java使用Spring、Hibernate和JSP创建下拉列表
应用程序:Hibernate、Spring 3.0 MVC、JSP(使用Spring表单)
要求:使用Hibernate从数据库中选择一个表数据,并使用Spring MVC在JSP页面中将其显示为下拉列表
代码: Hibernate/Dao代码是
烹饪课
@Entity
@Table(name = "cuisine")
public class Cuisine {
@Id
@Column(name = "id")
private int id;
@Column(name = "name")
private String name;
.. getters and setters
RecipeDaoImpl类
public List<Cuisine> getCuisine() {
String hql = "SELECT id, name FROM Cuisine";
return getSession().createQuery(hql).list();
}
春季MVC
@Controller
public class RecipeController {
...
@RequestMapping("/add")
public String newRecipe(Map<String, Object> map) {
/* Get cuisine list and new object for cuisine */
List<Cuisine> cuisines = recipeServices.getCuisine();
System.out.println(cuisines);
map.put("cuisineList", cuisines);
map.put("cuisine", new Cuisine());
return "recipes/new";
}
JSP页面:
<%@ taglib prefix="sf" uri="http://www.springframework.org/tags/form"%>
<tr>
<th><sf:label path="cuisine">Cuisine</sf:label></th>
</tr>
<tr>
<td><sf:select path="${cuisineList}">
<sf:options items="${cuisine}"></sf:options>
</sf:select></td>
</tr>
在这样做时,我得到了以下错误:
org.springframework.beans.NotReadablePropertyException: Invalid property '[Cuisine [id=1, name=Continental][id=2, name=Italian]' of bean class [com.recipe.tables.Recipe]: Bean property '[Cuisine [id=1, name=Continental][id=2, name=Italian]' is not readable or has an invalid getter method: Does the return type of the getter match the parameter type of the setter?
org.springframework.beans.BeanWrapperImpl.getPropertyValue(BeanWrapperImpl.java:729)
org.springframework.beans.BeanWrapperImpl.getPropertyValue(BeanWrapperImpl.java:721)
org.springframework.validation.AbstractPropertyBindingResult.getActualFieldValue(AbstractPropertyBindingResult.java:99)
org.springframework.validation.AbstractBindingResult.getFieldValue(AbstractBindingResult.java:219)
org.springframework.web.servlet.support.BindStatus.<init>(BindStatus.java:120)
有人能建议如何解决这个问题吗?我查阅了几篇文章,并试图复制它们,但运气不佳
共 (0) 个答案