使用Spring Expression Language以编程方式评估bean表达式

Question

我有一个简单的Spring Bean Expression,当我在应用程序上下文文件中定义它时,它会很好地评估:

<bean id="myConfigBean" class="com.example.myBeanConfigBean">
    <property name="myProperty" value="#{ someOtherBean.getData() }"/>
</bean>

现在,我想以编程方式进行相同的评估.我使用了以下代码:

final ExpressionParser parser = new SpelExpressionParser();
final TemplateParserContext templateContext = new TemplateParserContext();
Expression expression = parser.parseExpression("#{ someOtherBean.getData() }", templateContext);
final String value = (String) expression.getValue();

这引发了一个异常:

EL1007E:(pos 22): Field or property 'someOtherBean' cannot be found on null

我想我必须以某种方式设置一个根对象,允许配置的bean像属性一样.但我还没有让它工作.任何人,已经做过这个并且可以给出一个提示吗？

Answer 1

实现BeanFactoryAware以获取对bean工厂的引用; 然后...

StandardEvaluationContext context = new StandardEvaluationContext();
context.setBeanResolver(new BeanFactoryResolver(this.beanFactory));
Expression expression = parser.parseExpression("@someOtherBean.getData()"); 
// or "@someOtherBean.data"
final String value = expression.getValue(context, String.class);

编辑

回答下面的评论.该@触发器使用bean工厂解析器来访问一个bean; 另一种方法是BeanExpressionContextAccessor在评估上下文中添加一个并使用a BeanExpressionContext作为评估的根对象...

final ExpressionParser parser = new SpelExpressionParser();
StandardEvaluationContext context = new StandardEvaluationContext();
context.setBeanResolver(new BeanFactoryResolver(beanFactory));
context.addPropertyAccessor(new BeanExpressionContextAccessor());
Expression expression = parser.parseExpression("someOtherBean.getData()");
BeanExpressionContext rootObject = new BeanExpressionContext(beanFactory, null);

...

String value = expression.getValue(context, rootObject, String.class);