grails.gorm.transactions.Transactional不回滚

时间:2018-11-08 17:11:46

标签: spring hibernate grails integration-testing gorm

我在通过以下方式回滚服务层中的事务时遇到问题:

Grails 3.3.8

GORM 6.1.10。发布

我有以下服务方法:

val-*

我还创建了以下集成测试:

import grails.gorm.transactions.Transactional

@Transactional(rollbackFor = Exception.class)
class TestingService {

    void testServiceMethod(List<Factory> factories) {
        try {
            factories.each {
                if (it.name == 'second') {
                    throw new Exception('second')
                }
                it.testField = 'Edited'
                it.save()
                println(it.name + ' saved')
            }
        } catch (Exception e) {
            println('Exception Caught ' + e)
        }
    }
}

我还有以下控制器方法:

@Integration
@Rollback
class TestServiceIntSpec extends Specification {

    @Autowired
    TestingService testingService

    def setup() {
    }

    def cleanup() {
    }

    void "test something"() {
        when:
        Factory factoryOne = new Factory(name: "first").save(flush: true)
        Factory factoryTwo = new Factory(name: "second").save(flush: true)
        List<Factory> factories = [factoryOne, factoryTwo]
        testingService.testServiceMethod(factories)

        then:
        factoryOne.testField == null
        factoryTwo.testField == null
    }
}

我本来希望测试通过,因为我认为在抛出新异常后事务将回滚,但是it.testField仍然存在吗?另外,当我对TestController进行ping操作时,它会将factoryOne.testField输出为“已编辑”。我是否从文档中正确地理解了这一点?

“服务启用事务划分,这是定义要使哪些方法具有事务性的声明方式。要在服务上启用事务,请使用事务转换: 结果是所有方法都包装在事务中,并且如果方法抛出异常(Checked或Runtime异常)或错误,则会发生自动回滚。“

来源:https://docs.grails.org/latest/guide/services.html#declarativeTransactions

我也看不出自己在做什么与其他Stackoverflow答案不同:

https://stackoverflow.com/a/25739582/6887293

可以通过拉出以下Github项目并运行/factory/factory/src/integration-test/groovy/com/mycompany/myapp/TestServiceIntSpec.groovy或ping / factory / factory / grails-app / controllers来重新创建此问题/com/mycompany/myapp/TestController.groovy

https://github.com/georgy3k/IntegrationTestRollBack/tree/8addd2b95a8ffa4570e70eccb3b023b0ccfef5aa

预先感谢...

1 个答案:

答案 0 :(得分:0)

在您的catch块中,您需要重新引发异常。

catch (Exception e) {
        println('Exception Caught ' + e)
        throw e;
}

我所看到的问题是,异常永远不会逃脱方法。