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.
Related
I need a help with mass search and replace using regex.
I have a longer strings where I need to look for any number and particular string - e.g. 321BS and I need to replace just the text string that I was looking for. So I need to look for BS in "gf test test2 321BS test" (the pattern is always the same just the position differs) and change just BS.
Can you please help me to find particular regex for this?
Update: I need t keep the number and change just the text string. I will be doing this notepad++. However I need a general funcion for this if possible. I am a rookie in regex. Moreover, is it possible to do it in Trados SDL Studio? Or how am i able to do it in excel file in bulk?
Thank you very much!
Your question is a bit vague, however, as I understand it you want to match any digits followed by BS, ie 123BS. You want to keep 123 but replace BS?
Regex: (\d+)BS matches 123BS
In notepad++ you can:
match (\d+)BS
replace \1NEWTEXT
This will replace 123BS with 123NEWTXT.
\1 will substitue the capture group (\d+). (which matches 1 or more digits.
You could do this in Trados Studio using an app. The SDLXLIFF Toolkit may be the most appropriate for you. The advantage over Notepad++ is that it's controlled and will only affect the translatable text and not anything that might break the integrity of the file if you make a mistake. You can also handle multiple files, or even multiple Trados Studio projects in one go.
The syntax would be very similar to the suggestion above... you would:
match (\d+)BS
replace $1NEWTEXT
I have a string, let's say, jkdfkskjak some random string containing a desired word
I want to check if the given string has a word from a set of words, say {word1, word2, word3} in latex.
I can easily do it in Java, but I want to achieve it using regex. I am very new to regular expressions.
if you want only to recognise the words as part of a word, then use:
(word1|word2|...|wordn)
(see first demo)
if you want them to appear as isolated words, then
\b(word1|word2|...|wordn)\b
should be the answer (see second demo)
I am not able to understand the complete context like what kind of text you have or what kind of words will this be but I can offer you a easy solution the literal way programmatically you can generate this regex (dormammu|bargain) and then search this in text like this "dormammu I come to bargain". I have no clue about latex but I think that is not your question.
For more information you can tinker with it at [regex101][1].
If you are having trouble understanding it [regexone][2] this is the place to go. For beginners its a good start.
[1]: http://regex101.com [2]: https://regexone.com/
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 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
I am looking to find anything that matches this pattern, the beginning word will be:
organism aogikgoi egopetkgeopt foprkgeroptk 13
So anything that starts with organism needs to be found using regex.
^organism will match anything starting with "organism".
^organism(.*) will also capture everything that follows, into the variable that contains the first match (which varies according to language -- in Perl it's $1).
Also just wanna add for others newbies like me and their various circumstances, you can do it in various ways depending on your text and what you are tryna do.
Like here's an Example where I wanna delete everything after ?spam so I could use .?spm.+ or .?spm.+ or any other ways as long you are creative about it lol.
This might come in handy, here's a Link | Link where you can find some basic necessary regex and their meanings.