低效的Ehcache性能

Question

使用放入系统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); 

的Ehcache是​​没有效率为相同的查询，

问题是起作用实现QueryCache类的namedParameters.hashCode（）相关的，它产生一个不同的HashCode相同的查询！

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; 
} 

这是关系到类

org.hibernate.type.AbstractType 

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

它产生一个不同的（新）hachCode对于相同Array对象[01，1]！

这hashCode方法应该是递归的阵列

Answer 1

递归版本全部工作

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)); 
    }