这次来分析下切面的执行过程。
1.怎么看?
怎么开始看源码呢?就直接从被增强的方法调用那里打断点,看看怎么执行的:
然后就来到了这:
2.初步分析
里面有段:
if (this.advised.exposeProxy) {
// Make invocation available if necessary.
oldProxy = AopContext.setCurrentProxy(proxy);
setProxyContext = true;
}
就是上篇文章讲到的注解配置暴露代理对象,放到AopContext的ThreadLocal里去,之后就可以随时用 AopContext.currentProxy())取到代理对象。
接下来有段重要的:
List<Object> chain = this.advised.getInterceptorsAndDynamicInterceptionAdvice(method, targetClass);
获取拦截器链,就是把这次相关的增强器转化成拦截器获取出来
然后:
if (chain.isEmpty()) {
// We can skip creating a MethodInvocation: just invoke the target directly
// Note that the final invoker must be an InvokerInterceptor so we know it does
// nothing but a reflective operation on the target, and no hot swapping or fancy proxying.
Object[] argsToUse = AopProxyUtils.adaptArgumentsIfNecessary(method, args);
retVal = AopUtils.invokeJoinpointUsingReflection(target, method, argsToUse);
}
else {
// We need to create a method invocation...
invocation = new ReflectiveMethodInvocation(proxy, target, method, args, targetClass, chain);
// Proceed to the joinpoint through the interceptor chain.
retVal = invocation.proceed();
}
这里就是判断拦截器链有没有东西,如果是空的就直接通过反射调用,不是空就进行else逻辑了,那else是重点了,即invocation.proceed();
3.invocation.proceed()
public Object proceed() throws Throwable {
// We start with an index of -1 and increment early.
if (this.currentInterceptorIndex == this.interceptorsAndDynamicMethodMatchers.size() - 1) {
return invokeJoinpoint();
}
Object interceptorOrInterceptionAdvice =
this.interceptorsAndDynamicMethodMatchers.get(++this.currentInterceptorIndex);
if (interceptorOrInterceptionAdvice instanceof InterceptorAndDynamicMethodMatcher) {
// Evaluate dynamic method matcher here: static part will already have
// been evaluated and found to match.
InterceptorAndDynamicMethodMatcher dm =
(InterceptorAndDynamicMethodMatcher) interceptorOrInterceptionAdvice;
if (dm.methodMatcher.matches(this.method, this.targetClass, this.arguments)) {
return dm.interceptor.invoke(this);
}
else {
// Dynamic matching failed.
// Skip this interceptor and invoke the next in the chain.
return proceed();
}
}
else {
// It's an interceptor, so we just invoke it: The pointcut will have
// been evaluated statically before this object was constructed.
return ((MethodInterceptor) interceptorOrInterceptionAdvice).invoke(this);
}
}
第一个if是递归的终止条件,明显是根据下标进行终止的条件
后面进行前++,又调用了
((MethodInterceptor) interceptorOrInterceptionAdvice).invoke(this);
然而这里面是:
public Object invoke(MethodInvocation mi) throws Throwable {
MethodInvocation oldInvocation = invocation.get();
invocation.set(mi);
try {
return mi.proceed();
}
finally {
invocation.set(oldInvocation);
}
}
又跑回proceed方法去了,递归。
第二次来到的时候下标就是0了(第一次是-1,默认的),前++为1的下标的话,取出来的东西继续调用invoke发现进的是AspectJAfterThrowingAdvice的invoke了(第一次是ExposeInvocationInterceptor的invoke,记录下MethodInvocation供后面执行链获取)
这个AspectJAfterThrowingAdvice的invoke的源码如下(不一样):
public Object invoke(MethodInvocation mi) throws Throwable {
try {
return mi.proceed();
}
catch (Throwable ex) {
if (shouldInvokeOnThrowing(ex)) {
invokeAdviceMethod(getJoinPointMatch(), null, ex);
}
throw ex;
}
}
继续递归,不过这次把调用链try起来了,出异常就走异常增强通知invokeAdviceMethod
继续debug,又是递归到invoke,但这次是AfterReturningAdviceInterceptor的:
public Object invoke(MethodInvocation mi) throws Throwable {
Object retVal = mi.proceed();
this.advice.afterReturning(retVal, mi.getMethod(), mi.getArguments(), mi.getThis());
return retVal;
}
提一下啊,递归是栈结构!所以我们先看到了异常调用代码和返回通知代码!
继续递归proceed到了后置通知AspectJAfterAdvice类的invoke:
public Object invoke(MethodInvocation mi) throws Throwable {
try {
return mi.proceed();
}
finally {
invokeAdviceMethod(getJoinPointMatch(), null, null);
}
}
看到没,后置通知类的invokeAdviceMethod调用是用的finally,所以后置通知始终执行!
继续递归
跳到了前置通知类MethodBeforeAdviceInterceptor的invoke
public Object invoke(MethodInvocation mi) throws Throwable {
this.advice.before(mi.getMethod(), mi.getArguments(), mi.getThis());
return mi.proceed();
}
注意这里不一样了!它是先调用自己再调执行链,这也就是为什么前置通知早于方法执行
before方法执行完之后,进proceed了,递归即将结束:
所有的增强器取出来了,并执行了before 这里递归就结束了,调用目标方法:invokeJoinpoint
然后是后置通知各种,有异常就走之前try finally那里。
至此aop具体逻辑结束!
总结下易翻车点:
1.注意看自己的源码是哪个类,不然很懵逼,因为调了很多个类的同名方法invoke。
2.注意看是递归,和递归结束条件
3.注意invoke的实现,对于不同的增强器的逻辑是不一样的
4.增强器那个集合是有顺序好的
原文地址:https://www.cnblogs.com/chz-blogs/p/12006222.html