Question

从jsp我需要在用户点击提交表单的搜索按钮时传输下拉列表中选择的值。

    <form:form modelAttribute="firstVO" method="post"> 

    <form:select path="secondVO.status" 
    id="status" > 
    <form : option value="all" label="All" /> 
    <form : option value="new" label="New" /> 
    </form:select> 

    <input type="button" name="search" value="Search" onclick="filterData();" /> 

    </form:form> 

    function filterData(){ 

    document.forms[0].action    =   Path+"/filter.do"; 
    document.forms[0].submit(); 
    } 
    ------------------------------------------------- 

    public class FirstVO { 

    private SecondVO secondVO; 
    } 


    public class SecondVO { 

    private String status; 
    } 

    public class DPSController { 

    @RequestMapping(value = "/filter.do") 
    public ModelAndView filterDashboard(HttpServletRequest request, @ModelAttribute SecondVO secondVO) { 
    System.out.println("secondVO.getStatus:"+secondVO.getStatus()); 
    } 
    }

注意：DPSController是打印状态选定值的动作类。

Any idea please?

Answer 1

您的Model属性是firstVO而不是secondVO。如下所示更改filterDashboard签名应该有效。

更改

   @RequestMapping(value = "/filter.do") 
        public ModelAndView filterDashboard(HttpServletRequest request, @ModelAttribute SecondVO secondVO) { 
        System.out.println("secondVO.getStatus:"+secondVO.getStatus()); 
    }

到

@RequestMapping(value = "/filter.do") 
    public ModelAndView filterDashboard(HttpServletRequest request, @ModelAttribute FirstVO firstVO) { 
   System.out.println("firstVO.getSecondVO().getStatus:"+firstVO.getSecondVO().getStatus()); 
    }

DropDown选择要在spring中传输的值

1 个答案: