嗨，我完全来自移动背景，所以 Spring 对我来说是新的，我目前有一个名为 BusinessForm 的表单这里是内容
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)
}
}
这是我的观点（我将只显示表格以避免显示其他所有内容）
我的视图.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 是 form 的代码：options items 属性是“Business”对象的列表，Business 对象具有私有变量 businessName 和 businessId 以及 getter 和 setter
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，它实际上会显示一个列表，但那个 String 只不过是 Business 类的 toString() 函数。所以我的下拉项目看起来像 com.xxx.Business@c291000。如果不覆盖 Business 类的 toString()， 我如何让我的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：选择并显示另一个业务名称列表。请帮忙。谢谢。