request.url returns me this : http://localhost:3000/page?foo=bar.
request.url返回给我：http：// localhost：3000 / page？foo = bar。
Is there a method I can call to get http://localhost:3000/page , or do I have to parse the string to strip get parameters?
有没有我可以调用的方法来获取http：// localhost：3000 / page，还是我必须解析字符串以去除get参数？
3 个解决方案
#1
43
request.path should return what you're looking for if you're not concerned about the hostname. Otherwise you might try:
如果您不关心主机名，request.path应该返回您要查找的内容。否则你可能会尝试：
url_for(:only_path => false, :overwrite_params=>nil)
#2
8
I use the following:
我使用以下内容：
request.original_url.split('?').first
It does not regenerate the path, and therefore gives you exactly the url that the end-user sees in their browser, minus query parameters.
它不会重新生成路径，因此会为您提供最终用户在浏览器中看到的网址，减去查询参数。
#3
8
To get the request URL without any query parameters.
获取没有任何查询参数的请求URL。
def current_url_without_parameters
request.base_url + request.path
end
#1
43
request.path should return what you're looking for if you're not concerned about the hostname. Otherwise you might try:
如果您不关心主机名，request.path应该返回您要查找的内容。否则你可能会尝试：
url_for(:only_path => false, :overwrite_params=>nil)
#2
8
I use the following:
我使用以下内容：
request.original_url.split('?').first
It does not regenerate the path, and therefore gives you exactly the url that the end-user sees in their browser, minus query parameters.
它不会重新生成路径，因此会为您提供最终用户在浏览器中看到的网址，减去查询参数。
#3
8
To get the request URL without any query parameters.
获取没有任何查询参数的请求URL。
def current_url_without_parameters
request.base_url + request.path
end