Question

嗨，我完全来自移动背景，所以Spring对我来说是新手，我目前有一个名为as的表单这里是BusinessForm的内容

public class BusinessForm 
{
 private String selectedBusinessId; //getters and setters included
 private List businessNameList; //getters and setters included - list of Business class Objects
  private List businessIdList; //getters and setters included  - lisy of Business class Objects
//Business class defined below 
}

这是我的控制器

@Controller 
public class HomeController{
    @RequestMapping(value = "/showHome", method = RequestMethod.GET)
    @ExceptionHandler({CutsomException.class})
    public ModelAndView showHome()
    {
         //Init BusinessForm class, its defined above...
         //set values of businessNameList...
         //set values of businessIdList...
         BusinessForm businessForm = new BusinessForm();
         businessForm.setBusinessNameList(....);
         businessForm.setBusinessIdList(....);
         return ModelAndView("MyView","businessForm", businessForm)
    }
}

以下是我的观点（我将仅显示表格，以避免显示其他所有内容）

MyView.jsp

<form:form action="blah" method="post" modelAttribute="businessForm">
<form:select path="selectedBusinessId">
<form:option value="">Select ID</form:option>
<form:options items="${businessForm.businessIdList}" item/>
</form:select>              
</form:form>

所以现在我看到的businessIdList高于form的代码：options items属性是＆＃34; Business＆＃34;对象和Business对象具有私有变量businessName和businessId with getter and setters

public class Business
{
  private String businessId; //with getters and setter 
  private String businessName; //with getters and setters 
}

所以在上面的表格中，一旦我打开下拉列表，它实际上显示了一个列表，但该String只是Business类的toString（）函数。所以我的下拉项目看起来像com.xxx.Business@c291000。 忽略业务类的toString（），如何让表单在表单列表的下拉列表中显示实际的businessId。原因是我想获得另一种形式：选择并显示另一个businessName列表。请帮忙。感谢。

Answer 1

您需要更改

<form:select path="selectedBusinessId">
    <form:option value="">Select ID</form:option>
    <form:options items="${businessForm.businessIdList}" item/>
</form:select>

到

<form:select path="selectedBusinessId">
    <form:option value="">Select ID</form:option>
    <form:options items="${businessForm.businessIdList}" itemValue="businessId" itemLabel="businessName" />
</form:select>

SpringMVC - 以表单形式显示对象：选项

1 个答案: