JPA 2.1中带有枚举的@ConstructorResult [英] @ConstructorResult with Enum in JPA 2.1

问题描述

我不知道如何在使用@SqlResultSetMapping的@ConstructorResult时在@ColumnResult类型中使用Enum

I am not having any idea how to use Enum in @ColumnResult Type while using @ConstructorResult of @SqlResultSetMapping

@SqlResultSetMapping(name="DetailAndResult",
        classes={
                @ConstructorResult(targetClass=DetailAndResult.class, columns={
                        @ColumnResult(name="id", type= String.class),
                        @ColumnResult(name="runId", type=Integer.class),
                        @ColumnResult(name="subRunId", type=Integer.class),
                        @ColumnResult(name="transactionId", type=Integer.class),
                        @ColumnResult(name="referenceNumber", type=String.class),
                        @ColumnResult(name="customerName", type=String.class),
                        @ColumnResult(name="transactionType", type=TransactionType.class),
                        @ColumnResult(name="transactionResultStatus", type=String.class)

                })
        }
)

在上述配置中，名称"transactionType"的名称为TransactionType枚举.在这里使用Enum的正确方法是什么.

in above configuration, name 'transactionType' is of TransactionType Enum. What is the correct way to use Enum here.

如果以上是正确的方法，那么我将收到此异常(如果我将删除Enum字段，则没有异常)，因此认为应该有另一种使用此方法的方法.

if above is the correct way then I am getting this exception (If I will remove the Enum field then there is no exception) so thinking that there should be another way to use this.

Caused by: javax.persistence.PersistenceException: org.hibernate.type.SerializationException: could not deserialize
    at org.hibernate.jpa.spi.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1763) ~[hibernate-entitymanager-4.3.6.Final.jar:4.3.6.Final]
    at org.hibernate.jpa.spi.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1677) ~[hibernate-entitymanager-4.3.6.Final.jar:4.3.6.Final]
    at org.hibernate.jpa.internal.QueryImpl.getResultList(QueryImpl.java:458) ~[hibernate-entitymanager-4.3.6.Final.jar:4.3.6.Final]
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method) ~[na:1.7.0_51]
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:57) ~[na:1.7.0_51]
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43) ~[na:1.7.0_51]
    at java.lang.reflect.Method.invoke(Method.java:606) ~[na:1.7.0_51]
    at org.springframework.orm.jpa.SharedEntityManagerCreator$DeferredQueryInvocationHandler.invoke(SharedEntityManagerCreator.java:333) ~[spring-orm-4.0.5.RELEASE.jar:4.0.5.RELEASE]
    at com.sun.proxy.$Proxy146.getResultList(Unknown Source) ~[na:na]

使用hibernateTemplate，我们使用的是sqlquery.addscalar，并且有一种方法可以使用org.hibernate.type.Type和

With hibernateTemplate, we were using sqlquery.addscalar and there was a way to use Enum there using org.hibernate.type.Type and

TypeLocatorImpl(new TypeResolver()).custom(EnumType.class, params)

请建议是否将类似的东西用于@SqlResultSetMapping和@ConstructorResult

Please suggest if something like this will be used for @SqlResultSetMapping and @ConstructorResult

推荐答案

我也遇到了这个问题，经过相当广泛的搜索后没有发现任何东西.我已经深入研究了源代码，并且据我所知，根本没有任何类型的枚举处理(尽管我当然会遗漏一些东西).

I've also run into this problem and haven't found anything after a fairly extensive search. I've gone as far as looking at the source code, and as far as I could tell there wasn't any kind of handling for enums at all (although of course I could be missing something).

我最终要做的是创建一个替代构造函数，该构造函数接受枚举类型的String，然后将其传递给枚举的valueOf()方法.

What I ended up doing was creating an alternative constructor that took in a String for the enum type, and then passed it into the enum's valueOf() method.

例如

更改您的@SqlResultSetMapping以执行以下操作:

change your @SqlResultSetMapping to do this:

@ColumnResult(name="transactionType", type=String.class),

然后在您的类的构造函数中:

And then in the constructor of your class:

public DetailAndResult(..., String transactionType, ...) {
    ...
    this.transactionType = TransactionType.valueOf(transactionType);
    ...
}

我们必须这样做很烦人，但是只要您的枚举以字符串形式存储在db中(即，您实体上的列用@Enumerated(EnumType.STRING)注释)，它就可以工作.

It's annoying that we have to do this, but so long as your enum is stored as a String in the db (i.e. the column on your entity is annotated with @Enumerated(EnumType.STRING)), it works.

这篇关于JPA 2.1中带有枚举的@ConstructorResult的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！