This question already has an answer here:
Reference - What does this regex mean?
(1 answer)
Closed 2 years ago.
How would I match anything after asad= but before the next &?
Test Strings:
https://teststore.com/products/mens-sandals-black?asad=253485_e1c6ae3ad&ol_color=13968
https://teststore.com/products/womens-sandals-tan-?asad=252485_c1c63c01d&variant=2770725251
https://teststore.com/products/mens-shoes-blue?asad=254325_c1c63c01d
https://teststore.com/collections/men/products/mens-sneakers?variant=310637539&asad=204207_e1c1756d5
Expected Extraction:
253485_e1c6ae3ad
252485_c1c63c01
254325_c1c63c01d
204207_e1c1756d5
You can use lookahead:
x(?=y) – positive lookahead (matches 'x' when it's followed by 'y')
x(?!y) – negative lookahead (matches 'x' when it's not followed by 'y')
and you can also use lookbehinds:
(?<=y)x – positive lookbehinds (matches 'x' when it's precede by 'y')
(?<!y)x – negative lookbehinds (matches 'x' when it's not precede by 'y')
These are examples for your need,
if you have text with many lines, use this:
(?<=asad=).*?(?=(?:&|\n))
or if you have an array with multiple strings use that
(?<=asad=).*?(?=(?:&|$))
Example on https://regex101.com/r/8LvPtZ/1
You want non-greedy matching, and probably want to match either a & or the end of the line as a terminator:
/asad=(.*?)(?:&|$)/
This tells the regex engine:
We're looking for the sequence "asad="
Then, capture everything until...
We see either a "&" or the end of the line ($).
If you use greedy matching (.*) then it'll capture everything until it can't match terminators anymore.
With the test string:
https://teststore.com/collections/men/products/mens-sneakers?variant=310637539&asad=204207_e1c1756d5&foo=bar
The greedy match /asad=(.*)(?:&|$)/ would capture "asad=204207_e1c1756d5&foo=bar", because it can keep greedily matching up until the end-of-line ($), whereas the non-greedy match /asad=(.*)(?:&|$)/ captures just what you want.
Related
This question already has answers here:
Regex how to match an optional character
(5 answers)
Closed 9 months ago.
I want to get all Strings that start with a "$" sign followed by an integer or have exactly two digits after the decimal point.
e.g. $7.26 and $48.49 and $17
but not $.49 and $192.9
That's my regular expression so far: ^[$]\d+**[.][0-9][0-9]**
I want the marked part to be optional or the string has to end.
Also, how could I use [0-9]{2} instead of [0-9][0-9]?
Try this pattern: ^\$\d+(?:\.\d{2})?$
See Regex Demo
Explanation
^: Start of the string.
\$: Match with the character $.
\d+: Match with one or more digits between 0-9.
(?:: Start of the non-captured group.
\.: Match with the dot character.
\d{2}: Match exactly two digits.
): End of the group.
?: Make everything in the group optional.
$: End of the string.
Note: the $ and . character in regex means respectively end of the string and everything, so if we want to capture exactly the $ character (not the end of the string) we should escape those characters.
If I understand you correctly, you need to escape characters, which are used as part of syntax of RegEx. Try used backslash
^$\d+(?:.[0-9][0-9])?$
Or used example in first answer - with {2} - it's more correct
This question already has an answer here:
Regex match numbers not followed by a hyphen
(1 answer)
Closed 1 year ago.
I am trying to capture groups in a text that only match when the match is not followed by a specific character, in this case the opening parentheses "(" to indicate the start of a 'function/method' rather than a 'property'.
This seems pretty straightforward so I tried:
TEXT
$this->willMatch but $this->willNot()
RESULT
RegExp pattern: \$this->[a-zA-Z0-9\_]+(?<!\()
Expected: $this->willMatch
Actual: $this->willMatch, $this->willNot
RegExp pattern: \$this->[a-zA-Z0-9\_]+[^\(]
Expected: $this->willMatch
Actual: $this->willMatch, $this->willNot
RegExp pattern: \$this->[a-zA-Z0-9]+(?!\()
Expected: $this->willMatch
Actual: $this->willMatch, $this->willNo
My intuition says i need to add ^ and $ but that wont work for multiple occurrences in a text.
Curious to meet the RegExp wizard that can solve this!
Answer from The fourth bird definitely works and it is well explained as well.
As an alternative to using word boundary one can use possessive quantifier i.e. ++ to turn off backtracking thus improving efficiency further.
\$this->\w++(?!\()
RegEx Demo
Please note use of \w instead of equivalent [a-zA-Z0-9_] here.
Like a greedy quantifier, a possessive quantifier repeats the token as many times as possible. Unlike a greedy quantifier, it does not give up matches as the engine backtracks.
The (?<!\() will always be true as the character class does not match a (
Note that you don't have to escape the \_
You can use a word boundary after the character class to prevent backtracking, and turn the negative lookbehind into a negative lookahead (?!\() to assert not ( directly to the right.
\$this->[a-zA-Z0-9_]+\b(?!\()
Regex demo
This question already has answers here:
RegExp exclusion, looking for a word not followed by another
(3 answers)
Closed 3 years ago.
I'm trying to match the string "this" followed by anything (any number of characters) except "notthis".
Regex: ^this.*(?!notthis)$
Matches: thisnotthis
Why?
Even its explanation in a regex calculator seems to say it should work. The explanation section says
Negative Lookahead (?!notthis)
Assert that the Regex below does not match
notthis matches the characters notthis literally (case sensitive)
The negative lookahead has no impact in ^this.*(?!notthis)$ because the .* will first match until the end of the string where notthis is not present any more at the end.
I think you meant ^this(?!notthis).*$ where you match this from the start of the string and then check what is directly on the right can not be notthis
If that is the case, then match any character except a newline until the end of the string.
^this(?!notthis).*$
Details of the pattern
^ Assert start of the string
this Match this literally
(?!notthis)Assert what is directly on the right is notnotthis`
.* Match 0+ times any char except a newline
$ Assert end of the string
Regex demo
If notthis can not be present in the string instead of directly after this you could add .* to the negative lookahead:
^this(?!.*notthis).*$
^^
Regex demo
See it in a regulex visual
Because of the order of your rules. Before your expression would get to negative lookahead, prior rules has been fulfilled, there is nothing left to match.
If you wish to match everything after this, except for notthis, this RegEx might also help you to do so:
^this([\s\S]*?)(notthis|())$
which creates an empty group () for nothing, with an OR to ignore notthis:
^this([\s\S]*?)(notthis|())$
You might remove (), ^ and $, and it may still work:
this([\s\S]*?)(notthis|)
This question already has answers here:
My regex is matching too much. How do I make it stop? [duplicate]
(5 answers)
Closed 5 years ago.
https://regex101.com/r/rnr6SC/2
I am writing a regex to find any instances of words that start with & and end with ;. There could be multiple copies on the same line, on different lines, or whatever. I just want to find anything that looks like
&
&dog;
&cat; and &dog;
or similar. I think my check for the ; is too greedy in my example. How do I fix my regex to find the specific words I need without selecting anything else?
This regex should work for you:
(?<=&)[^;&]+(?=;)
RegEx Demo
(?<=&): Lookbehind to assert previous position has &
[^;&]+: Match 1+ character that is not ; and not &
(?=;): Lookahead to assert next position has ;
Lookarounds just make sure that a matching text is surrounded by & and ;, if you want these markers also to be part of match then remove lookarounds and use:
&[^;&]+;
You might remove the lookaheads (?= and make the .* non greedy .*?
&.*?;
That would match:
ampersand &,
.*? any character zero or more times non greedy
a semicolon ;
This question already has answers here:
Regex: match everything but a specific pattern
(6 answers)
Closed 5 years ago.
I want to match any word that starts/ends or not contain with word "end" but ignore word "end", for example:
hello - would match
boyfriend - would match
endless - would match
endend - would match
but
end - would NOT match
I'm using ^(?!end)).*$ but its not what I want.
Sorry for my english
Try this:
^(?!(end)$).+$
This will match everything except end.
You can use this \b(?!(?:end\b))[\w]+
Components:
\b -> Start of the word boundary for each words.
(?! Negative lookahead to eliminate the word end.
(?:end\b) Non capturing parenthesis with the word end and word boundary.
) Closing tag for negative lookahead.
[\w]+ character class to capture words.
Explanation: The regex search will only look for locations starting with word boundaries, and will remove matches with end as only word. i.e [WORD BOUNDARY]end[END OF WORD BOUNDARY]. \w will capture rest of the word. You can keep incrementing this character class if you wish to capture some special characters like $ etc.
So you want to match any word, but not "end" ?
Unless I'm misunderstanding, a conditional statement is everything that is needed... In pseudocode:
if (word != "end") {
// Match
}
If you want to match all the words in a text that are not "end" you could just remove all the non-alpha characters, replace pattern (^end | end | end$) by an empty string, and then do a string split.
The other answers with a single regex might be better then, because regex matches are O(n), no matter of the pattern.