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Regex, how to detect value in double curly braces? [closed]

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Sorry i can't speak english :(
I write Supplant I have a regex
({{([']).*\2}})
needs regex to detect what is in a rectangle
I need it rigidly between {{' - '}} because there can be any character between the quotation marks (e.g. one { )
best to use it \1 because regex is more longer
Thanks for the help. Regards

This can work for you:
/\{\{'((?:(?!\{\{').)*?)'\}\}/g
\{\{' - start by matching {{'.
( ... ) - if you're writing a template engine you probably care about what's inside the curly braces and quotes. This captures the string inside the quotes so you'll be able to use it.
(?: ... ) - a non-captureing group (the value here will not be used).
(?!\{\{'). - match anything, except if we're seeing another {{'.
(?!\{\{').)*? - *? is a lazy match, so we'll stop at the first '}}
and finally, the closing '}}
as code it will looks something like this - I included a function to set the replaced value, because typically that's what you'd do in a template engine:
let s = "n {{'a{123456789}'}} n {{\"a{123456789}\"}} n {{'a{1234{{'a{12345{{'a{123456789}'}}6789}'}}56789}'}} ";
s = s.replace(/\{\{'((?:(?!\{\{').)*?)'\}\}/g, (wholeMatch, capturedKey) => {
console.log('captured key:', capturedKey);
return "REPLACED " + capturedKey;
});
console.log(s);
if you want to support double quotes it becomes a bit more complicated:
s = s.replace(/\{\{(["'])((?:(?!\{\{["']).)*?)\1\}\}/g,
(wholeMatch, quote, capturedKey) => { ... }
);

Try this:/(?<={{')(?!.+?{{)[^']+/gm
This basically looks for the {{' which doesn't have any other {{ in front of it then, matches everything till the first '.
Test here: https://regex101.com/r/rw6kkP/2
var myString = `{{'a{content 1}a'}}
{{'a{every{{'a{everysign}a'}}sign}a'}}
{{'a{every{{'a{every{{'a{inside content}a'}}sign}a'}}sign}a'}} `;
var myRegexp = /(?<={{')(?!.+?{{)[^']+/gm;
console.log(myString.match(myRegexp))
// [ "a{content 1}a", "a{everysign}a", "a{inside content}a" ]

How to match numbers followed by any combination of special characters and spaces using regex [closed]

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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)));

Javascript - Regex Does not contain some character [closed]

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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);

Split operating without concating [closed]

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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.

Get regex rules for a capture group [closed]

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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 /(\([^\)\))+/