java - 带有包含 URL 的字符串参数的 RequestMapping

Question

我有一个带有两个参数 long 和 String 的 Spring Controller :

@RequestMapping(value = "/webpage")
@Controller
public class WebpageContentController {
//...
    @RequestMapping(value = "{webpageId}/{webpageAddress}", method = RequestMethod.GET)
    public String contentWebpageById(@PathVariable long webpageId, @PathVariable String webpageAddress) {
        System.out.println("webpageId=" + webpageId);
        System.out.println("webpageAddress=" + webpageAddress);
        //...
    }
//...

如果我这样调用它:

http://localhost:8080/webarch/webpage/1/blahblah

一切都很好:

webpageId=1
webpageAddress=blahblah

但是如果我用斜杠传递字符串参数(在本例中是 URL 地址):

http://localhost:8080/webarch/webpage/1/https://en.wikipedia.org/wiki/Main_Page

我得到一个错误:

org.springframework.web.servlet.PageNotFound.noHandlerFound No mapping found for HTTP request with URI [/webarch/webpage/1/https://en.wikipedia.org/wiki/Main_Page] in DispatcherServlet with name 'appServlet'

如何传递这样的参数？

Answer 1

这个错误是由 spring Controller 映射引起的，当 spring 看到这样的 url 时

http://localhost:8080/webarch/webpage/1/https://en.wikipedia.org/wiki/Main_Page

它“不知道”“https://en.wikipedia.org/wiki/Main_Page” ' 应该作为参数映射到“{webpageId}/{webpageAddress}”映射，因为每个斜杠都被解释为更深层次的 Controller 方法映射。它寻找像 (webpage/1/http:{anotherMapping}/wiki{anotherMapping}/Main_Page{anotherMapping}) 这样的 Controller 方法映射，这种映射显然不是由“{webpageId}/”处理的{网页地址}”

编辑

根据您的评论，您可以尝试这样的操作

@RequestMapping(value = "/{webpageId}/**", method = RequestMethod.GET)
public String contentWebpageById(HttpServletRequest request) {

    String pattern = (String) request.getAttribute(HandlerMapping.BEST_MATCHING_PATTERN_ATTRIBUTE);  

    String extractedPathParam = pathMatcher.extractPathWithinPattern(pattern, request.getServletPath());
    extractedPathParam = extractedPathParam.replace("http:/", "http://");
    extractedPathParam = extractedPathParam.replace("https:/", "https://");
    //do whatever you want with parsed string..
}

使用 Spring 4.2.1

SomeParsing 应该使用一些正则表达式来仅提取 URL“变量”