请求帮助实现仅显示网页当前子目录URL并解决空格编码问题

Hey there! Let's work through this together—since you're new to coding, I'll keep this simple and break down every step.

What we need to fix: #

1. Extract only the subdirectory part (like /dir1/ instead of the full URL)
2. Convert %20 back to regular spaces so they display correctly

Solution 1: Get the full subdirectory path (with spaces fixed) #

Replace your existing script with this code:

<p3>
  <script>
    // Grab the part of the URL after the domain (e.g., "/dir1" from "example.com/dir1")
    let path = location.pathname;
    // Decode URL-encoded characters (turns "%20" into a regular space)
    let decodedPath = decodeURI(path);
    // Add a trailing slash if it's not already there (and we're not at the root)
    if (decodedPath !== '/' && !decodedPath.endsWith('/')) {
      decodedPath += '/';
    }
    // Write the final result to the page
    document.write(decodedPath);
  </script>
</p3>

How this works:

• location.pathname is a built-in browser property that directly gives you the path after the domain—no need to mess with the full URL.
• decodeURI() handles all URL-encoded characters (not just %20, it works for things like %23 for # too).
• The if statement makes sure we always have a trailing slash, matching your example of /dir1/.

Solution 2: Get only the first-level subdirectory (if that's what you need) #

If you want to show just the top-level subdirectory (e.g., /dir1/ even if the URL is example.com/dir1/subdir), use this instead:

<p3>
  <script>
    let path = location.pathname;
    let decodedPath = decodeURI(path);
    // Split the path into parts and remove empty strings
    let pathSegments = decodedPath.split('/').filter(segment => segment !== '');
    // If there's a subdirectory, format it with slashes; otherwise show root
    let firstSubdir = pathSegments.length > 0 ? `/${pathSegments[0]}/` : '/';
    document.write(firstSubdir);
  </script>
</p3>

Just copy whichever code fits your needs into your <p3> tag, and it should work exactly how you want!

内容的提问来源于stack exchange，提问作者Kazankoph