Spring表达式语言(SpEL)不适用于JPA/hibernate实体

Ste*_*sen 5 spring hibernate jpa spring-el spring-data-jpa

我试图使用SpEL模板从实体生成文件名.我有两个看起来类似的实体:

@Entity
public class Invoice implements Serializable {
    private String invoicenumber;
    private Customer customer;

    @Column(name = "invoicenumber", nullable = false, length = 20)
    public String getInvoicenumber() {
        return this.invoicenumber;
    }

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "fk_customer", nullable = false)
    public Customer getCustomer() {
        return this.customer;
    }
}

@Entity
public class Customer implements Serializable {
    private String firstname;
    private String lastname;

    @Column(name = "firstname", nullable = false, length = 20)
    public String getFirstname() {
        return this.firstname;
    }

    @Column(name = "lastname", nullable = false, length = 20)
    public String getLastname() {
        return this.lastname;
    }
}
Run Code Online (Sandbox Code Playgroud)

和这个类似的SpEL模板:

String template = "invoicenumber + '-' + customer.firstname + ' ' + customer.lastname";
Run Code Online (Sandbox Code Playgroud)

然后我使用SpEL从模板生成带有发票对象的文件名

public String generateFilename(String filenameTemplate, Object dataObject) {
    ExpressionParser parser = new SpelExpressionParser();
    Expression expression = parser.parseExpression(filenameTemplate);
    return expression.getValue(dataObject, String.class);
}
Run Code Online (Sandbox Code Playgroud)

此测试有效:

String testTemplate = "invoicenumber + '-' + customer.firstname + ' ' + customer.lastname";
Invoice invoice = new Invoice();
invoice.setInvoicenumber("BF2016-06-ABCDEF");
invoice.setCustomer(new Customer());
invoice.getCustomer().setFirstname("Hans");
invoice.getCustomer().setLastname("Hansen");
assertEquals("BF2016-06-ABCDEF-Hans Hansen", generator.generateFilename(testTemplate, invoice));
Run Code Online (Sandbox Code Playgroud)

此测试不会:

Invoice invoice = invoiceRepository.findOne(4);

String template = "invoicenumber + '-' + customer.firstname + ' ' + customer.lastname";
String filename = filenameGenerator.generateFilename(template, invoice);
assertEquals("12344-201601-Heinrich Jahnke", filename);
Run Code Online (Sandbox Code Playgroud)

这个测试实际上导致了"12344-201601-",这使我得出这样的假设,即用于延迟加载客户对象的hibernate代理是问题所在.firstname和lastname字段在从数据库加载之前为null,这将解释呈现的文件名.

有想法该怎么解决这个吗?我已经尝试过的一些事情:

Hibernate.initialize(invoice);
Hibernate.initialize(invoice.getCustomer());
System.out.println(invoice.getCustomer().getFirstname());
Run Code Online (Sandbox Code Playgroud)
  • 在表达式中使用"customer.getFirstname()"而不是"customer.firstname"
  • 将@Transactional添加到我的FilenameGenerator类

Ste*_*sen 1

问题出在其他地方,SpEL 和 JPA/Hibernate 一起工作得很好。对此感到抱歉!

我的真实表情是这样的:

"invoicenumber + '-' + (customer.company == null ? customer.fname + ' ' + customer.sname : customer.company)"
Run Code Online (Sandbox Code Playgroud)

遗憾的是,从数据库加载的客户确实有一家公司,一家空公司......使用以下表达式,一切正常:

"invoicenumber + '-' + (customer.company == null or customer.company.isEmpty() ? customer.fname + ' ' + customer.sname : customer.company)"
Run Code Online (Sandbox Code Playgroud)