1
I’m trying to build a regular expression and it’s a little difficult.
I would like you to help me, and if possible explain how the computer works in relation to this expression that I am asking for.
I need to take everything at the beginning of a string, until a string group is found, for example:
I want to pick from the beginning of the string until it is found "2.8" or "6v" or "2p".
Thank you.
5
/^(.)+(?=([\d+]\.?[\d+]|\d\w))/g
^(.)+
(?=
(
[\d+]\.?[\d+]
|
\d\w
)
)
This expression will get any text that comes before "0.0" or "0x".
Explanation of the expression:
^(.)+ any text from the beginning of the string(?=) the text shall be followed by the following expression:[\d+]\.?[\d+] digit digit digit digit, or...\d\w digit and a letterYour solution was the one that worked the way I wanted, thank you!
0
/((^[\s\S]+)2\.8)|((^[\s\S]+)6v)|((^[\s\S]+)2p)/g
Explaining:
[\s\S] picks up everything.
+ takes all characters that meet the above condition, so "all of everything".
^ indicates the beginning of String, so I put before searching for "all".
After that, just put where you want the search to stop. The parentheses are for organizing purposes, to leave everything separated.
| is an "or" operator. I just put the same code 3 times, changing only the stop condition.
\. is to search by the point. I had forgotten to add but edited.
Here’s an example: http://regexr.com/3evn9
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http://www.regexpal.com/ this tool helps too much in the construction of regular expressions.
– relaxeaza
Are those "2.8" or "6v" or "2p" strings dynamic right? are they in an array? or will you have to type by hand?
– Sergio
No! I’m going to set something around 3 or 4 groups of Strings that will always be these
– Edvaldo Szymonek