如何用正则表达式排除空p标签，仅移除其他空HTML标签？

Great question! To preserve empty <p> tags while stripping all other empty elements from your HTML string, you can modify your regex with a negative lookahead to exclude the p tag. Here's the adjusted code:

let s = "<h1>test</h1><h1></h1><p>a</p><p></p><h2></h2>";
s = s.trim().replace( /<(?!p\b)(\w*)\s*[^\/>]*>\s*<\/\1>/g, '' )
console.log(s); // Output: "<h1>test</h1><p>a</p><p></p>"

How it works: #

The key change is adding (?!p\b) immediately after the opening < in the regex. This is a negative lookahead that tells the regex:

• "Do NOT match this opening tag if it starts with p followed by a word boundary (\b)"
• The word boundary ensures we don't accidentally exclude similar tags like <pre> or <progress>

The rest of the regex behaves as before:

• (\w*) captures the tag name to ensure we match the corresponding closing tag
• \s*[^\/>]* accounts for optional whitespace and attributes in the opening tag
• \s*<\/\1> matches the closing tag (allowing for any whitespace inside the empty tag)

This modification will keep your empty <p> tags intact while removing all other empty elements like the empty <h1> and <h2> in your sample input.

内容的提问来源于stack exchange，提问作者totalnoob