I have a string that looks like:
this is a string [[and]] it is [[awesome|amazing]]
I have the following regular expression so far:
(?<mygroup>(?<=\[\[).+?(?=\]\]))
I am basically trying to capture everything inside the brackets. However, I need to add another condition that says: If the matched result contains a pipe delimiter then only return the word to the right of the pipe delimiter. If there is no pipe then just return everything inside the brackets.
The parsing result I am looking for given the example above should look like:
and
amazing
Any input is appreciated.
(?<mygroup>(?<=\[\[)([^|\]]*|)?([^|]+?)(?=\]\]))
You could use this regex:
(?<=\[\[[^\]]*?)(?!\w+\|)\w+(?=\]\])
it matches both and and amazing words in your test example. You could check it out, I created a test app on Ideone.
From the regex info page:
The tremendous power and expressivity
of modern regular expressions can
seduce the gullible — or the foolhardy
— into trying to use regexes on every
string‐related task they come across.
My advice: Just grab what is between the brackets and parse it after.
Regular expressions are not the answer to everything. May those who follow after you be spared from deciphering the regex you come up with.
Related
I have a simple issue: Using Alteryx, I want to take a string, match a certain pattern and return the matched pattern.
This is my current approach:
Regex_replace("CP:ConsumerProducts&Retail</td><td><strong><fontcl","[^\<]+","$1")
According to various sources and tools like regex101, the first matched sequence should be "CP:ConsumerProducts&Retail". However, Alteryx returns
<<<<
Alteryx uses the Perl RegEx Syntax (https://help.alteryx.com/2018.2/boost/syntax_perl.html), therefore, it should have no problem with the pattern itself.
I believe I am missing something obvious but I cannot figure it out.
I have received a reply through a different forum. A solution that works for me is to use the following pattern: ([^\<]+).*
You can try the following workflow:
I am working with regular expressions, I need to create an expression for validating strings against the following scenario:
Solution.<word1|word2|word3>.<word4|word5>.anyword.(any word containing proj in it)
I tried
Solution.\b(word1|word2|word3)\b.\b(word4|word5)\b.(.*).\b(.*proj)\b
But this allows strings like Solution.word1.word4.blabla.blabla.csproj, meaning it allows anything before the proj because of the .*.
Can someone help me with this??
Looks like you need this regex:
Solution\.(word1|word2|word3)\.(word4|word5)\.([^.]+)\..*?\bproj\b
RegEx Demo
You might want to try (need to escape the . and allow capturing group to have chars except .):
Solution\.\b(word1|word2|word3)\b\.\b(word4|word5)\b\.([^\.]*)\.\b([^\.]*proj)\b
It's hard to consider the actual strings you want to allow without more clarification.
You can try the following regular expression.
Solution\.word[123]\.word[45]\.\w+\.\w*proj\b
Regular Expressions are incredible. I'm in my regex infancy so help solving the following would be greatly appreciated.
I have to search through a string to match for a P character that's not surrounded by operators, power or negative signs. I then have to insert a multiplication sign. Case examples are:
33+16*55P would become 33+16*55*P
2P would become 2*P
P( 33*sin(45) ) would become P*(33*sin(45))
I have written some regex that I think handles this although I don't know how using regex I can insert a character:
The reg is I've written is:
[^\^\+\-\/\*]?P+[^\^\+\-\/\*]
The language where the RegEx will be used is ActionScript 3.
A live example of the regex can be seen at:
http://www.regexr.com/39pkv
I would be massively grateful if someone could show me how I insert a multiplication sign in middle of the match ie P2, becomes P*2, 22.5P becomes 22.5P
ActionScript 3 has search, match and replace functions that all utilise regular expressions. I'm unsure how I'd use string.replace( expression, replaceText ) in this context.
Many thanks in advance
Welcome to the wonder (and inevitable frustration that will lead to tearing your hair out) that is regular expressions. You should probably read over the documentation on using regular expressions in ActionScript, as well as this similar question.
You'll need to combine RegExp.test() with the String.replace() function. I don't know ActionScript, so I don't know if it will work as is, but based on the documentation linked above, the below should be a good start for testing and getting an idea of what the form of your solution might look like. I think #Vall3y is right. To get the replace right, you'd want to first check for anything leading up to a P, then for anything after a P. So two functions is probably easier to get right without getting too fancy with the Regex:
private function multiplyBeforeP(str:String):String {
var pattern:RegExp = new RegExp("([^\^\+\-\/\*]?)P", "i");
return str.replace(pattern, "$1*P");
}
private function multiplyAfterP(str:String):String {
var pattern:RegExp = new RegExp("P([^\^\+\-\/\*])", "i");
return str.replace(pattern, "P*$1");
}
Regex is used to find patterns in strings. It cannot be used to manipulate them. You will need to use action script for that.
Many programming languages have a string.replace method that accepts a regex pattern. Since you have two cases (inserting after and before the P), a simple solution would be to split your regex into two ([^\^\+\-\/\*]?P+ and P+[^\^\+\-\/\*] for example, this might need adjustment), and switch each pattern with the matching string ("*P" and "P*")
I would like to remove hundreds on onmouseover events from my code. the evt all pass different variables and I want to be able to use dreamwaever to find and replace all the strings with nothing.
Here is an example
onmouseover="parent.mv_mapTipOver(evt,'Wilson');"
onmouseover="parent.mv_mapTipOver(evt,'Harris');"
onmouseover="parent.mv_mapTipOver(evt,'Walker');"
I want to run a search that will identify all of these and replace/remove them.
I have tried seemingly infinite permutations of things like:
onmouseover="parent.mv_mapTipOver(evt,'[^']');"
or
onmouseover="parent.mv_mapTipOver(evt,'[^']);"
or
onmouseover="parent.mv_mapTipOver(evt,[^']);"
or
onmouseover="parent.mv_mapTipOver(evt,'[^']+');"
And many more. I cannot find the regular expression that will work.
Any/all help would be appreciated.
Thanks a ton!
"." and "(" have special meaning in regular expressions, so you need to escape them:
onmouseover="parent\.mv_mapTipOver\(evt,'[^']+'\);"
I'm not sure if this is correct dreamweaver regex syntax, but this stuff is standard enough.
Try this one:
onmouseover="parent\.mv_mapTipOver\(evt,'.+?'\);"
And see it in action here.
When using reg expressions you have to be very careful about how you handle white space. For example the following piece of code will not get caught by most of the reg expressions mentioned so far because of the space after the comma and equals sign, despite the fact that it is most likely valid syntax in the language you are using.
onmouseover= "parent.mv_mapTipOver(evt, 'Walker');"
In order to create regexp that ignore white space you must insert /s* everywhere in the regexp that white space might occur.
The following regexp should work even if there is additional white space in your code.
onmouseover\s*=\s*"parent\.mv_mapTipOver\(\s*evt\s*,\s*'[A-Za-z]+'\s*\);"
I have statements like this all over my code:
LogWrite (String1,
String2,
L"=======format string======",
...
);
I want to change each of these to:
LogWrite (String1,
String2,
L"format string",
...
);
I'm trying to write the regexp required to do this using the Emacs function query-replace-regexp, but not much success yet. Help please!
UPDATE:
1) In case it is not clear, this question is emacs specific.
2) I would like to match the entire code chunk starting from Log... ending at );
3) I used the following reg-exp to match the code chunk:
L.*\n.*\n.*==.*;
I used re-builder to match this regexp. the \n is used because I found that otherwise emacs would stop matching at the new line. The problem is that I don't know how to select the format string and save it to use it in the replace regexp - hence the ==.* part in the regexp. That needs to be modified to save the format string.
If you don't have multiple (or escaped) double quotes in those format string lines, you can
//replace
L"=+(.*)=+"
//with
L"\1"
Update: Removed the lazy quantifier (thanks #tim). Make sure that the regex is not multiline; the greedy * will lead to pretty bad results if . matches new lines
A great tool to figure out emacs regular expressions is:
M-x re-builder
A brief description from the documentation:
When called up re-builder' attaches
itself to the current buffer which
becomes its target buffer, where all
the matching is done. The active
window is split so you have a view on
the data while authoring the RE. If
the edited expression is valid the
matches in the target buffer are
marked automatically with colored
overlays (for non-color displays see
below) giving you feedback over the
extents of the matched (sub)
expressions. The (non-)validity is
shown only in the modeline without
throwing the errors at you. If you
want to know the reason why RE Builder
considers it as invalid call
reb-force-update' ("\C-c\C-u") which
should reveal the error.
It comes built into Emacs (since 21)
And for the syntax of Emacs regular expressions, you can read these info pages:
Syntax of Regular Expressions
Backslash in Regular Expressions
/={7}(.*)={6}/\1/
this should do.