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I am looking for a method to get the regex rules, for a capture group.
So if I had: /(\w) (\d)/ I would want $1 = /\w/ and $2 = /\d/.
Is there a method for this?
Provided that the regex is valid and no nested groups are used, then you can try this out.
var arr = (regex + "").match(/\(.*?\)/g).map(function(rule){
return rule.replace(/[()]/g, "");
});
Now, arr[0] will have rule of group1, arr[1] will be group2 and so on.
Although I don't know of any direct way to do this on a regex rule given by /(rule1)(rule2)/ you could build the regex rule using strings and specify your grouping in different strings.
var group1 = '([A-Za-z]+)',
group2 = '(\\d+)',
r = new RegExp(group1+group2);
r.test('hello1234');
To get the groups from a regex string you could run a regex on that regex string to extract the groups. If your regex string is "(\w+)(\d+)" then you'd have a regex to extract the groups as /(\([^\)\))+/
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I am a newbie with regex and I am having trouble trying to match some parts of a string so I can remove it from a piece of entered text. I want to match digits followed by a sequence of a combination of any special characters + spaces. There could also be non Latin characters that should not be removed inside the sequence (for example Ñ).
for example inputted it may look like:
11#- &-.text
11 $ab*cÑ .somewords123
outputted I would expect
text
abcÑsomewrods123
I am using javascript replaceall method with regex to find it. So far I have something basic like this regex
.replaceAll(/\d+(\#|\s)+(\-|\$)+(\s|\&)+(\&)+(\-)+(\.)/g, '');
Is there a way to write this more efficiently so that it captures any special characters since the text can contain more different special chars than in the examples? Or is this a situation better handled with pure JS?
Thanks in advance for your help.
You should ether have blacklist of what you calling special characters, or whitelist of the allowed characters.
for blacklist it gonna look like:
const blacklist = "!##$%^&*()_+.";
const exampleInputs = [
"te&*st+_1.",
"te%%st^*2",
"t###es*(*(*t3"
];
function removeSpecialChars(str) {
const reg = new RegExp(`[${blacklist}]`, "g");
return str.replace(reg, "");
}
exampleInputs.forEach(input => console.log(removeSpecialChars(input)));
for whitelist it gonna looks like:
const whitelist = "0-9a-zA-Z";
const exampleInputs = [
"te&*st+_1.",
"te%%st^*2",
"t###es*(*(*t3"
];
function removeSpecialChars(str) {
const reg = new RegExp(`[^${whitelist}]`, "g");
return str.replace(reg, "");
}
exampleInputs.forEach(input => console.log(removeSpecialChars(input)));
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I try to regex for not contain some character.
I need to show /%7(.*?);/g which dose not contain "=".
I try to input
?!xx=1
and change to
?!( (.?)=(.?) )
But it dose not work.
Please help. Thanks.
//Here is my simple regex
reg = /%7((?!xx=1).*?);/g ;
//Here is my string
str = "%7aa; %7bb=11; %7cc=123; %7xx=1; %7yy; %7zz=2;"
//I need
%7aa; and %7yy;
Instead of using a negative lookahead, try using a ^ block:
const reg = /%7([^=;]+);/g;
The ([^=;]+) bit matches any non-=, the condition you're looking for, and non-;, the character at the end of your regex.
I left the capture group in since your question's regex also contains it.
const reg = /%7([^=;]+);/g;
const str = "%7aa; %7bb=11; %7cc=123; %7xx=1; %7yy; %7zz=2;"
const matches = str.match(reg);
console.log(matches);
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I have a string 'OR-xxxxxxxx-001-01'.
I need to split it as 'OR-xxxxxxx-001' and '01'. Is it possible without split by '-' and concatenating again ??
In Java, use lastIndexOf:
String string = "OR-xxxxxxxx-001-01";
int lastDash = string.lastIndexOf('-');
String prefix = string.substring(0, lastDash); // OR-xxxxxxx-001
String suffix = string.substring(lastDash + 1); // 01
This solution is for JavaScript (Java and JavaScript is not the same thing).
Since you don't want to split by - it's possible to do it with indexes. You haven't specified if it's of fixed length or not (which this solution depends on), but this should do the trick:
t = 'OR-xxxxxxxx-001-01'
str = t.slice(0, 12) + t.slice(16);
Will give str = 'OR-xxxxxxxx-01'
You could use a split with a positive look-ahead:
String str = "OR-xxxxxxxx-001-01";
String[] split = str.split("-(?=[^-]+$)");
[^-]: any character other than -.
+: repeated one or more times
$: at the end of the String
Then the (?= ... ) is used for the positive look-ahead, so it won't be removed by the split. This means only the - outside the parenthesis act as delimiter to split on.
Try it online.
PS: In Java.
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I have a string like as shown below:
var String = "xxxxxxxx-xxxx-xxxx-xxxx-xxxxxxxxxxxx:some string without comma,xxxxxxxx-xxxx-xxxx-xxxx-xxxxxxxxxxxx:some string,with comma,xxxxxxxx-xxxx-xxxx-xxxx-xxxxxxxxxxxx:some string,xxxxxxxx-xxxx-xxxx-xxxx-xxxxxxxxxxxx:some string without comma,xxxxxxxx-xxxx-xxxx-xxxx-xxxxxxxxxxxx:some string , with comma,xxxxxxxx-xxxx-xxxx-xxxx-xxxxxxxxxxxx:String,with comma"
Where xxxxxxxx-xxxx-xxxx-xxxx-xxxxxxxxxxxx represents an alphanumeric generated Id and after the colon is a string related to that Id.The string can be a string with comma or without comma.What I wanted was that I wanted to split the string such that I get an array with ID:its corresponding string , just like shown below.
["xxxxxxxx-xxxx-xxxx-xxxx-xxxxxxxxxxxx:some string without comma","xxxxxxxx-xxxx-xxxx-xxxx-xxxxxxxxxxxx:some string,with comma","xxxxxxxx-xxxx-xxxx-xxxx-xxxxxxxxxxxx:some string","xxxxxxxx-xxxx-xxxx-xxxx-xxxxxxxxxxxx:some string without comma",
"xxxxxxxx-xxxx-xxxx-xxxx-xxxxxxxxxxxx:some string , with comma","xxxxxxxx-xxxx-xxxx-xxxx-xxxxxxxxxxxx:String,with comma"]
HOW I ACCOMPLISHED THIS
I used the javascript split function where i split the string by comma followed by 36 characters (for ID) and colon.
String.split(/,(?=.{36}:)/);
PS: I apologize as previously I was not able to ask the question in the correct manner.Hope this time people understand it.
You could use String#split by comma and a look ahead for numbers and colon.
var x = "123456:a,b,c,435213:r,567876:e,363464:t,y,u";
array = x.split(/,(?=\d+:)/);
console.log(array);
For alphanumeric values
var x = "1A3456:a,b,c,43Y213:r,567W76:e,363x64:t,y,u";
array = x.split(/,(?=[a-z0-9]+:)/i);
console.log(array);
You can use the method .split(). I used the "$" as a split sign instead of "," because i thought you would like to keep them.
var values = "123456:a,b,c$435213:r$567876:e$363464:t,y,u".split("$");
var x = values[0];
console.log(values)
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So I have a string that will have basically any kind of text in it and I want to capture all the text with regular expression and replace it.
var img_url = "http://grfx.domain.com/photos/dir/school/dir/m-dir/auto_original/123456.jpeg?123456";
img_url.replace(/regexp/,newvalue);
Just trying to capture everything in the img_url var and replace it.
Thanks!
Well, if you want to replace everything, you don't need regex. Simply write:
img_url = newvalue;
If you are really determined to do this with a regular expression, this could help:
img_url = img_url.replace(/.*/, newvalue);
If you're reassigning, just use =
If you want to replace all matches in one go, use the g flag, in which case this question is a duplicate of this post
If that expression needs to be dynamic:
var expr = new RegExp(replaceVar,'gi');//globally, and case-insensitively
var newUrl = oldUrl.replace(expr,replacement);
Note that, here, all back-slashes like you'd use for \b have to be escaped, too: \\b and \\/ for forward slashes. and \\\\ for an escaped backslash