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 对于相同的查询效率不高,
问题与 QueryCache 类的函数 namedParameters.hashCode() 有关,它为同一个查询生成不同的 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;
}
与类有关
org.hibernate.type.AbstractType
public int getHashCode(Object x) {
return x.hashCode();
}
它为同一个数组对象 [01, 1] 生成不同的(新)hachCode!
这个哈希代码方法对于数组应该是递归的
递归版本 完全工作
Class 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 ) );
}