问题描述
我使用freemarker模板作为视图部分创建了spring mvc应用程序.在这个尝试使用表单添加模型的过程中,我也在使用Spring Security这是代码
I created a spring mvc application using freemarker template as view part. In this tried to add a model using forms.I am also using spring securityHere is the code
<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";
}
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>
Spring-security.xml
<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>
点击提交按钮时,它会返回错误`
When click submit button it returns error `
`我在ftl和controller上都给出了POST方法.那为什么会这样呢?
`I gave POST method on both ftl and controller. Then why would this happen?
推荐答案
我不确定是否有帮助,但是我遇到了同样的问题.
I am not sure if this helps but I had the same problem.
您正在使用具有CSRF保护的springSecurityFilterChain.这意味着通过POST请求发送表单时必须发送令牌.尝试将下一个输入添加到您的表单中:
You are using springSecurityFilterChain with CSRF protection. That means you have to send a token when you send a form via POST request. Try to add the next input to your form:
<input type="hidden"
name="${_csrf.parameterName}"
value="${_csrf.token}"/>
这篇关于HTTP状态405-具有Spring Security的Spring MVC不支持请求方法'POST'的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!