Bash脚本中提取网页图片链接ID的正则表达式求助

Hey there! Let's work through extracting those image IDs from your fetched HTML in Bash. First, a quick syntax fix for your variable setup—Bash doesn't allow spaces around the equals sign when assigning variables, and quoting your URL avoids issues with special characters:

currentURL="https://www.example.com"
currentPage=$(wget -q -O - "$currentURL")

情况1：提取<img>标签的id属性值 #

If you're targeting the id attribute directly on <img> elements (like <img id="hero-banner" src="...">), this Perl-compatible regex with grep will do the trick:

echo "$currentPage" | grep -oP '(?<=<img[^>]*id=")[^"]+'

Let's break this down:

• (?<=<img[^>]*id="): A positive lookbehind that finds an <img tag, skips any attributes before id=", and stops right at the opening quote of the ID value.
• [^"]+: Matches every character until the closing quote—this is your actual ID.

This handles those random &l... HTML entities you mentioned too, since it only cares about the content inside the quotes for the id attribute.

情况2：提取图片URL中的ID参数 #

If your "image link ID" refers to a query parameter in the image's src (like <img src="/photos?img_id=1234&size=large">), use this adjusted regex:

echo "$currentPage" | grep -oP '(?<=<img[^>]*src="[^?]*img_id=)[^&"]+'

Here's what's happening:

• (?<=<img[^>]*src="[^?]*img_id=): Looks for the src attribute in an <img> tag, skips the base URL, and stops right after img_id=.
• [^&"]+: Grabs everything until the next & (other parameter) or closing quote—your target ID.

兼容单双引号的通用版本 #

If some tags use single quotes for attributes (like <img id='sidebar-photo' ...>), use this flexible regex that works with both quote types:

echo "$currentPage" | grep -oP '(?<=<img[^>]*id=([\'"]))(?(1)[^\1]+)'

A Quick Note #

Regex works great for simple HTML, but for complex, nested pages, tools like pup (a command-line HTML parser) or xmllint are more reliable. But since you asked for regex, the above should cover your use case!

内容的提问来源于stack exchange，提问作者Mr. C