小编Abh*_*sal的帖子

我目前正在使用Spring框架中的博客.我正在实现Spring Security以进行登录.一切都按预期工作,直到我提交始终返回404代码的登录凭据.

这是我的web.xml cod e

<?xml version="1.0" encoding="UTF-8"?>
<web-app id="WebApp_ID" version="2.4"
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">

<display-name>avispring</display-name>          

<error-page>
    <error-code>404</error-code>
    <location>/404.html</location>
</error-page>

<context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>/WEB-INF/spring-database.xml</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>/admin/*</url-pattern>
</filter-mapping>

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

<servlet>
    <servlet-name>spring</servlet-name>
    <servlet-class>
        org.springframework.web.servlet.DispatcherServlet
    </servlet-class>
    <load-on-startup>1</load-on-startup>        
</servlet>
<servlet-mapping>
    <servlet-name>spring</servlet-name>
    <url-pattern>/</url-pattern>        
</servlet-mapping>

这是我的春季安全码:

<bean id="dataSource"
    class="org.springframework.jdbc.datasource.DriverManagerDataSource">
    <property name="driverClassName" value="com.mysql.jdbc.Driver"/>
    <property name="url" value="jdbc:mysql://localhost:3306/avispring"/>
    <property name="username" value="root"/>
    <property name="password" value=""/>
</bean>
<security:debug/>
<security:http auto-config="true">
    <security:intercept-url pattern="/admin/**" access="hasRole('ROLE_ADMIN')"/>
    <security:intercept-url pattern="/j_spring_security_check" access="permitAll"/>
    <security:form-login        
    login-page="/login.html"
    authentication-failure-url="/login?login_error=1"
    default-target-url="/admin/home.html"/>     
</security:http>

<security:authentication-manager>
    <security:authentication-provider> …

我一直在 Spring MVC 应用程序中工作并尝试使用 Twitter Typeahead 来显示建议。然而问题是，虽然建议是从服务器正确获取的，但建议框根本没有显示。相同的代码以前工作过，但现在不工作了。

我的 javascript 代码是：

var skillSuggestions=new Bloodhound({
        datumTokenizer:Bloodhound.tokenizers.obj.whitespace('value'),
        queryTokenizer:Bloodhound.tokenizers.whitespace,

        remote:{
            url:"/tags/get.html/?searchTerm=%QUERY",
            filter:function(x){
                return $.map(x,function(item){
                    return{value:item.name};
                });
            },
            wildcard:'%QUERY',

        }
    });

    skillSuggestions.initialize();

    $('#skill-name').typeahead({
        hint:true,
        highlight:true,
        minLength:1
    },{
        name:'value',
        displayKey:'value',
        source:skillSuggestions.ttAdapter()
    })

对于输入“j”，接收到的 json 响应是：

[{"id":"56d546f5535a3c819f080558","name":"Java","category":"Information Technology","subCategory":"Programming/Software","createdDate":1456817909648,"updatedDate":null,"createdBy":"bob","modifiedBy":null},{"id":"56d93f8e535a773c1f8cc846","name":"Javascript","category":"Information Technology","subCategory":"Programming / Software","createdDate":1457078158043,"updatedDate":null,"createdBy":"bob","modifiedBy":null},{"id":"56d93fa2535a773c1f8cc847","name":"JQuery","category":"Information Technology","subCategory":"Programming / Software","createdDate":1457078178030,"updatedDate":null,"createdBy":"bob","modifiedBy":null},{"id":"56d93fb7535a773c1f8cc848","name":"JavaSE","category":"Information Technology","subCategory":"Programming / Software","createdDate":1457078199012,"updatedDate":null,"createdBy":"bob","modifiedBy":null},{"id":"56e226a47b49d4215eaefa9a","name":"javahhhh","category":"hhhhh","subCategory":"jjjjjj","createdDate":1457661604324,"updatedDate":null,"createdBy":null,"modifiedBy":null}]

我想name在建议中显示该字段。我错过了什么吗？请帮忙。我看了这里，这里，这里等等，但不知道出了什么问题。