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如何:配置Spring WS以用''发布WSDL文件?WSDL样式的URL?

mule spring-ws web-services

Darren Bishop · 技术社区 · 15 年前

我正在尝试使用Mule ESB配置web服务代理。

我正试图使用Mule的WSProxyService来实现这一点,但是在遍历了相应的代码(使用调试器)之后,很明显这个类替换了端点地址。

http://xxxx/xxxx.wsdl ,但WSProxyService需要 http://xxxx/xxxx?wsdl 或 http://xxxx/xxxx&wsdl . 它用本地WSDL地址替换远程端点地址;它在问号处剪切远程WSDL地址,即?WSDL'是要被切掉的,所以要创建搜索项。但是由于Spring WS,这不起作用。

要分解它:

http://xxxx/xxxx.wsdl

替换

http://xxxx/xxxx

具有

http://yyyy/yyyy

有人注意到/解决过这个问题吗??

4 回复 | 直到 15 年前

Axel Fontaine 14 年前

在springwservlet前面的一个servlet过滤器检查url和参数如何?

如果您绝对必须拥有WS+的url,那么这应该很容易实现,并且可以满足您的需要?附加到它的WSDL。

代码如下:

public class WsdlQueryCompatibilityFilter implements Filter {
    @Override
    public void init(final FilterConfig filterConfig) throws ServletException {
        // do nothing
    }

    @Override
    public void doFilter(final ServletRequest request, final ServletResponse response, final FilterChain chain)
            throws IOException, ServletException {
        final HttpServletRequest httpServletRequest = (HttpServletRequest) request;
        if ("GET".equals(httpServletRequest.getMethod())
                && "wsdl".equalsIgnoreCase(httpServletRequest.getQueryString())) {
            httpServletRequest.getSession().getServletContext().getRequestDispatcher("/ws.wsdl")
                    .forward(request, response);
        } else {
            chain.doFilter(request, response);
        }
    }

    @Override
    public void destroy() {
        // do nothing
    }
}

Dave 15 年前

这可能适用于:

http://forum.springsource.org/showpost.php?p=101967&postcount=4

属于一个阶级。在SWS中,服务不是

jlucasps 9 年前

您可以创建一个实现 Filter ,创建 doFilter 方法拦截请求并获取URL(如果它以 "wsdl" ,然后将请求转发到Spring默认URL ".wsdl"

@Component
public class QuestionMarkFilter implements Filter {

    @Override
    public void doFilter(ServletRequest servletRequest, ServletResponse servletResponse, FilterChain chain) throws IOException, ServletException {

        HttpServletRequest request = (HttpServletRequest) servletRequest;

        if (request.getQueryString() != null && request.getQueryString().toLowerCase().endsWith("wsdl")) {
            request.getRequestDispatcher("/your/service/here.wsdl?").forward(request, servletResponse);
        } else {
            chain.doFilter(servletRequest, servletResponse);
        }
    }

    @Override
    public void init(FilterConfig filterConfig) { }

    @Override
    public void destroy() { }
}

Peter Knuts 8 年前

您还可以使用HttpServletRequestWrapper来处理?wsdl扩展并让服务器处理请求:

@Component
public class WsdlQuestionMarkSuffixFilter implements javax.servlet.Filter {
    @Override
    public void init(FilterConfig filterConfig) throws ServletException {

    }

    @Override
    public void doFilter(ServletRequest request, ServletResponse response, FilterChain chain) throws IOException, ServletException {
        HttpServletRequest httpRequest = (HttpServletRequest) request;
        if ("wsdl".equalsIgnoreCase(httpRequest.getQueryString())) {
            HttpServletRequestWrapper requestWrapper = new HttpServletRequestWrapper(httpRequest) {
                @Override
                public String getQueryString() {
                    return null;
                }

                @Override
                public String getRequestURI() {
                    return super.getRequestURI() + ".wsdl";
                }
            };
            chain.doFilter(requestWrapper, response);
        } else {
            chain.doFilter(request, response);
        }
    }

    @Override
    public void destroy() {

    }
}