Question

这里有两个实体，一个是Customer，另一个是Order，一个客户可以获得多个Order，最后我使用springmvc在控制器中返回一个Customer对象，但它总是给我sqlgrammer错误，jackson-bind是否可以处理一对多的问题？这是控制器中的代码片段：

package src;
  import javax.xml.parsers.DocumentBuilderFactory;
    import javax.xml.parsers.DocumentBuilder;
    import org.w3c.dom.Document;
    import org.w3c.dom.NodeList;
    import org.w3c.dom.Node;
    import org.w3c.dom.Element;
    import java.io.File;

    public class ReadXMLFile {
      public static void main(String argv[]) {
     try {
    File fXmlFile = new File("/Users/mykong/Documents/staff.xml");
        DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
        DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
        Document doc = dBuilder.parse(fXmlFile);
    //optional, but recommended
        //read this - http://stackoverflow.com/questions/13786607/normalization-in-dom-parsing-with-java-how-does-it-work
        doc.getDocumentElement().normalize();
    System.out.println("Root element :" + doc.getDocumentElement().getNodeName());
    NodeList nList = doc.getElementsByTagName("staff");
    System.out.println("----------------------------");
    for (int temp = 0; temp < nList.getLength(); temp++) {
    Node nNode = nList.item(temp);
    System.out.println("\nCurrent Element :" + nNode.getNodeName());
    if (nNode.getNodeType() == Node.ELEMENT_NODE) {
    Element eElement = (Element) nNode;
        System.out.println("Staff id : " + eElement.getElementsByTagName("id").item(0).getTextContent());
                System.out.println("First Name : " + eElement.getElementsByTagName("UploadURL").item(0).getTextContent());
    //          System.out.println("Last Name : " + eElement.getElementsByTagName("lastname").item(0).getTextContent());
    //          System.out.println("Nick Name : " + eElement.getElementsByTagName("nickname").item(0).getTextContent());
    //          System.out.println("Salary : " + eElement.getElementsByTagName("salary").item(0).getTextContent());
        }
        }
        } catch (Exception e) {
        e.printStackTrace();
        }
      }
     }

和错误跟踪

org.springframework.web.util.NestedServletException：Request 处理失败;嵌套异常是 org.hibernate.exception.SQLGrammarException：无法执行声明

客户实体

@RequestMapping(value = "/getCustomer", method = RequestMethod.GET)
public @ResponseBody Customer getCustomer(){

    Customer customer = new Customer();
    customer.setRecord("Crabime");
    customer.setGender("male");
    Order order = new Order("vegitable", 12.1);
    Order order1 = new Order("fruit", 3.2);
    Set<Order> set = new HashSet<Order>();
    set.add(order);
    set.add(order1);
    customer.setOrders(set);

    customerService.save(customer);
    return customer;
}

这是订单：

     @Entity
    @Table(name = "customer")
    public class Customer {
        @Id
        @GeneratedValue(strategy = GenerationType.IDENTITY)
        private Integer id;
        private String record;
        private String gender;
        @OneToMany(fetch = FetchType.EAGER,cascade = {CascadeType.ALL}, mappedBy = "customer")
        private Set<Order> orders;
        public Customer(){}
...omit setter and getter...

**更新：**也许这是tomcat缓存问题，我记得当时重启了服务器而没有任何改变，但在关闭项目几个小时后重新打开它，好吧！

Answer 1

对于这个错误，建议的方法是添加spring.jpa.show-sql=true来打印sql语句，然后你可以检查sql。典型的它包含关键字，例如“命令”

org.springframework.web.util.NestedServletException：嵌套异常是org.hibernate.exception.SQLGrammarException：无法执行语句

1 个答案: