fn_blah - No Match
If you're searching for hits within a larger text, you don't want to use ^ and $ as some other responders have said; those match the beginning and end of the text. Try this instead:
\bdbo\.\w+_fn\b
\b is a word boundary: it matches a position that is either preceded by a word character and not followed by one, or followed by a word character and not preceded by one. This regex will find what you're looking for in any of these strings:
dbo.functionName_fn
foo dbo.functionName_fn bar
(dbo.functionName_fn)
...but not in this one:
foodbo.functionName_fnbar
\w+ matches one or more "word characters" (letters, digits, or _). If you need something more inclusive, you can try \S+ (one or more non-whitespace characters) or .+? (one or more of any characters except linefeeds, non-greedily). The non-greedy +? prevents it from accidentally matching something like dbo.func1_fn dbo.func2_fn as if it were just one hit.
^dbo\..*_fn$
This should work you.
Well, the simple regex is this:
/^dbo\..*_fn$/
It would be better, however, to use the string manipulation functionality of whatever programming language you're using to slice off the first four and the last three characters of the string and check whether they're what you want.
\bdbo\..*fn
I was looking through a ton of java code for a specific library: car.csclh.server.isr.businesslogic.TypePlatform (although I only knew car and Platform at the time). Unfortunately, none of the other suggestions here worked for me, so I figured I'd post this.
Here's the regex I used to find it:
\bcar\..*Platform
Scanner scanner = new Scanner(System.in);
String part = scanner.nextLine();
String line = scanner.nextLine();
String temp = "\\b" + part + "|" + part + "\\b";
Pattern pattern = Pattern.compile(temp.toLowerCase());
Matcher matcher = pattern.matcher(line.toLowerCase());
System.out.println(matcher.find() ? "YES" : "NO");
If you need to determine if any of the words of this text start or end with the sequence, you can use this regex: \bsubstring|substring\b:
anythingsubstring
substringanything
anythingsubstringanything
The simplest thing that you can do is:
dbo.*_fn$
It searches with dbo, followed by any characters, and then ends with _fn.
If you can identify what’s the right next character after n if it’s space, you can replace $ with space .
I have been struggle to write regex that matches words longer than a given length within parentheses. First I thought I could do this with \(\w{a,}\) but I realize that it doesn't match with words with white space (ab cd ef). All I want to do is find out any characters within parentheses longer than, for instance, 3 characters. How can I resolve this problem ?
What is a word with white space?
if you want to match any character then use .
\(.{3,}\)
. matches any character except newlines
But be careful, this is greedy. it will match for example also
(a)123(b)
To avoid this you could do something like
\([^)]{3,}\)
See it here online on Regexr
[^)] means any character except a )
You could use a character class that includes both \w and \s:
\([\w\s]{a,}\)
Maybe do you mean?
\([\w\s]{a,}\)
if it has a space in it it's not a word anymore.
is matching any characters fine \(.{a,}\)? Or you just need the whitespace \(\(\w|\s\){a,}\)?