Lets say i have this string §cHi there
I want to use regex to match §c so I can extract the message.
The way I want it to work is it looks for § and then match the next and the previous character.
So far I am able to match the next character by this expression [§].{1}.
My question is how to match the privious character which is Â.
First match .[§]., then take the first and third character of the match.
Note: your example regex [§].{1} is the same as [§].
. matches any character (sometimes with the exception of the newline character)
Related
I just started learning regex and I'm trying to understand how it possible to do the following:
If I have:
helmut_rankl:20Suzuki12
helmut1195:wasserfall1974
helmut1951:roller11
Get:
helmut_rankl:20Suzuki1
helmut1195:wasserfall197
helmut1951:roller1
I tried using .$ which actually match the last character of a string, but it doesn't match letters and numbers.
How do I get these results from the input?
You could match the whole line, and assert a single char to the right if you want to match at least a single character.
.+(?=.)
Regex demo
If you also want to match empty strings:
.*(?=.)
This will do what you want with regex's match function.
^(.*).$
Broken down:
^ matches the start of the string
( and ) denote a capturing group. The matches which fall within it are returned.
.* matches everything, as much as it can.
The final . matches any single character (i.e. the last character of the line)
$ matches the end of the line/input
using: this tool to evaluate my expression
My test string: "Little" Timmy (tim) McGraw
my regex:
^[()"]|.["()]
It looks like I'm properly catching the characters I want but my matches are including whatever character comes just before the match. I'm not sure what, or if anything, I'm doing wrong to be catching the preceding characters like that? The goal is to capture characters we don't want in the name field of one of our systems.
Brief
Your current regex ^[()"]|.["()] says the following:
^[()"]|.["()] Match either of the following
^[()"] Match the following
^ Assert position at the start of the line
[()"] Match any character present in the list ()"
.["()] Match the following
. Match any character (this is the issue you were having)
["()] Match any character present in the list "()
Code
You can actually shorten your regex to just [()"].
Ultimately, however, it would be much easier to create a negated set that determines which characters are valid rather than those that are invalid. This approach would get you something like [^\w ]. This means match anything not present in the set. So match any non-word and non-space characters (in your sample string this will match the symbols ()" since they are not in the set).
I think that there is a match,but there are two.That's strange.I want to know why
Why are you surprised? .* matches any number of characters, including 0.
So you get one match that contains the entire line, and a second match that contains the empty string between the first match and the end of the string.
Regular expressions don't just deal with characters, but also with positions between characters (known as anchors). For example ^ matches the position before the first character, $ matches the position after the last character in a string.
A regex engine "walks through" a string, starting from the position before the first character. It then steps forward one character at a time.
For example, when applying the regex .* to "Hello", the regex engine starts before the H. It then matches Hello - after that .* can't match any more characters, so the regex engine returns "Hello" as the first match. The regex engine is now positioned after the o. If you call it again and ask it to match, it will succeed in returning a match because you're asking it to match any string, even an empty one, from the current position - and that's possible.
Why doesn't the regex engine return an infinite number of empty strings, then? It checks whether the last match was started from the end of the string, and if it was, no further matches will be attempted.
Some languages don't even try a regex match once from the final position in a string (Ruby seems to be one example), but I'd say it's more correct to return two matches.
Since it appears more clarification is necessary: The regex engine steps through the string along the positions visualized by |s below:
"|H|e|l|l|o|"
^ Position before the first character
^ Position after the last character
I have some long string where i'm trying to catch a substring until a certain character is met.
Lets suppose I have the following string, and I would like to get the text until the first ampersand.
abc.8965.aghtj&hgjkiyu5.8jfhsdj
I would like to extract what is present before the ampersand so: abc.8965.aghtj
W thought this would work:
grep'^.*&{1}'
I would translate it as
^ start of string
.* match whatever chars
&{1} until the first ampersand is matched
Any advice?
I'm afraid this will take me weeks
{1} does not match the first occurrence; instead it means "match exactly one of the preceding pattern/character", which is identical to just matching the character (&{3} would match &&&).
In order to match the first occurrence of &, you need to use .*?:
grep'^.*?&'
Normally, .* is greedy, meaning it matches as much as possible. This means your pattern would match the last ampersand rather than the first one. .*? is the non-greedy version, matching as little as possible while fulfilling the pattern.
Update: That syntax may not be supported by grep. Here is another option:
'^[^&]*&'
It matches anything that is not an ampersand, up to the first ampersand.
You also may have to enable extended regular expression in grep (-E).
Try this one:
^.*?(?=&)
it won't get ampersand sign, just a text before it
I have this regex:
(?:\S)\++(?:\S)
Which is supposed to catch all the pluses in a query string like this:
?busca=tenis+nike+categoria:"Tenis+e+Squash"&pagina=4&operador=or
It should have been 4 matches, but there are only 3:
s+n
e+c
s+e
It is missing the last one:
e+S
And it seems to happen because the "e" character has participated in a previous match (s+e), because the "e" character is right in the middle of two pluses (Teni s+e+S quash).
If you test the regex with the following input, it matches the last "+":
?busca=tenis+nike+categoria:"Tenis_e+Squash"&pagina=4&operador=or
(changed "s+e" for "s_e" in order not to cause the "e" character to participate in the match).
Would someone please shed a light on that?
Thanks in advance!
In a consecutive match the search for the next match starts at the position of the end of the previous match. And since the the non-whitespace character after the + is matched too, the search for the next match will start after that non-whitespace character. So a sequence like s+e+S you will only find one match:
s+e+S
\_/
You can fix that by using look-around assertions that don’t match the characters of the assumption like:
\S\++(?=\S)
This will match any non-whitespace character followed by one or more + only if it is followed by another non-whitespace character.
But tince whitespace is not allowed in a URI query, you don’t need the surrounding \S at all as every character is non-whitespace. So the following will already match every sequence of one or more + characters:
\++
You are correct: The fourth match doesn't happen because the surrounding character has already participated in the previous match. The solution is to use lookaround (if your regex implementation supports it - JavaScript doesn't support lookbehind, for example).
Try
(?<!\s)\++(?!\s)
This matches one or more + unless they are surrounded by whitespace. This also works if the plus is at the start or the end of the string.
Explanation:
(?<!\s) # assert that there is no space before the current position
# (but don't make that character a part of the match itself)
\++ # match one or more pluses
(?!\s) # assert that there is no space after the current position
If your regex implementation doesn't support lookbehind, you could also use
\S\++(?!\s)
That way, your match would contain the character before the plus, but not after it, and therefore there will be no overlapping matches (Thanks Gumbo!). This will fail to match a plus at the start of the string, though (because the \S does need to match a character). But this is probably not a problem.
You can use the regex:
(?<=\S)\++(?=\S)
To match only the +'s that are surrounded by non-whitespace.