regexr | composing regular expressions without the need | Regex library

by trusktr JavaScript Version: Current License: No License

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kandi X-RAY | regexr Summary

regexr is a JavaScript library typically used in Utilities, Regex, Nodejs applications. regexr has no bugs, it has no vulnerabilities and it has low support. You can install using 'npm i regexr' or download it from GitHub, npm.

Easily compose regular expressions. Doing this with strings would otherwise be tedious due to having to double-escape things. (Note that int is an instance of RegExp and can be composed into the template string, and the resulting USD is also a RegExp). Regexr provides an ES6 template tag function that makes it easy to compose RegExps using template strings without double-escaped hell.

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Quality

Security

License

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Support

regexr has a low active ecosystem.

It has 54 star(s) with 1 fork(s). There are 1 watchers for this library.

It had no major release in the last 6 months.

There are 1 open issues and 3 have been closed. On average issues are closed in 352 days. There are no pull requests.

It has a neutral sentiment in the developer community.

The latest version of regexr is current.

Quality

regexr has 0 bugs and 0 code smells.

Security

regexr has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.

regexr code analysis shows 0 unresolved vulnerabilities.

There are 0 security hotspots that need review.

License

regexr does not have a standard license declared.

Check the repository for any license declaration and review the terms closely.

Without a license, all rights are reserved, and you cannot use the library in your applications.

Reuse

regexr releases are not available. You will need to build from source code and install.

Deployable package is available in npm.

Installation instructions are not available. Examples and code snippets are available.

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regexr Key Features

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regexr Examples and Code Snippets

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Community Discussions

Trending Discussions on regexr

Using regex on Python to find any numerical value in an expression

PLSQL - help on my regex phone number with area code

Capture from 1st occurance of x to nth occurance of y

Regex to find all placeholder occurrences in text

Regex, split at groups and get numbers

Karate - XML Assertion with regex for a number

regExr pattern for get char between to special char

Why Regular Expression is not working correctly?

get some char between two spacial char without space with regex

SyntaxError: (irb):4: invalid pattern in look-behind (positive look-behind/ahead)

QUESTION

Using regex on Python to find any numerical value in an expression

Asked 2021-Jun-15 at 04:23

I am trying to get all numerical value (integers,decimal,float,scientific notation) from an expression and want to differentiate them from digits that are not realy number but part of a name. For example in the expression below.

...

ANSWER

Answered 2021-Jun-15 at 04:23

This should take care of it. (All the items are strings)

Source https://stackoverflow.com/questions/67979579

QUESTION

PLSQL - help on my regex phone number with area code

Asked 2021-Jun-14 at 21:00

I have difficulties to find the right regex under PL/SQL, but my regex is normally good

I have a phone number like this :

+44 (0)22 3333 4444 from the text that should not be there

And I want to get this:

+4402233334444

So I made the following regex:

/[^+\d]|\s/g

It works very well on the site https://regexr.com/ but not in my PL/SQL query, it gives me the same result

I tried to use the oracle doc, but without success https://www.techonthenet.com/oracle/regexp_like.php

...

ANSWER

Answered 2021-Jun-14 at 20:57

The \d and other shorthand character classes should not be used inside a bracket expression.

You can use

Source https://stackoverflow.com/questions/67977099

QUESTION

Capture from 1st occurance of x to nth occurance of y

Asked 2021-Jun-10 at 18:11

I have a regular expression problem that seems fairly do-able, but i can't seem to be able to get correct.

Give strings that look like:

...

ANSWER

Answered 2021-Jun-10 at 18:11

You get the error because you are using a lookbehind, (?<=\|), but did not use the perl=TRUE argument to enable the PCRE regex engine that supports lookbehinds. The default TRE regex engine does not support lookarounds.

You need to match the whole string:

Source https://stackoverflow.com/questions/67926318

QUESTION

Regex to find all placeholder occurrences in text

Asked 2021-Jun-03 at 09:57

Im struggling to create a Regex that finds all placeholder occurrences in a given text. Placeholders will have the following format:

...

ANSWER

Answered 2021-Jun-02 at 14:33

string input = "[{PRE.Word1.Word2}]";

// language=regex
string pattern = @"\[{ PRE \. (?'group1' .{1,15}? ) \. (?'group2' .{1,64}? ) }]";

var match = Regex.Match(input, pattern, RegexOptions.IgnorePatternWhitespace);

Console.WriteLine(match.Groups["group1"].Value);
Console.WriteLine(match.Groups["group2"].Value);

Source https://stackoverflow.com/questions/67806697

QUESTION

Regex, split at groups and get numbers

Asked 2021-Jun-02 at 20:53

https://regexr.com/5u8j6

Expression: (?<=\/)(\d*)

...

ANSWER

Answered 2021-Jun-02 at 19:37

My two cents to solving your problem:

Source https://stackoverflow.com/questions/67811075

QUESTION

Karate - XML Assertion with regex for a number

Asked 2021-May-26 at 14:29

For the following data,

...

ANSWER

Answered 2021-May-26 at 14:29

You have a list of 2 strings, please pay attention to the error message.

Try this:

Source https://stackoverflow.com/questions/67706368

QUESTION

regExr pattern for get char between to special char

Asked 2021-May-24 at 22:03

for get *created by:* and *M* and don't get *proudly * or * proudly*(have space near * or * ) with regex in javascript.

...

ANSWER

Answered 2021-May-24 at 17:28

The pattern that you tried matches:

\* Match *
( Capture group 1
- [^*]* Match 0+ times any char except * (So also match a space and therefore also match * proudly*)
- [^*\s] Match a single char other than * or a whitespace char
) Close group
\* Match *

To not get a space after the opening * or before the closing * you can start the match after the asterix with a char other than an asterix or whitespace char.

Then optionally repeat matching a space and again any char except an asterix or whitespace char to not match a space at the end.

Source https://stackoverflow.com/questions/67676141

QUESTION

Why Regular Expression is not working correctly?

Asked 2021-May-24 at 20:03

I was trying to solve Valid Number problem on leetcode. My choice of language for this problem is Java.

Here is the problem statement if you do not wish to visit website.

Problem Statement

A valid number can be split up into these components (in order):

A decimal number or an integer.
(Optional) An 'e' or 'E', followed by an integer.

A decimal number can be split up into these components (in order):

(Optional) A sign character (either '+' or '-').
One of the following formats:
1. At least one digit, followed by a dot '.'.
2. At least one digit, followed by a dot '.', followed by at least one digit.
3. A dot '.', followed by at least one digit.

An integer can be split up into these components (in order):

(Optional) A sign character (either '+' or '-').
At least one digit.

For example, all the following are valid numbers:

["2", "0089", "-0.1", "+3.14", "4.", "-.9", "2e10", "-90E3", "3e+7", "+6e-1", "53.5e93", "-123.456e789"]

while the following are not valid numbers:

["abc", "1a", "1e", "e3", "99e2.5", "--6", "-+3", "95a54e53"]

Given a string s, return true if s is a valid number.

My Approach

I tried to match the given string against the following regular expression.

((\\+|-)?(\\d*\\.\\d+|\\d)[eE]?)

I analysed the whole regular expression on RegExr in Javascript. It's working fine in Javascript when I check it against inputs. I just simply escaped the escape character \ to make my regular expression work in Java.

Regular Expression in Javascript :- /\+|-?(\d*\.\d+|\d)[eE]?/g

Below is the whole code :-

...

ANSWER

Answered 2021-May-24 at 20:03

Do not use regex to solve it. Try parsing the string using Double#parseDouble and if it fails, declare that the string does not represent a number.

Demo:

Source https://stackoverflow.com/questions/67677354

QUESTION

get some char between two spacial char without space with regex

Asked 2021-May-24 at 17:19

for example : i wanna get *created by:* and *M* and don't get *proudly * or * proudly*(have space near * or * ) with regex in javascript.

...

ANSWER

Answered 2021-May-24 at 12:45

First, match all result that is in between * and then filter the result that doesn't contain space before last (*)

Source https://stackoverflow.com/questions/67672139

QUESTION

SyntaxError: (irb):4: invalid pattern in look-behind (positive look-behind/ahead)

Asked 2021-May-22 at 00:18

I am trying to write a regex-replace pattern in order to replace a number in a hash like such:

regexr link

...

ANSWER

Answered 2021-May-22 at 00:18

The reason is that Ruby's Onigmo regex engine does not support infinite-width lookbehind patterns.

In a general case, positive lookbehinds that contain quantifiers like *, + or {x,} can often be substituted with a consuming pattern followed with \K:

Source https://stackoverflow.com/questions/67644969

Community Discussions, Code Snippets contain sources that include Stack Exchange Network

Vulnerabilities

No vulnerabilities reported

Install regexr

You can install using 'npm i regexr' or download it from GitHub, npm.

Support

For any new features, suggestions and bugs create an issue on GitHub. If you have any questions check and ask questions on community page Stack Overflow .

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