Take this regular expression: /^[^abc]/
. This will match any single character at the beginning of a string, except a, b, or c.
If you add a *
after it – /^[^abc]*/
– the regular expression will continue to add each subsequent character to the result, until it meets either an a
, or b
, or c
.
For example, with the source string "qwerty qwerty whatever abc hello"
, the expression will match up to "qwerty qwerty wh"
.
But what if I wanted the matching string to be "qwerty qwerty whatever "
...In other words, how can I match everything up to (but not including) the exact sequence "abc"
?
This question is related to
regex
The $
marks the end of a string, so something like this should work: [[^abc]*]$
where you're looking for anything NOT ENDING in any iteration of abc
, but it would have to be at the end
Also if you're using a scripting language with regex (like php or js), they have a search function that stops when it first encounters a pattern (and you can specify start from the left or start from the right, or with php, you can do an implode to mirror the string).
If you're looking to capture everything up to "abc":
/^(.*?)abc/
Explanation:
( )
capture the expression inside the parentheses for access using $1
, $2
, etc.
^
match start of line
.*
match anything, ?
non-greedily (match the minimum number of characters required) - [1]
[1] The reason why this is needed is that otherwise, in the following string:
whatever whatever something abc something abc
by default, regexes are greedy, meaning it will match as much as possible. Therefore /^.*abc/
would match "whatever whatever something abc something ". Adding the non-greedy quantifier ?
makes the regex only match "whatever whatever something ".
This will make sense about regex.
("(.*?)")/g
Here, we can get the exact word globally which is belonging inside the double quotes. For Example, If our search text is,
This is the example for "double quoted" words
then we will get "double quoted" from that sentence.
On python:
.+?(?=abc)
works for the single line case.
[^]+?(?=abc)
does not work, since python doesn't recognize [^] as valid regex.
To make multiline matching work, you'll need to use the re.DOTALL option, for example:
re.findall('.+?(?=abc)', data, re.DOTALL)
As @Jared Ng and @Issun pointed out, the key to solve this kind of RegEx like "matching everything up to a certain word or substring" or "matching everything after a certain word or substring" is called "lookaround" zero-length assertions. Read more about them here.
In your particular case, it can be solved by a positive look ahead: .+?(?=abc)
A picture is worth a thousand words. See the detail explanation in the screenshot.
What you need is look around assertion like .+? (?=abc)
.
See: Lookahead and Lookbehind Zero-Length Assertions
Be aware that [abc]
isn't the same as abc
. Inside brackets it's not a string - each character is just one of the possibilities. Outside the brackets it becomes the string.
try this
.+?efg
Query :
select REGEXP_REPLACE ('abcdefghijklmn','.+?efg', '') FROM dual;
output :
hijklmn
I ended in this stackoverflow question after looking for help to solve my problem but found no solution to it :(
So I had to improvise... after some time I managed to reach the regex I needed:
As you can see, I needed up to one folder ahead of "grp-bps" folder, without including last dash. And it was required to have at least one folder after "grp-bps" folder.
Edit
Text version for copy-paste (change 'grp-bps' for your text):
.*\/grp-bps\/[^\/]+
For regex in Java, and I believe also in most regex engines, if you want to include the last part this will work:
.+?(abc)
For example, in this line:
I have this very nice senabctence
select all characters until "abc" and also include abc
using our regex, the result will be: I have this very nice senabc
Test this out: https://regex101.com/r/mX51ru/1
I believe you need subexpressions. If I remember right you can use the normal ()
brackets for subexpressions.
This part is From grep manual:
Back References and Subexpressions
The back-reference \n, where n is a single digit, matches the substring
previously matched by the nth parenthesized subexpression of the
regular expression.
Do something like ^[^(abc)]
should do the trick.
Source: Stackoverflow.com