低效率的EhCache性能

Question

使用这些JPA属性

props.put( "hibernate.cache.use_query_cache", "true" );
props.put( "hibernate.cache.use_second_level_cache", "true" );
props.put("hibernate.temp.use_jdbc_metadata_defaults", "false");
props.put( "hibernate.cache.region.factory_class", "org.hibernate.cache.ehcache.SingletonEhCacheRegionFactory" );
props.put( "javax.persistence.sharedCache.mode", SharedCacheMode.ALL );

Ehcache对于同一查询效率不高，

问题与QueryCache类的namedParameters.hashCode（）函数有关，它为同一查询生成了不同的HashCode！

private int generateHashCode() {
        int result = 13;
        result = 37 * result + ( firstRow==null ? 0 : firstRow.hashCode() );
        result = 37 * result + ( maxRows==null ? 0 : maxRows.hashCode() );
        for ( int i=0; i< positionalParameterValues.length; i++ ) {
            result = 37 * result + ( positionalParameterValues[i]==null ? 0 : positionalParameterTypes[i].getHashCode( positionalParameterValues[i] ) );
        }
        result = 37 * result + ( namedParameters==null ? 0 : namedParameters.hashCode() );
        result = 37 * result + ( filterKeys ==null ? 0 : filterKeys.hashCode() );
        result = 37 * result + ( customTransformer==null ? 0 : customTransformer.hashCode() );
        result = 37 * result + ( tenantIdentifier==null ? 0 : tenantIdentifier.hashCode() );
        result = 37 * result + sqlQueryString.hashCode();
        return result;
}

与班级有关

org.hibernate.type.AbstractType

public int getHashCode(Object x) {
    return x.hashCode();
}

它将为同一Array对象[01，1]生成一个不同的（新）hachCode！

对于数组，此hashCode方法应该是递归的

Answer 1

递归版本完全正常

类org.hibernate.type.AbstractType

public int getHashCode(Object x) {      
        if (x instanceof Object[]){
            int result = 1;
            for (Object element : (Object[]) x)
                result = 31 * result + (element == null ? 0 : getHashCode(element));
            return result;
        }
        return x.hashCode();
    }

和

public static boolean arraysEquals(Object[] a, Object[] a2) {
            if (a==a2)
                return true;
            if (a==null || a2==null)
                return false;

            int length = a.length;
            if (a2.length != length)
                return false;

            for (int i=0; i<length; i++) {
                Object o1 = a[i];
                Object o2 = a2[i];
                if (o1==null){
                    if (o2!=null)                   
                        return false;
                }else{
                    if (o2==null)
                        return false;
                    if (o1 instanceof Object[]){
                        if (!(o2 instanceof Object[]))
                            return false;
                        else
                            if (!arraysEquals( (Object[]) o1, (Object[]) o2))
                                return false;
                    }else
                        if (!o1.equals(o2))
                            return false;
                }                           
            }
            return true;
    }
    public static boolean equals(final Object x, final Object y) {
        if (x!=null && x instanceof Object[] && y!=null && y instanceof Object[] )
            return arraysEquals((Object[])x, (Object[])y);
        return x == y || ( x != null && y != null && x.equals( y ) );
    }