I have the following string 3}HFB}4AF4}1 -M}1.
I have searched for this string using the regex :
([0-9])(\})([A-Z]{3})(\})([0-9][A-Z]{2}[0-9])(\})([0-9])(\s\-)([A-Z])(\})([0-9]).
I want to replace the } with 0. The Result I am looking for is 30HFB04AF401-M01, any assistance is appriciated. The tool I am using is Regex Buddy
A possible solution
Problem solved? In JavaScript at least :-)
"3}HFB}4AF4}1 -M}1".replace(/\}/g, "0");
// "30HFB04AF401 -M01"
I'm missing the point, right?
Assuming the language is JavaScript, we can write something like
"dfghj456783}HFB}4AF4}1 -M}1fghjkl8765".replace(/(?:[\d\w\s]+)([0-9]}[A-Z]{3}}[0-9][A-Z]{2}[0-9]}[0-9] -[A-Z]}[0-9])(?:[\d\w\s]+)/g, function () {
return arguments[1].replace(/}/g, "0");
});
What's possible in other languages though may be a different story.
Try the home of RegexBuddy for details.
So you've already got an expression to find instances of the string. Now you can either use groups to replace the characters, or you can use a separate regular expression over the string you found, simply replacing the } character within group(0) (which is the entire matched part of the input). I would certainly prefer the latter.
Fred seems to have created the replacement method for you already, so I won't repeat it here.
I have managed to find a solution to the formating in the JGSoft Lanugage used by Regex Buddy, thanks to all that provided suggestions that helped me channel my thoughts in the right direction.
Solution(I am still a beginner with Regex hence the syntax might not be efficent, but it does the job!!)
Using Group Names instead of Regex assiging groups with backreference and $ syntax.
Hence to replace 0 for } in the string 3}HFB}4AF4}1 -M}1 or any similar string. I used the following search and replacement syntax
Search : (?<Gp1>([0-9]))(?:})(?<Gp2>([A-Z]){3})(?:})(?<Gp3>([0-9])([A-Z]{2})([0-9]))(?:})(?<Gp4>([0-9]))(?:\s-)(?<Gp5>([A-Z]))(?:})(?<Gp6>[0-9])
Replace : ${Gp1}0${Gp2}0${Gp3}0${Gp4}-${Gp5}0${Gp6}
Result : 30HFB04AF401-M01
Related
I'm upgrading a Symfony app with VSCode and I have numerous occurences of this kind of string :
#Template("Area:Local:delete.html.twig")
or
#Template("Group:add.html.twig")
In this case, I want to replace all the : with / to have :
#Template("Area/Local/delete.html.twig")
I can't think of doing it manually, so I was looking for a regular expression for a search/replace in the editor.
I've been toying with this fearsome beast without luck (i'm really dumb in regexp) :
#Template\("([:]*)"
#Template\("(.*?)"
#Template\("[a-zA-Z.-]{0,}[:]")
Still, I think there should be a "simple" regexp for this kind of standard replacement.
Anyone has any clue ? Thank you for any hint
You can use this regex with a capture group: (#Template.*):.
And replace with this $1/.
But you'll have to use replace all until there's no : left, that won't take long.
Just explaining a lit bit more, everything between the parenthesis is a capture group that we can reference later in replace field, if we had (tem)(pla)te, $1 would be tem and $2 would be pla
Regex!
You can use this regex #Template\("(.[^\(\:]*)?(?:\:)(.[^\(\:]*)?(?:\:)?(.[^\(\:]*)?"\) and replacement would simply be #Template\("$1/$2/$3
You can test it out at https://regex101.com/r/VfZHFa/2
Explanation: The linked site will give a better explanation than I can write here, and has test cases you can use.
I need something besides [^0-9\n], I want a regex(but dont know how to make one), that captures anything in a pattern of numbers like this, "0000000000" or "000-000-0000" or basically any numbers that exist with spaces and or special characters right before or in between.
so any number, even like these (626*) 34a2- 4387) should convert to 6263424387
How can this be accomoplished? Im thinking its too hard?
You can search for all non-digits using:
\D+
OR
[^0-9]+
And replace by empty string.
RegEx Demo
I always forget this one myself and go hunting the internet for the answer. Here it is, for my future reference, and the rest of the internet
const rawNumber = '(555) 123-4567';
const strippedNumber = rawNumber.replace(/\D+/g, '');
Of course, my above example is JavaScript specific, but it can be adapted to other languages easily. The original post didn't specify language.
This is a better approche. it's includes decimal numbers as well.
"$9.0".replace(/[^0-9.]+/g, '');
JavaScript :)
Given this string: hello"C07","73" (quotes included) I want to get "C07". I'm using (?:hello)|(?<=")(?<screen>[a-zA-Z0-9]+)?(?=") to try to do this. However, it consistently matches "73" as well. I've tried ...0-9]+){1}..., but that doesn't work either. I must be misunderstanding how this is supposed to work, but I can't figure out any other way.
How can I get just the first set of characters between quotes?
EDIT: Here's a link to show my problem.
EDIT: Ok, here's exactly what I'm trying to do:
Basically, what I'm trying to get is this: 1) a positive match on "hello", 2) a group named "screen" with, in this case, "C07" in it and 3) a group named "format" with, in this case, "73" in it.
Both the "C07" and "73" will vary. "hello" will always be the same. There may or may not be an extra comma between "hello" and the first double-quote.
For you initial question of how to stop after the first match either removing the global search, or searching from the start of the string would accomplish that.
For the latter question you can name your groups and just keep extending the pattern throughout the line(s).
hello"(?<screen>[^"]+)","(?<format>[^"]+)"
Demo: http://regex101.com/r/PBXe8l/1
Based on your regex example, why not:
^(?:hello)"([a-zA-Z\d]+)"
Regex Demo
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 want to create a regex that would start with a integer number and then it might have a colon followed by a string. For example, it should pass for:
123
123:e43e
123:444+:343
I tried using the regex as:
String timeZoneRegex = "^\\d+[:(=[a-zA-Z+-:0-9]+)]*";
This does not work; appreciate any help here.
I have to say that some regexp features depend on the regexp engine, but try with:
\d+(\:[a-zA-Z0-9\-+]+)*
I've given a look to your express, you've made some mistake, maybe the most relevat one is the use of embeded [], you should know that inside the squared brackets the behaviour of symbols intepretation is a little different. This is a very good source if you want to learn them. Cheers.