I want to get all matches of #[SomeText] pattern in a string.
For example, for this string:
here is #[text1] some text #[text2]
I want #[text1] and #[text2].
I'm using Regex Hero to check my pattern matching online,
and my pattern works fine when there's one expression to match,
For example:
here is #[text1] text
but with more then one, I get both matches with the text in the middle.
This is my regex:
#\[.*\]
I would appreciate assistance in isolating the occurrences.
The problem here is that you are using greedy quantifier (*). To capture all you need, you should use lazy quantifier (*?) with a global modifier:
/(#\[.*?\])/g
Take a look here https://regex101.com/r/pH0gA5/1
This should work :
#\[(.*?)\]
Details :
(.*?) : match everything in a non-greedy way and capture it.
Because the *? quantifier is lazy (non-greedy), it matches as few characters as possible to allow the overall match attempt to succeed, i.e. text1. For the match attempt that starts at a given position, a lazy quantifier gives you the shortest match.
.* is greedy by default, so it only finds one match, treating "text1] and #[text2" as the text between the two square brackets.
If you add a questions mark after the .* then it will find the minimum number of characters before reaching a ].
So the regex \#[.*?] do what you want.
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.
String is given below from which i want to extract the text.
String:
Hello Mr John and Hello Ms Rita
Regex
Hello(.*?)Rita
I am try to get text between 2 strings which "Hello" and "Rita" I am using the above given regex, but its is giving me
Mr John and Hello Ms
which is wrong. I need only "Ms" Can anyone help me out to write proper regex for this situation?
Use a tempered greedy token:
Hello((?:(?!Hello|Rita).)*)Rita
^^^^^^^^^^^^^^^^^^^
See regex demo here
The (?:(?!Hello|Rita).)* is the tempered greedy token that only matches text that is not Hello or Rita. You may add word boundaries \b if you need to check for whole words.
In order to get a Ms without spaces on both ends, use this regex variation:
Hello\s*((?:(?!Hello|Rita).)*?)\s*Rita
Adding the ? to * will form a lazy quantifier *? that matches as few characters as needed to find a match, and \s* will match zero or more whitespaces.
To get the closest match towards ending word, let a greedy dot in front of the initial word consume.
.*Hello(.*?)Rita
See demo at regex101
Or without whitespace in captured: .*Hello\s*(.*?)\s*Rita
Or with use of two capture groups: .*(Hello\s*(.*?)\s*Rita)
Your (.*?) is picking up too much text because .* matches any string of characters. So it grabs everything from the first "Hello" to "Rita" at the end.
One easy way you could get what you want is with this regular expression:
Hello (\S+) Rita
\S matches any non-whitespace character, so \S+ matches any consecutive string of non-whitespace characters, i.e. a single word.
This would be a bit more robust, allowing for multiple spaces or other whitespace between the words:
Hello\s+(\S+)\s+Rita
Demo
you can use lookahead and lookbehind (?<=Hello).*?(?=Rita)
[SOME_WORDS:200:1000]
Trying to match just the last 1000 part. Both numbers are variable and can contain an unknown number of characters (although they are expected to contain digits, I cannot rule out that they may also contain other characters). The SOME_WORDS part is known and does not change.
So I begin by doing a positive lookbehind for [SOME_WORDS: followed by a positive lookahead for the trailing ]
That gives us the pattern (?<=\[SOME_WORDS:).*(?=])
And captures the part 200:1000
Now because I don't know how many characters are after SOME_WORDS:, but I know that it ends with another : I use .*: to indicate any character any amount of time followed by :
That gives us the pattern (?<=\[SOME_WORDS:.*:).*(?=])
However at this point the pattern no longer matches anything and this is where I become confused. What am I doing wrong here?
If I assume that the first number will always be 3 characters long I can replace .* with ... to get the pattern (?<=\[SOME_WORDS:...:).*(?=]) and this correctly captures just the 1000 part. However I don't understand why replacing ... with .* makes the pattern not capture anything.
EDIT:
It seems like the online tool I was using to test the regex pattern wasn't working correctly. The pattern (?<=\[SOME_WORDS:.*:).*(?=]) matches the 1000 with no issues when actually done in .net
You usually cannot use a + or a * in a lookbehind, only in a lookahead.
If c# does allow these than you could use a .*? instead of a .* as the .* will eat the second :
Try this:
(?<=\[SOME_WORDS:)(?=\d+:(\d+)])
The match wil be in the first capture group
Quote from http://www.regular-expressions.info/lookaround.html
The bad news is that most regex flavors do not allow you to use just any regex inside a lookbehind, because they cannot apply a regular expression backwards. The regular expression engine needs to be able to figure out how many characters to step back before checking the lookbehind. When evaluating the lookbehind, the regex engine determines the length of the regex inside the lookbehind, steps back that many characters in the subject string, and then applies the regex inside the lookbehind from left to right just as it would with a normal regex.
As Robert Smit mentions this is due to the * being a greedy operator. Greedy operators consume as many characters as they possibly can when they are matched first. They only give up characters if the match fails. If you make the greedy operator lazy(*?), then matching consumes as little number of characters as possible for the match to succeed, so the : is not consumed by *. You can also use [^:]* which is match any character other than :.
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 need a help with regular expression as I do not have good knowledge in it.
I have regular expression as:
Regex myregex = new Regex("testValue=\"(.+?)\"");
What does (.+?) indicate?
The string it matches is "testValue=123e4567" and returns 123e4567 as output.
Now I need help in regular expression to match a string "<helpMe>123e4567</helpMe>" where I need 123e4567 as output. How do I write a regular expression for it?
This means:
( Begin captured group
. Match any character
+ One or more times
? Non-greedy quantifier
) End captured group
In the case of your regex, the non-greedy quantifier ? means that your captured group will begin after the first double-quote, and then end immediately before the very next double-quote it encounters. If it were greedy (without the ?), the group would extend to the very last double-quote it encounters on that line (i.e., "greedily" consuming as much of the line as possible).
For your "helpMe" example, you'd want this regex:
<helpMe>(.+?)</helpMe>
Given this string:
<div>Something<helpMe>ABCDE</helpMe></div>
You'd get this match:
ABCDE
The value of the non-greedy quantifier is evident in this variation:
Regex: <helpMe>(.+)</helpMe>
String: <div>Something<helpMe>ABCDE</helpMe><helpMe>FGHIJ</helpMe></div>
The greedy capture would look like this:
ABCDE</helpMe><helpMe>FGHIJ
There are some useful interactive tools to play with these variations:
Regex Tester
Regex Pal
Ken Redler has a great answer regarding your first question. For the second question try:
<(helpMe)>(.*?)</\1>
Using the back reference \1 you can find values between the set of matching tags. The first group finds the tag name, the second group matches the content itself, and the \1 back reference re-uses the first group's match (in this case the tag name).
Also, in C# you can use named groups, like: <(helpMe)>(?<value>.*?)</\1> where now match.Groups["value"].Value contains your value.
What does (.+?) indicate?
It means match any character (.) one or more times (+?)
A simple regex to match your second string would be
<helpMe>([a-z0-9]+)<\/helpMe>
This will match any character of a-z and any digit inside <helpme> and </helpMe>.
The pharanteses are used to capture a group. This is useful if you need to reference the value inside this group later.