I have this for example:
<#445288012218368010>
And I want to get from between <# > symbols the value.
I tried so:
string.replace(/^(?:\<\#)(?:.*)(?:\>)$/gim, '');
But then I don't get any result. It will delete/remove the whole string.
I want only this part: 445288012218368010 (it will be dynamic, so yeah it will be not the same numbers).
Anyway it is for the discord chat bot and I know that there is other methods for check the mentioned names but I want to do that in regex because which I am trying to do can't go the common method.
So yeah how can I get the value from between those symbols?
I need this in node.js regex.
You can use String#match which will return regular expression matches for the string (in this case the RegExp would be <#(\d+)> (the parenthesis around the \d+ make \d+ become its own group). This way you can use <string>.match(/<#(\d+)>/) to get the regular expression results and <string>.match(/<#(\d+)>/)[1] to get the first group of the regex (in this case the number).
You regex matches but you use a non capturing group (?:.*) so you get the full match and replace that with an empty string. Note that you could omit the first and the third non capturing group and use <# and > instead.
You could match what is between the brackets using a capturing group ([^>]+) or (\d+) and use replace and refer the first capturing group $1 in the replacement.
console.log("<#445288012218368010>".replace(/^<#([^>]+)>$/gim, '$1'));
Related
There is one condition where I have to split my string in the manner that all the alphabetic characters should stay as one unit and everything else should be separated like the example shown below.
Example:
Some_var='12/1/20 Balance Brought Forward 150,585.80'
output_var=['12/1/20','Balance Brought Forward','150,585.80']
Yes, you could use some regex to get over this.
Some_var = '12/1/20 Balance Brought Forward 150,585.80'
match = re.split(r"([0-9\s\\\/\.,-]+|[a-zA-Z\s\\\/\.,-]+)", Some_var)
print(match)
You will get some extra spaces but you can trim that and you are good to go.
split isn't gonna cut it. You might wanna look into Regular Expressions (abbreviated regex) to accomplish this.
Here's a link to the Python docs: re module
As for a pattern, you could try using something like this:
([0-9\s\\\/\.,-]+|[a-zA-Z\s\\\/\.,-]+)
then trim each part of the output.
I have an example text string text_var = 'ndTail7-40512-1' and I want to split the first time I see a number followed by a - BUT I want to keep the number. Currently, I have print(re.split('\d*(?=-)',text_var,1)) and my output is ['ndTail', '-40512-1']. But I want to keep that number which is the trigger so it should look like ['ndTail', '7-40512-1']. Any help?
We can try using re.findall here:
text_var = 'ndTail7-40512-1'
matches = re.findall(r'(.*?)(\d-.*$)', text_var)
print(matches[0])
This prints:
('ndTail', '7-40512-1')
Sometimes it can be easier to use re.findall rather than re.split.
The regex pattern used here says to:
(.*?) match AND capture all content up to, but including
(\d-.*$) the first digit which is followed by a hyphen;
match and capture this content all the way to the end of the input
Note that we are using re.findall which typically has the potential to return multiple matches. However, in this case, our pattern matches to the end of the input, so we are left with just a single tuple containing the two desired capture groups.
I'm looking for a regex to match this: a_*_*#example.com where the * is any text of any length. Doing this in NodeJS
Additionally I'm looking for a regex that matches any string not including the # symbol.
a_.*_.*#example\.com for the first
^[^#]*$ for the second
I wanted to correct the automatically created Linux scripts. I use findAll(String, String) function to change "$APP_ARGS" for something else.
I have tried variants:
replaceAll('"$APP_ARGS"', 'simulators ' + '"\\\\$APP_ARGS"') - doesn't find
replaceAll('\"\$APP_ARGS\"',... - doesn't find
replaceAll('"\$APP_ARGS"',... - doesn't find
replaceAll('\\"\\$APP_ARGS\\"',... - editor warning - excessive escape
replaceAll('"\\\\$APP_ARGS"',... - doesn't find
replaceAll('\\\\"\\\\$APP_ARGS\\\\"',... - doesn't find
replaceAll($/"$$APP_ARGS"/$, ...) - does not find
replaceAll('"[$]APP_ARGS"', 'something simple') - finds.
replaceAll('"[$]APP_ARGS"', '"\\\\$APP_ARGS"') - fails.
As you see, if I use the regex format, the finding works ok. But is there a way to make an escaping work? For I need that $ in the replacing string, too.
According to Groovy manuals, /../ string needn't escaping for anything except slashes themselves. But
replaceAll(/"$APP_ARGS"/,...
fails, too, with a message: Could not get unknown property 'APP_ARGS'.
It seems that behaviour of that function has no logic and we have to find the correct solution by experiments.
replaceAll('"\\$APP_ARGS"', 'simulators ' + '"\\$APP_ARGS"')
The additional possible problem is that \\ before $ should be in the both strings, replacing and replaced.
The first argument of replaceAll is always treated as an regexp, so we need to quote $ (line end). The second param may contain backreferences to groups from the regexp, which start with a $, so that one must be quoted too.
A saner way is to use replace instead of replaceAll, which already quotes/escapes both params according to that useage.
I have a string like hello /world today/
I need to replace /world today/ with /MY NEW STRING/
Reading the manual I have found
newString = string.match("hello /world today/","%b//")
which I can use with gsub to replace, but I wondered is there also an elegant way to return just the text between the /, I know I could just trim it, but I wondered if there was a pattern.
Try something like one of the following:
slashed_text = string.match("hello /world today/", "/([^/]*)/")
slashed_text = string.match("hello /world today/", "/(.-)/")
slashed_text = string.match("hello /world today/", "/(.*)/")
This works because string.match returns any captures from the pattern, or the entire matched text if there are no captures. The key then is to make sure that the pattern has the right amount of greediness, remembering that Lua patterns are not a complete regular expression language.
The first two should match the same texts. In the first, I've expressly required that the pattern match as many non-slashes as possible. The second (thanks lhf) matches the shortest span of any characters at all followed by a slash. The third is greedier, it matches the longest span of characters that can still be followed by a slash.
The %b// in the original question doesn't have any advantages over /.-/ since the the two delimiters are the same character.
Edit: Added a pattern suggested by lhf, and more explanations.