employee.ftl

<fieldset>
    <legend>Add Employee</legend>
  <form name="employee" action="addEmployee" method="post">
    Firstname: <input type="text" name="name" /> <br/>
    Employee Code: <input type="text" name="employeeCode" />   <br/>
    <input type="submit" value="   Save   " />
  </form>

employeeController.java

@RequestMapping(value = "/addEmployee", method = RequestMethod.POST)
    public String addEmployee(@ModelAttribute("employee") Employee employee) {
        employeeService.add(employee);
        return "employee";
    }

的web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app 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">

<!-- Spring MVC -->
    <servlet>
        <servlet-name>appServlet</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>appServlet</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>

    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>

    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>
            /WEB-INF/spring/appServlet/servlet-context.xml,
            /WEB-INF/spring/springsecurity-servlet.xml
        </param-value>
    </context-param>

    <!-- Spring Security -->
    <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>


</web-app>

弹簧security.xml文件

<beans:beans xmlns="http://www.springframework.org/schema/security"
    xmlns:beans="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-3.2.xsd">

    <http security="none" pattern="/resources/**"/>
    <!-- enable use-expressions -->
    <http auto-config="true" use-expressions="true">
        <intercept-url pattern="/login" access="isAnonymous()"/>
        <intercept-url pattern="/**" access="hasRole('ROLE_ADMIN')" />

        <!-- access denied page -->
        <access-denied-handler error-page="/403" />
        <form-login 
            login-page="/login" 
            default-target-url="/"
            authentication-failure-url="/login?error" 
            username-parameter="username"
            password-parameter="password" />
        <logout logout-success-url="/login?logout" />
        <!-- enable csrf protection -->
        <csrf />
    </http>

    <authentication-manager>
        <authentication-provider user-service-ref="userDetailsService" >
            <password-encoder hash="bcrypt" />    
        </authentication-provider>
    </authentication-manager>

</beans:beans>

单击“提交”按钮时，将返回错误“

HTTP状态405 - 请求方法＆＃39; POST＆＃39;不支持

` 我在ftl和controller上都给了POST方法。那为什么会发生这种情况呢？

Answer 1

我不确定这是否有帮助，但我遇到了同样的问题。

您正在使用带有CSRF保护的springSecurityFilterChain。这意味着您必须在通过POST请求发送表单时发送令牌。尝试将下一个输入添加到表单中：

<input type="hidden"
name="${_csrf.parameterName}"
value="${_csrf.token}"/>

Answer 2

据我所知，上述解决方案并不适用于最新的SpringSecurity。您可以通过以下操作网址发送它，而不是通过隐藏传递：

<form method="post" action="doUpload?${_csrf.parameterName}=${_csrf.token}" enctype="multipart/form-data">

Answer 3

我找到了解决方案。这是因为春季安全跨站请求伪造（CSRF）保护。它阻止了网址。所以我在表单中添加了一个额外的字段。

<input type="hidden" name="${_csrf.parameterName}" value="${_csrf.token}"/>

现在它正常运作。

Answer 4

尝试替换：

action="addEmployee"

使用：

action="${pageContext.request.contextPath}/addEmployee"

除非您使用的是Spring 3.2

看到XML后编辑：

尝试将servlet-context.xml移动到WEB-INF目录并将其重命名为“appServlet-context.xml”。然后删除该行：

/WEB-INF/spring/appServlet/servlet-context.xml,

来自web.xml中的contextConfigLocation。

约定是上下文xml文件名为'[servlet-name] -context.xml'，其中[servlet-name]是DispatcherServlet的名称。

还尝试在表单操作中添加“/”，所以：

action="/addEmployee"

Answer 5

这对我有用：

public static boolean isFragmentShowing(Activity activity, String tag) throws Exception {
    FragmentManager fragmentManager = FragmentUtils.getFragmentManagerInstance(activity);

    if (tag.equals("")) {
        throw new NullPointerException("isFragmentShowing: fragment tag passed is empty");
    }

    Fragment fragment = fragmentManager.findFragmentByTag(tag);

    return fragment != null && fragment.isVisible();
}

另一种解决方案（但每种形式）

.and().csrf().disable();

HTTP状态405 - 请求方法＆＃39; POST＆＃39; Spring MVC不支持

employee.ftl

employeeController.java

的web.xml

弹簧security.xml文件

5 个答案: