Spring MVC POST提交不起作用

时间:2015-03-06 10:44:53

标签: java spring spring-mvc weblogic weblogic12c

我的应用服务器是Weblogic 12C,它正在apache代理后面工作。

当我通过spring MVC表单进行GET提交时,值会回到结尾,但是当我进行POST提交时,所有值在后端都是null。

这是我的档案。

JSP文件

    <form:form method="post" modelAttribute="userForm" action="createUser">

            <form:input path="userInfo.number" readonly="true" maxlength="100"/>
            <form:input path="userInfo.firstName" readonly="true" maxlength="500"/>
            <form:input path="userInfo.lastName" readonly="true" maxlength="500"/>
            <input type="submit" style="font-weight: bold;"                                             value="Send"></td>
</form:form>

控制器

@Controller
@RequestMapping(value = "/")
@SessionAttributes("user")
public class TestController {

@RequestMapping(method = RequestMethod.POST, value="createUser")
    public String createUser(@Valid UserForm userForm,BindingResult result,ModelMap modelMap) {
        LOGGER.debug("Handling POST description" + userForm.getFirstName());
}
}

的web.xml

<web-app>

    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/applicationContext.xml</param-value>
    </context-param>

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

    <listener> 
        <listener-class>org.springframework.web.context.request.RequestContextListener</listener-class>
    </listener>

    <filter>
        <filter-name>characterEncodingFilter</filter-name>
        <filter-class>org.springframework.web.filter.CharacterEncodingFilter</filter-class>
        <init-param>
            <param-name>encoding</param-name>
            <param-value>UTF-8</param-value>
        </init-param>
        <init-param>
            <param-name>forceEncoding</param-name>
            <param-value>true</param-value>
        </init-param>
    </filter>

    <filter-mapping>
        <filter-name>characterEncodingFilter</filter-name>
        <url-pattern>/*</url-pattern>
    </filter-mapping>

    <servlet>
        <servlet-name>userFormServlet</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>
                /WEB-INF/spring-beans/servlet-context.xml
            </param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>userFormServlet</servlet-name>
        <url-pattern>/apps/*</url-pattern>
    </servlet-mapping>

1 个答案:

答案 0 :(得分:1)

根据此行userForm.getFirstName()的外观,似乎属性 firstName lastName number 是属性 userForm 对象,在这种情况下,您的表单应为

<form:form method="post" modelAttribute="userForm" action="createUser">
            <form:input path="number" readonly="true" maxlength="100"/>
            <form:input path="firstName" readonly="true" maxlength="500"/>
            <form:input path="lastName" readonly="true" maxlength="500"/>
            <input type="submit" style="font-weight: bold;"                                             value="Send"></td>
</form:form>

可能会略有不同,具体取决于您的 UserForm 类的结构,但您应该从此处获得解决方案的要点