XML 스키마 네임 스페이스에 대한 Spring NamespaceHandler를 찾을 수 없습니다. [http://www.springframework.org/schema/security]
봄 보안에서 첫 번째 응용 프로그램을 개발 중입니다. 내 applicationContext-security.xml 파일은 다음과 같습니다.
<?xml version="1.0" encoding="UTF-8"?>
<!--
- Namespace-based OpenID configuration
-->
<b:beans xmlns="http://www.springframework.org/schema/security"
xmlns:b="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/security http://www.springframework.org/schema/security/spring-security-3.0.xsd">
<http>
<intercept-url pattern="/**" access="ROLE_USER"/>
<intercept-url pattern="/index.xhtml*" filters="none"/>
<logout/>
<openid-login login-page="/index.xhtml" authentication-failure-url="/index.xhtml?login_error=true">
<attribute-exchange>
<openid-attribute name="email" type="http://schema.openid.net/contact/email" required="true" count="2"/>
<openid-attribute name="name" type="http://schema.openid.net/namePerson/friendly" />
</attribute-exchange>
</openid-login>
<remember-me token-repository-ref="tokenRepo"/>
</http>
<b:bean id="tokenRepo"
class="org.springframework.security.web.authentication.rememberme.InMemoryTokenRepositoryImpl" />
<authentication-manager alias="authenticationManager"/>
<user-service id="userService">
<user name="http://user.myopenid.com/" authorities="ROLE_SUPERVISOR,ROLE_USER" />
</user-service>
</b:beans>
Web.xml 파일은 다음과 같습니다.
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://java.sun.com/xml/ns/j2ee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd" version="2.4">
<display-name>Spring Security OpenID Demo Application</display-name>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/applicationContext-security.xml
</param-value>
</context-param>
<context-param>
<param-name>log4jConfigLocation</param-name>
<param-value>/WEB-INF/classes/log4j.properties</param-value>
</context-param>
<context-param>
<param-name>webAppRootKey</param-name>
<param-value>openid.root</param-value>
</context-param>
<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>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<listener>
<listener-class>org.springframework.web.util.Log4jConfigListener</listener-class>
</listener>
<welcome-file-list>
<welcome-file>faces/index.xhtml</welcome-file>
</welcome-file-list>
</web-app>
응용 프로그램을 정리하고 빌드하는 데 성공했지만 응용 프로그램 jetty 7을 배포하려고하면 다음과 같은 오류가 발생합니다.
SEVERE: Context initialization failed
org.springframework.beans.factory.parsing.BeanDefinitionParsingException: Configuration problem: Unable to locate Spring NamespaceHandler for XML schema namespace [http://www.springframework.org/schema/security]
Offending resource: ServletContext resource [/WEB-INF/applicationContext-security.xml]
at org.springframework.beans.factory.parsing.FailFastProblemReporter.error(FailFastProblemReporter.java:68)
at org.springframework.beans.factory.parsing.ReaderContext.error(ReaderContext.java:85)
at org.springframework.beans.factory.parsing.ReaderContext.error(ReaderContext.java:80)
Tried everything but can't solve this error. Any help would be appreciated.
EDIT I tried added 3.0.2 version of the Spring-Security and got this:
Context initialization failed
org.springframework.beans.factory.xml.XmlBeanDefinitionStoreException: Line 13 in XML document from ServletContext resource [/WEB-INF/applicationContext-security.xml] is invalid;
nested exception is org.xml.sax.SAXParseException; lineNumber: 13; columnNumber: 11; cvc-complex-type.2.4.c: The matching wildcard is strict, but no declaration can be found for element 'http'. at org.springframework.beans.factory.xml.XmlBeanDefinitionReader.doLoadBeanDefinitions(XmlBeanDefinitionReader.java:396)
at org.springframework.beans.factory.xml.XmlBeanDefinitionReader.loadBeanDefinitions(XmlBeanDefinitionReader.java:334
You need a spring-security-config.jar
on your classpath.
The exception means that the security:
xml namescape cannot be handled by spring "parsers". They are implementations of the NamespaceHandler
interface, so you need a handler that knows how to process <security:
tags. That's the SecurityNamespaceHandler
located in spring-security-config
I had the same problem. The only thing that solved it was merge the content of META-INF/spring.handler and META-INF/spring.schemas of each spring jar file into same file names under my META-INF project.
This two threads explain it better:
- maven assembly plugin and spring namespace handlers
- Thread: unable to locate Spring NamespaceHandler for XML schema namespace
In my case, this was caused by custom manifest entries added by the maven-jar-plugin.
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-jar-plugin</artifactId>
<version>2.6</version>
<configuration>
<archive>
<index>true</index>
<manifest>
<addClasspath>true</addClasspath>
</manifest>
<manifestEntries>
<git>${buildNumber}</git>
<build-time>${timestamp}</build-time>
</manifestEntries>
</archive>
</configuration>
</plugin>
Removing the following entries fixed the problem
<index>true</index>
<manifest>
<addClasspath>true</addClasspath>
</manifest>
ReferenceURL : https://stackoverflow.com/questions/7188719/unable-to-locate-spring-namespacehandler-for-xml-schema-namespace-http-www-sp
'programing' 카테고리의 다른 글
레이아웃을 깨지 않고 CSS Div 너비 비율 및 패딩 (0) | 2021.01.14 |
---|---|
Application Loader 용 업로드 파일을 만드는 방법은 무엇입니까? (0) | 2021.01.14 |
클릭시 버튼 텍스트 변경 (0) | 2021.01.14 |
RequireJs-정의 대 요구 (0) | 2021.01.14 |
버튼의 텍스트와 아이콘의 수직 정렬 (0) | 2021.01.14 |