I'm very new at regex, and to be completely honest it confounds me. I need to grab the string after a certain character is reached in said string. I figured the easiest way to do this would be using regex, however like I said I'm very new to it. Can anyone help me with this or point me in the right direction?
For instance:
I need to check the string "23444:thisstring" and save "thisstring" to a new string.
If this is your string:
I'm very new at regex, and to be completely honest it confounds me
and you want to grab everything after the first "c", then this regular expression will work:
/c(.*)/s
It will return this match in the first matched group:
"ompletely honest it confounds me"
Try it at the regex tester here: regex tester
Explanation:
The c is the character you are looking for
.* (in combination with /s) matches everything left
(.*) captures what .* matched, making it available in $1 and returned in list context.
Regex for deleting characters before a certain character!
You can use lookahead like this
.*(?=x)
where x is a particular character or word or string.{using characters like .,$,^,*,+ have special meaning in regex so don't forget to escape when using it within x}
EDIT
for your sample string it would be
.*(?=thisstring)
.* matches 0 to many characters till thisisstring
Here is a one-line solution for matching everything after "before"
print $1."\n" if "beforeafter" =~ m/before(.*)/;
Edit:
While using lookbehind is possible, it's not required. Grouping provides an easier solution.
To get the string before : in your example, you have to use [^:][^:]*:\(.*\). Notice that you should have at least one [^:] followed by any number of [^:]s followed by an actual :, the character you are searching for.
Related
Let's say I have the text a123456. I want a string of b123456 to match. So essentially, 'match if all characters are the same except for the first character'. Am I asking for the impossible with regex?
Use the dot (.) to match any character. So, a possible Regex would be:
/^.123456$/
If you want to use zero length assertion with regex, you can have lookbehind approach in following way :
(?<=\w)your_value$ // your_value should be text which you want to check
I think you can figure it out on your own. This ain't tough, just needs some understanding between you and Regex. Why don't you go through the following links and try to make a regex on your own.
https://www.talentcookie.com/2015/07/regular-expressions/
https://www.talentcookie.com/2015/07/lets-practice-regular-expression/
https://www.talentcookie.com/2016/01/some-useful-regular-expression-terminologies/
I'm using an API that sometimes truncates links inside the text that it returns and instead of "longtexthere https://fancy.link" I get "longtexthere https://fa…".
I'm trying to get to match the link only if it's complete, or in other words does not contain "…" character.
So far I am able to get links by using the following regex:
((?:https?:)?\/\/\S+\/?)
but obviously it returns every link including broken ones.
I've tried to do something like this:
((?:https?:)?\/\/(?:(?!…)\S)+\/?)
Although that started to ignore the "…" character it was still returning the link but just without including the character, so with the case of "https://fa…" it returned "https://fa" whereas I simply want it to ignore that broken link and move on.
Been fighting this for hours and just can't get my head around it. :(
Thanks for any help in advance.
You can use
(?:https?:)?\/\/[^\s…]++(?!…)\/?
See the regex demo. The possessive quantifier [^\s…]++ will match all non-whitespace and non-… characters without later backtracking and then check if the next character is not …. If it is, no match will be found.
As an alternative, if your regex engine allow possessive quantifiers, use a negative lookahead version:
(?!\S+…)(?:https?:)?\/\/\S+\/?
See another regex demo. The lookahead (?!\S+…) will fail the match if 1+ non-whitespace characters are followed with ….
You can try following regex
https?:\/\/\w+(?:\.\w+\/?)+(?!\.{3})(\s|$)
See demo https://regex101.com/r/bS6tT5/3
Try:
((?:https?:)?\/\/\S+[^ \.]{3}\/?)
Its the same as your original pattern.. you just tell it that the last three characters should not be '.' (period) or ' ' (space)
UPDATE: Your second link worked.
and if you tweak your regex just slightly it will do what you want:
((?:https?:)?\/\/\S+[^ …] \/?)
Yes it looks just like what you had in there except I added a ' ' (space) after the part we do not want.. this will force the regular expression to match up until and including the space which it cannot with a url that has the '...' character. Without the space at the end it would match up until the not including the '...' which was why it was not doing what we wanted ;)
Please try:
https?:\/\/[^ ]*?…|(https?:\/\/[^ ]+\.[^ ]+)
Here is the demo.
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,}\)?
I'm searching the pattern (.*)\\1 on the text blabl with regexec(). I get successful but empty matches in regmatch_t structures. What exactly has been matched?
The regex .* can match successfully a string of zero characters, or the nothing that occurs between adjacent characters.
So your pattern is matching zero characters in the parens, and then matching zero characters immediately following that.
So if your regex was /f(.*)\1/ it would match the string "foo" between the 'f' and the first 'o'.
You might try using .+ instead of .*, as that matches one or more instead of zero or more. (Using .+ you should match the 'oo' in 'foo')
\1 is the backreference typically used for replacement later or when trying to further refine your regex by getting a match within a match. You should just use (.*), this will give you the results you want and will automatically be given the backreference number 1. I'm no regex expert but these are my thoughts based on my limited knowledge.
As an aside, I always revert back to RegexBuddy when trying to see what's really happening.
\1 is the "re-match" instruction. The question is, do you want to re-match immediately (e.g., BLABLA)
/(.+)\1/
or later (e.g., BLAahemBLA)
/(.+).*\1/