Regex lookahead and behind? - regex

Question

So I have a unordered list that looks like:
<ul class='radio' id='input_16_5'>
<li>
<input name='input_5' type='radio' value='location_1' id='choice_16_5_0' />
<label for='choice_16_5_0' id='label_16_5_0'>Location 1</label></li>
<li>
<input name='input_5' type='radio' value='location_2' id='choice_16_5_1' />
<label for='choice_16_5_1' id='label_16_5_1'>Location 2</label></li>
<li>
<input name='input_5' type='radio' value='location_3' id='choice_16_5_2' />
<label for='choice_16_5_2' id='label_16_5_2'>Location 3</label></li>
</ul>
I would like to pass a value (ie. location_2) to a regular expression that will then capture the whole list item that it's a part of in order to remove it. So if I pass it location_2 it will match the to the (including) <li> and the </li> of the list item that it's in.
I can match up to the end of the list item with /location_3.+?(?=<li|<\/ul)/ but is there something I can do to match before and not capture other items?

Answer 1

This should get what you want
<li>(?:(?!<li>)[\S\s])+location_1[\S\s]+?<\/li>
Exaplanation
<li>: open li tag,
(?:(?!<li>)[\S\s])+: match for any characters including a newline and use negative look ahead to make sure that your highlight will not consume two or more <li> tags,
location_1: keyword that you use for highlight the whole <li> tag,
[\S\s]+?: any characters including a newline. (Here, thanks #Tensibai for your comment that make this regex be more simple with non-greedy)
<\/li> close li tag.
DEMO: https://regex101.com/r/cU4eC6/5
Additional information:
/<li>(?:(?!<li>).)+location_2.+?<\/li>/s
This regex is also work where you use modifier s to handle a newline instead of [\S\s]. (Thanks again to #Tensibai)