I would like to manipulate this following string: 20_hbeu50272359_402_21
so that I am left with only hbeu50272359 is there a way to how could I delete those outer numbers? Not sure how to go about it.
The strings may also take this format:
hbeu50272359_402_21
You can use split
"20_hbeu50272359_402_21".split('_')[1]
The above splits your string based on the _ char, then use [1] to select the part you want (in the returned array or ["20", "hbeu50272359", "402", "21"])
This of course would asume that your strings always follow the same format. if not you could use a regEx match with what appears to be a prefix of hbeu. Something like /(hbeu[0-9]+)/ But I'd need more info to give a better approach to the split method
UPDATE:
based on your update you could do a regex match like:
var stuffWeWant = "20_hbeu50272359_402_21".match(/(hbeu[0-9]+)/)[0];
// this will work regardless of the string before and after hbeu50272359.
"20_hbeu50272359_402_21".match(/(hbeu[0-9]+)/)[0]; in my console I get hbeu50272359
You could use string.replace function also. ie, replace
One or more digits plus the following underscore symbol at the start,
OR
underscore plus one or more digits followed by another underscore or end of the line anchor
with an empty string.
> "20_hbeu50272359_402_21".replace(/_\d+(?=_|$)|^\d+_/g, "")
'hbeu50272359'
> "hbeu50272359_402_21".replace(/_\d+(?=_|$)|^\d+_/g, "")
'hbeu50272359'
(?!\d)[^_]+ will work on both strings and it doesn't require the second string to start with hbeu.
20_hbeu50272359_402_21".match(/_(.*?)_/)[1]In general the split solution is better in performance compared to using a regular expression.
* is greedy, the engine repeats it as many times as it can, so the regex continues to try to match the _ with next characters, resulting with matching
_hbeu50272359_402_21
In order to make it ungreedy, you should add an ?:
_(.*?)_
Now it'll match the first _somecharacters_.
I would use String.replace() for that:
var s = "20_hbeu50272359_402_21";
var part = s.replace(/.*?_(.*?)_.*/, '$1');
The search pattern matches any char .* until the next occurrence of a _. Using the ungreedy quantifier ? makes sure it does not match any char until the last _ in the string. The it captures the text until the next _ into a capturing group. (.*?). Then the next _ and the remaining string is matched.
The replace pattern uses $1 to address the contents of the first capturing group and throws away the remaining match.
*is greedy..var s = '20_hbeu50272359_402_21'; /_([^_]+)_/.exec(s)[1];