Question

我从链接 https://stackoverflow.com/q/15784984/814074中读到了一个问题，并尝试了上面链接中给出的解决方案。但是，运行代码时出现以下错误：

Error creating bean with name 'JobArgs' defined in class path resource [pipelineJob.xml]: 
Initialization of bean failed; nested exception is org.springframework.beans.ConversionNotSupportedException: Failed to convert property value of type '$Proxy2 implementing java.io.Serializable,org.springframework.aop.scope.ScopedObject,org.springframework.aop.framework.AopInfrastructureBean,org.springframework.aop.SpringProxy,org.springframework.aop.framework.Advised' to required type 'com.test.genepanel.job.JobArguments' for property 'jobArguements'; nested exception is java.lang.IllegalStateException:
Cannot convert value of type [$Proxy2 implementing java.io.Serializable,org.springframework.aop.scope.ScopedObject,org.springframework.aop.framework.AopInfrastructureBean,org.springframework.aop.SpringProxy,org.springframework.aop.framework.Advised] to required type [com.test.genepanel.job.JobArguments] for property 'jobArguements': no matching editors or conversion strategy found

xml包含

   <?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:batch="http://www.springframework.org/schema/batch"
     xmlns:aop="http://www.springframework.org/schema/aop"
    xmlns:tx="http://www.springframework.org/schema/tx" xmlns:p="http://www.springframework.org/schema/p"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns:util="http://www.springframework.org/schema/util"
    xsi:schemaLocation="http://www.springframework.org/schema/beans
            http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
            http://www.springframework.org/schema/batch
            http://www.springframework.org/schema/batch/spring-batch-2.1.xsd
            http://www.springframework.org/schema/aop
            http://www.springframework.org/schema/aop/spring-aop-3.0.xsd
            http://www.springframework.org/schema/util
            http://www.springframework.org/schema/util/spring-util-2.5.xsd">

    <batch:job id="pipelineJob">
        <batch:step id="initializationStep" next="CleanUPStep">
            <batch:tasklet ref="initializationStepTasklet" />
        </batch:step>
        <batch:step id="CleanUPStep">
            <batch:tasklet ref="cleanupTaskLet" />
        </batch:step>
    </batch:job>



    <bean id="basicStep" class="com.test.mutation.steps.BasicStep" abstract="true">
        <property name="testJobArgs" ref="JobArgs"/>
    </bean>

    <bean id="JobArgs" class="com.test.mutation.application.TestJobArguements">
        <property name="jobArguements" ref="jobArg"> 
        </property>
    </bean>

    <bean id="jobArg" class="com.test.genepanel.job.JobArguments" scope="step">
        <constructor-arg value="#{jobParameters['jobOutputDir']}"/>
    </bean>

    <bean id="emptyTaskLet" class="com.test.mutation.steps.EmptyStep" scope="step" parent="basicStep" />

    <bean id="cleanupTaskLet" class="com.test.mutation.steps.CleanUpStep" scope="step" parent="basicStep">
    </bean>

    <bean id="initializationStepTasklet" class="com.test.mutation.steps.InitializationStep"  scope ="step" parent="basicStep">
    </bean> 
</beans>

我错过了什么吗？

Answer 1

使用步骤范围的简单方法如下：

<bean id="myReader" class="org.springframework.batch.item.file.FlatFileItemReader" scope="step">
    <property name="resource" value="file:#{jobParameters['input.file']}" />
    <property name="linesToSkip" value="0" />
    <property name="recordSeparatorPolicy" ref="simpleRecordPolicy" />
    <property name="lineMapper" ref="journalSicIemtLineMapper" />
</bean>

将步骤范围放在bean上会延迟他的创建，直到他所提到的步骤即将开始。这就是后期绑定的含义，因此您可以访问ExecutionContext中的变量。

正如StepScope中的文档所述：

使用StepScope的bean将是aop：scoped-proxy。这意味着代理绕过真实对象。

因此，当您定义常规Spring Bean（就像您使用jobArg一样）并将scope = step放在其上时。当你想在另一个bean中设置它时，你将不得不找到一种方法来检索这个代理中的对象（JobArgs）

Answer 2

我查看了所有Spring引用，并且我没有在构造函数参数中看到后期绑定，所以我不确定它是否有效。我假设您在JobArgs中设置了jobParameters，在这种情况下它应该具有相同的scope="step"

在Spring Batch中将jobParameters传递给bean

2 个答案: