Hibernate Don';I don’我不知道如何迭代提供的;项目“;弗雷奇
嗨,我在显示所选滑雪板的详细信息时出错: JSP代码:Hibernate Don';I don’我不知道如何迭代提供的;项目“;弗雷奇,hibernate,jsp,spring-mvc,crud,jsp-tags,Hibernate,Jsp,Spring Mvc,Crud,Jsp Tags,嗨,我在显示所选滑雪板的详细信息时出错: JSP代码: <c:forEach var="skis" items="${skis}"> <div class="container"> <div class="card"> <div class="container-fliud"> <div class="wrapper row">
<c:forEach var="skis" items="${skis}">
<div class="container">
<div class="card">
<div class="container-fliud">
<div class="wrapper row">
<div class="preview col-md-6">
<div class="preview-pic tab-content">
<div class="tab-pane active" id="pic-1"><img src="#" /></div>
</div>
</div>
<div class="details col-md-6">
<h3 class="product-title">${skis.company} - ${skis.model}</h3>
<p class="product-description">${skis.description}</p>
<h4 class="price">price per day: <span>12$</span></h4>
<h4>Day's reserved: </h4>
</c:forEach>
Controller:
@RequestMapping(value = "/ski/show-details/{skisId}" , method = RequestMethod.GET)
public String getShowDetailsPage(@PathVariable("skisId") Integer skisId, Model model) {
Skis skis = skisDAO.findOne(skisId);
model.addAttribute("skis", skis);
return "ski-details";
}
Model:
@Entity
@Table(name = "skis")
public class Skis {
@Id
@Column(name= "skisId")
@GeneratedValue
private Integer skisId;
@Column(name = "company", length = 20)
private String company;
@Column(name = "model", length = 20)
private String model;
@Column(name = "description", length = 200)
private String description;
@Lob
@Basic(fetch = FetchType.LAZY)
private byte[] photo;
@Repository
public interface SkisDAO extends CrudRepository<Skis, Integer> {
但这是同样的问题
有人能帮忙吗
Stacktrace如下所示:
javax.servlet.jsp.JspTagException:不知道如何迭代
在中提供的“项目”
org.apache.taglibs.standard.tag.common.core.ForEachSupport.toForEachIterator(ForEachSupport.java:274)
org.apache.taglibs.standard.tag.common.core.ForEachSupport.supportedTypeForEachIterator(ForEachSupport.java:238)
org.apache.taglibs.standard.tag.common.core.ForEachSupport.prepare(ForEachSupport.java:155)
javax.servlet.jsp.jstl.core.LoopTagSupport.doStartTag(LoopTagSupport.java:256)
org.apache.jsp.WEB_002dINF.ski_002details_jsp._jspx_meth_c_005fforEach_005f0(ski_002details_jsp.java:483)
org.apache.jsp.WEB\u 002dINF.ski\u 002details\u jsp.\u jsp服务(ski\u 002details\u jsp.java:317)org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:70)
javaservlet.http.HttpServlet.service(HttpServlet.java:729)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:438)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:396)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:340)
答案在评论中。jsp中不需要循环实体。删除
foreach
循环,findOne
方法返回一个对象。您可以像这样直接访问值:${skis.company}
我不确定,但重复的变量名可能不明确。使用var=“ski”并重试。仍然是相同的错误remove foreach loop,findOne方法返回一个对象?当我删除循环错误消失时,它不会在页面上显示数据库中的实体Yes one object,detailas关于缩略图列表中选定的滑雪板
@Query("select s from Skis s where s.skisId = :skisId")
Skis findBySkisId(@Param("skisId") Integer skisId);