为什么web.xml'context-param'不能有字符[children]?

问题描述 投票:6回答:4

我只是很好奇这个web.xml代码发生了什么,我在Eclipse(juno版本)cvc-complex-type.2.3: Element 'context-param' cannot have character [children], because the type's content type is element-only.视图中有这个错误Markers

这是代码:

<?xml version="1.0" encoding="UTF-8"?>
<web-app id="WebApp_ID" 
        version="2.5" 
        xmlns="http://java.sun.com/xml/ns/javaee" 
        xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
        xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">

  <display-name>Spring security web application (series)</display-name>

  <!-- to specifically stop trouble with multiple apps on tomcat -->
  <context-param>
    <param-name>webAppRootKey</param-name>
    <param-value>customauth_root</param-value>
  </context-param>

  <!-- Location of the XML file that defines the root application context
    applied by ContextLoaderListener. -->
  <context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>
        WEB-INF/applicationContext.xml
        WEB-INF/applicationContext-security.xml
        </param-value>
  </context-param>

  <!-- Loads the root application context of this web app at startup. The
    application context is then available via WebApplicationContextUtils.getWebApplicationContext(servletContext). -->
  <listener>
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
  </listener>

  <listener>
    <listener-class>
      org.springframework.security.web.session.HttpSessionEventPublisher</listener-class>
  </listener>

  <filter>
    <filter-name>springSecurityFilterChain</filter-name>
    <filter-class>org.springframework.web.filter.DelegatingFilterProxy</filter-class>
  </filter>

  <filter-mapping>
    <filter-name>springSecurityFilterChain</filter-name>
    <url-pattern>/*</url-pattern>
  </filter-mapping>

  <!-- Provides core MVC application controller. See customauth-servlet.xml. -->
  <servlet>
    <servlet-name>customauth</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <load-on-startup>1</load-on-startup>
  </servlet>

  <servlet-mapping>
    <servlet-name>customauth</servlet-name>
    <url-pattern>*.htm</url-pattern>
  </servlet-mapping>

  <servlet-mapping>
    <servlet-name>customauth</servlet-name>
    <url-pattern>*.do</url-pattern>
  </servlet-mapping>

  <welcome-file-list>
    <welcome-file>index.jsp</welcome-file>
  </welcome-file-list>

</web-app>
eclipse java-ee
4个回答
9
投票

对于初学者,您在XML节点param-value中使用非法字符

您不能在节点内使用“/”,“<”和其他字符,因为它会破坏XML文档的解析。

您应该在XML节点内围绕您的值使用CDATA或PCDATA包装器。

EG

    <param-value><![CDATA[
        WEB-INF/applicationContext.xml
        WEB-INF/applicationContext-security.xml]]>
    </param-value>

这就是造成问题的原因。


2
投票

某些不可见的字符不被视为XML空格,因此会显示错误。通过配置Eclipse“文本编辑器”首选项来显示所有空格字符,然后突出显示的是粗体字符。您还可以使用此工具验证XML文件:

xmllint --noout --schema http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd XML_FILE

0
投票

您正在评论结束标记,再次查看<filter-mapping>

 <filter-mapping>
 <filter-name>springSecurityFilterChain</filter-name>
 <url-pattern>      /*</url-pattern> 
 </filter-mapping>`

0
投票

此错误只表示您的xml中有拼写错误。尝试格式化您的XML,您的拼写错误将脱颖而出。

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