在 Python 中构建 URL 时如何连接路径的组件 [英] How to join components of a path when you are constructing a URL in Python

问题描述

例如，我想将前缀路径加入到/js/foo.js 等资源路径中.

For example, I want to join a prefix path to resource paths like /js/foo.js.

我希望生成的路径相对于服务器的根目录.在上面的例子中，如果前缀是media"，我希望结果是/media/js/foo.js.

I want the resulting path to be relative to the root of the server. In the above example if the prefix was "media" I would want the result to be /media/js/foo.js.

os.path.join 在这方面做得很好，但它如何连接路径取决于操作系统.在这种情况下，我知道我的目标是网络，而不是本地文件系统.

os.path.join does this really well, but how it joins paths is OS dependent. In this case I know I am targeting the web, not the local file system.

当您使用已知将在 URL 中使用的路径时，是否有最佳选择?os.path.join 能正常工作吗?我应该自己动手吗?

Is there a best alternative when you are working with paths you know will be used in URLs? Will os.path.join work well enough? Should I just roll my own?

推荐答案

由于从 OP 发布的评论来看，他似乎不想希望在连接中保留绝对 URL"(这是 urlparse.urljoin 的关键工作之一；-)，我建议避免这样做.os.path.join 也很糟糕，原因完全相同.

Since, from the comments the OP posted, it seems he doesn't want to preserve "absolute URLs" in the join (which is one of the key jobs of urlparse.urljoin;-), I'd recommend avoiding that. os.path.join would also be bad, for exactly the same reason.

所以，我会使用类似 '/'.join(s.strip('/') for s inpieces)(如果前导 / 必须也可以忽略——如果首部必须是特殊情况，那当然也是可行的；-)

So, I'd use something like '/'.join(s.strip('/') for s in pieces) (if the leading / must also be ignored -- if the leading piece must be special-cased, that's also feasible of course;-).

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