烧瓶404捕获要求的网址 [英] Flask 404 Catch requested url

问题描述

我使用此功能在网站上捕获404错误:

I'm catching 404 errors on my website with this function:

@app.errorhandler(404)
def page_not_found(e):
    logger.warning('User raised an 404: {error}'.format(error=str(e)))
    return render_template('404.html'), 404

但是我想知道用户尝试访问以提高404的网址，这是执行此操作的简单方法吗?

But I would like to know wich url the user tried to access to raise the 404, is they a simple way to do this ?

推荐答案

使用request.path知道哪个URL试图访问并引发404.

Use request.path to know which url tried to access and raised 404.

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