问题描述
以下面的URL为例。
example.com作为输出。我知道有一个方法
servletrequest.getServerName()
。它为我输出 test1.example.com 任何帮助表示感谢。
在,您可以使用以下方法获取URI的各个部分。您也可以使用它们逐个重建URL(以帮助调试或其他任务),如下所示:
/ /示例:http:// myhost:8080 / people?lastname = Fox& age = 30
String uri = request.getScheme()+://+ //http+ ://
request.getServerName()+ //myhost
:+ request.getServerPort()+ //:+8080
request.getRequestURI() + /// people
(request.getQueryString()!= null??+
request.getQueryString():); //?+lastname = Fox&年龄= 30
所以是最接近的我们得到你需要。
根域:
对于根域,你将不得不处理从 getServer返回的
。这是必要的,因为Servlet无法提前知道你所谓的主机或者只是像 String
名称() .com
这样的域(它可能是一台名为网络中的 com
- 而不仅仅是后缀 - 谁知道?)。
对于您提供的模式(三分之一+二级+ com / net),以下内容应该得到你所需要的:
String domain = request.getServerName( ).replaceAll(。* \\。(?=。* \\。),);
以上将给出以下输入/输出:
www.test.com - > test.com
test1.example.com - > example.com
a.b.c.d.e.f.g.com - > g.com
www.com - > www.com
com - > com
Consider the below URL as an example.http://www.test1.example.com
Is there any method by which I can get "example.com" as an output. I know there is a method servletrequest.getServerName()
. It gives me output as test1.example.com
Any help appreciated.
In HttpServletRequest
, you can get individual parts of the URI using the methods below. You could also use them to reconstruct the URL piece by piece (to help debugging, or other tasks), like this:
// Example: http://myhost:8080/people?lastname=Fox&age=30
String uri = request.getScheme() + "://" + // "http" + "://
request.getServerName() + // "myhost"
":" + request.getServerPort() + // ":" + "8080"
request.getRequestURI() + // "/people"
(request.getQueryString() != null ? "?" +
request.getQueryString() : ""); // "?" + "lastname=Fox&age=30"
So request.getServerName()
is the closest we got to you need.
The "root domain":
For the "root domain", you'll have to work through the String
returned from getServerName()
. This is necessary because the Servlet would have no way of knowing ahead of time what you call "host" or what is just a domain like .com
(it could be a machine called com
in your network - and not just a suffix -, who knows?).
For the pattern you gave (one third+secondlevel+com/net), the following should get what you need:
String domain = request.getServerName().replaceAll(".*\\.(?=.*\\.)", "");
The above will give the following input/outputs:
www.test.com -> test.com
test1.example.com -> example.com
a.b.c.d.e.f.g.com -> g.com
www.com -> www.com
com -> com
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