Regex to find pattern, return subpattern

» javascript » Regex to find pattern, return subpattern

By : Sonia Das

Date : November 24 2020, 03:41 PM

This might help you in other language regex support lookbehind by using regex like this (?<=url\()(([^)]+))
But unfortunately javascript regex doesn't support lookbehind like this example, so you must iterate it. Except you don't realy need that url word to be included.

code :

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Regex pattern with subpattern exceptions (Python)

By : Andrew Jones

Date : March 29 2020, 07:55 AM

it should still fix some issue I am using BeautifulSoup to extract tabledata tags from a table. The TD's have a class of either 'a','u','e','available-unavailable' or 'unavailable-available'. (Yes, I know quirky class names but hey...)

code :

table.findAll('td', attrs = {"class":re.compile(r'(^|\s)(a|unavailable-available)($|\s)')})

class="a"
class="a ui-xxx"
class="ui-xxx a"
class="ui-xxx a ui-yyy"
class="unavailable-available"
class="unavailable-available foo"

Is it possible to reuse a subpattern inside of a pattern in Regex

By : user2716191

Date : March 29 2020, 07:55 AM

Does that help I am trying to write a Regex in C# but I have similar constructions that I would like not to duplicate but to reuse.

code :

^([A-Z][a-z]+(?:\s+|$)){3}$

Find lines that match a pattern with parameterized subpattern, but keep only the first hit for each subpattern

By : Arash Aryani

Date : March 29 2020, 07:55 AM

I hope this helps . The title is a bit long. , This might work for you:

code :

sed '/^X/d' file | sort -uk1,1

Regular Expressions - match pattern but return subpattern

By : Marcin Janiszewski

Date : March 29 2020, 07:55 AM

wish help you to fix your issue You have just to use capturing groups by using parentheses in your regex:

code :

Amount: (\d*) - Class (\d*)
        ^--^-- Here---^---^

String str = "0 - Amount: 3 - Class 29\n1 - Amount: 2 - Class 21\n2 - Amount: 11 - Class 1";

Pattern pattern = Pattern.compile("Amount: (\\d*) - Class (\\d*)");
Matcher matcher = pattern.matcher(str);

StringBuffer sb = new StringBuffer("{");
while (matcher.find()){
    sb.append(String.format("{%s, %s},", matcher.group(1), matcher.group(2)));
}
sb.setLength(sb.length()-1); // remove last comma
sb.append("}");

System.out.println(sb.toString())

How to find a pattern that does not contains a subpattern?

By : Kushagra Kumar

Date : March 29 2020, 07:55 AM

hop of those help? Use a negative lookahead as I have already mentioned, just use a consuming subpattern instead of the \1 (that was consuming text in your original regex). Also, do not forget that to match a literal dot, you need to escape it.

code :

ps\.set.*\(3, c0\.get\((.*)\)\);(\n.*){3}.*ps\.set.*\(7, c1\.get\(t\)\.get\((?!\1\))[^()]*\)\);\n.*ps\.addBatch\(\);
                                                                            ^^^^^^^^^^^^