Question

我正在处理一个项目，但我收到错误“前缀”上下文“for element”上下文：component-scan“未绑定”在xml文件的一边我的文件代码是

 <?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"xmlns:security="http://www.springframework.org/schema/security"
xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
    http://www.springframework.org/schema/security http://www.springframework.org/schema/security/spring-security-3.0.xsd">

 <!--Here is i am getting error "The prefix "context" for element "context:component-scan" is not bound."-->
    <context:component-scan base-package="tudu.security"/>

<global-method-security secured-annotations="enabled" />

<HTTP>
    <INTERCEPT-URL PATTERN="/AJAX/**" ACCESS="ROLE_USER" />
    <INTERCEPT-URL PATTERN="/TUDU/WELCOME" ACCESS="ROLE_ANONYMOUS,ROLE_USER" />
    <INTERCEPT-URL PATTERN="/TUDU/RECOVERPASSWORD" ACCESS="ROLE_ANONYMOUS,ROLE_USER" />
    <INTERCEPT-URL PATTERN="/TUDU/REGISTER" ACCESS="ROLE_ANONYMOUS,ROLE_USER" />
    <INTERCEPT-URL PATTERN="/TUDU/ADMIN" ACCESS="ROLE_ADMIN" />
    <INTERCEPT-URL PATTERN="/TUDU/**" ACCESS="ROLE_USER" />
    <INTERCEPT-URL PATTERN="/SERVLET/**" ACCESS="ROLE_USER" />

    <FORM-LOGIN LOGIN-PROCESSING-URL="/TUDU/LOGIN" LOGIN-PAGE="/TUDU/WELCOME"
        AUTHENTICATION-FAILURE-URL="/TUDU/WELCOME?AUTHENTICATION=FAILURE"
        DEFAULT-TARGET-URL="/TUDU/LISTS" ALWAYS-USE-DEFAULT-TARGET="TRUE" />

    <REMEMBER-ME KEY="TUDUROCKS" />
    <LOGOUT LOGOUT-URL="/TUDU/LOGOUT" />
</HTTP>

<AUTHENTICATION-MANAGER ALIAS="AUTHENTICATIONMANAGER">
    <AUTHENTICATION-PROVIDER USER-SERVICE-REF="USERDETAILSSERVICE" />
</AUTHENTICATION-MANAGER>

Answer 1

你的xml名称空间需要一些更改

您缺少上下文命名空间
您的默认命名空间是bean，看起来您更喜欢将安全性作为默认值
如果您不使用bean作为默认名称空间，则应将前缀beans：添加到根元素
我更喜欢用小写字母保存xml名称和属性

这应该有用。

<beans:beans xmlns="http://www.springframework.org/schema/security"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns:context="http://www.springframework.org/schema/context"
    xmlns:beans="http://www.springframework.org/schema/beans"
    xsi:schemaLocation="http://www.springframework.org/schema/security http://www.springframework.org/schema/security/spring-security-3.0.xsd
        http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
        http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context.xsd">

    <!--Here is i am getting error "The prefix "context" for element "context:component-scan" 
        is not bound." -->
    <context:component-scan base-package="tudu.security" />

    <global-method-security secured-annotations="enabled" />
    <http>
        <intercept-url pattern="/AJAX/**" access="ROLE_USER" />
        <intercept-url pattern="/TUDU/WELCOME" access="ROLE_ANONYMOUS,ROLE_USER" />
        <intercept-url pattern="/TUDU/RECOVERPASSWORD" access="ROLE_ANONYMOUS,ROLE_USER" />
        <intercept-url pattern="/TUDU/REGISTER" access="ROLE_ANONYMOUS,ROLE_USER" />
        <intercept-url pattern="/TUDU/ADMIN" access="ROLE_ADMIN" />
        <intercept-url pattern="/TUDU/**" access="ROLE_USER" />
        <intercept-url pattern="/SERVLET/**" access="ROLE_USER" />

        <form-login login-processing-url="/TUDU/LOGIN" login-page="/TUDU/WELCOME"
            authentication-failure-url="/TUDU/WELCOME?AUTHENTICATION=FAILURE"
            default-target-url="/TUDU/LISTS" always-use-default-target="true" />

        <remember-me key="TUDUROCKS" />
        <logout logout-url="/TUDU/LOGOUT" />
    </http>
    <authentication-manager alias="AUTHENTICATIONMANAGER">
        <authentication-provider user-service-ref="USERDETAILSSERVICE" />
    </authentication-manager>
</beans:beans>

我正在使用STS提供的弹簧配置编辑器，这让我的生活更轻松。

Answer 2

命名空间可能存在一些问题...... 你做了什么，直到现在是正确的.....但只是你需要改变一件事即

将安全性放在默认名称空间中......

只需复制此行：

<beans xmlns="http://www.springframework.org/schema/security"

并在 bean开启标记

中过去

或者您也可以为所有安全标记使用前缀，例如：

<security:http>

Answer 3

这是正常的..

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:context="http://www.springframework.org/schema/context"
xmlns:security="http://www.springframework.org/schema/security"
xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.2.xsd
    http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.2.xsd
    http://www.springframework.org/schema/security http://www.springframework.org/schema/security/spring-security-3.1.xsd">

<context:component-scan base-package="tudu.security" />

<security:global-method-security
    secured-annotations="enabled" />

<security:http>
    <security:intercept-url pattern="/AJAX/**"
        access="ROLE_USER" />
    <security:intercept-url pattern="/TUDU/WELCOME"
        access="ROLE_ANONYMOUS,ROLE_USER" />
    <security:intercept-url pattern="/TUDU/RECOVERPASSWORD"
        access="ROLE_ANONYMOUS,ROLE_USER" />
    <security:intercept-url pattern="/TUDU/REGISTER"
        access="ROLE_ANONYMOUS,ROLE_USER" />
    <security:intercept-url pattern="/TUDU/ADMIN"
        access="ROLE_ADMIN" />
    <security:intercept-url pattern="/TUDU/**"
        access="ROLE_USER" />
    <security:intercept-url pattern="/SERVLET/**"
        access="ROLE_USER" />

    <security:form-login login-processing-url="/TUDU/LOGIN"
        login-page="/TUDU/WELCOME" authentication-failure-url="/TUDU/WELCOME?AUTHENTICATION=FAILURE"
        default-target-url="/TUDU/LISTS" always-use-default-target="true" />

    <security:remember-me key="TUDUROCKS" />
    <security:logout logout-url="/TUDU/LOGOUT" />
</security:http>

<security:authentication-manager alias="AUTHENTICATIONMANAGER">
<security:authentication-provider user-service-ref="USERDETAILSSERVICE"/>
</security:authentication-manager>

</beans>

application-context-security.xml文件中的错误

3 个答案: