shiro自定义异常无法被捕获总是抛出AuthenticationException解决方案
这个问题我也是出的莫名其妙,刚开始好好的,然后配置多realm之后出的。
现在直入主题
在继承了 org.apache.shiro.authc.pam.ModularRealmAuthenticator的类中重写doMultiRealmAuthentication方法
以下是重写的代码,判断是否存在异常。如果存在异常,则抛出。
public class MyModularRealmAuthenticator extends ModularRealmAuthenticator {
private static final Logger log = LoggerFactory.getLogger(ModularRealmAuthenticator.class);
@Override
protected AuthenticationInfo doMultiRealmAuthentication(Collection<Realm> realms, AuthenticationToken token) throws AuthenticationException {
AuthenticationStrategy strategy = getAuthenticationStrategy();
AuthenticationInfo aggregate = strategy.beforeAllAttempts(realms, token);
if (log.isTraceEnabled()) {
log.trace("Iterating through {} realms for PAM authentication", realms.size());
}
AuthenticationException authenticationException = null;
for (Realm realm : realms) {
aggregate = strategy.beforeAttempt(realm, token, aggregate);
if (realm.supports(token)) {
log.trace("Attempting to authenticate token [{}] using realm [{}]", token, realm);
AuthenticationInfo info = null;
try {
info = realm.getAuthenticationInfo(token);
} catch (AuthenticationException e) {
authenticationException = e;
if (log.isDebugEnabled()) {
String msg = "Realm [" + realm + "] threw an exception during a multi-realm authentication attempt:";
log.debug(msg, e);
}
}
aggregate = strategy.afterAttempt(realm, token, info, aggregate, authenticationException);
} else {
log.debug("Realm [{}] does not support token {}. Skipping realm.", realm, token);
}
}
if(authenticationException != null){
throw authenticationException;
}
aggregate = strategy.afterAllAttempts(token, aggregate);
return aggregate;
}
}
关键点在于
if(authenticationException != null){ throw authenticationException; }
有可能出现的问题。。。
可能会出现第二个realm会出现第一个realm的抛出的异常。楼主直接在subject.login(token);try处理了。。
如果想在第二个realm里throw new ,别抛出RuntimeException这个就行。。。
try { subject.login(token); }catch (LoginPhoneException e){ Map map = new HashMap(); map.put("code", CodeAndMsgEnum.INFO.getCode()); return map; }catch (RuntimeException e){ System.out.println("出现了这个异常。。。但是不管他,因为我也不知道怎么处理"); }
瞎搞一通,总之解决了问题,但是个人觉得有点不理想,算了,希望可以找到更好的解决方法。
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