Question

使用hibernate我将从样本表ANIMALS中提取结果列表。 hibernate方法如下所示：

...
List<Animals> animalList= null;
        try {
            Query query = session.createQuery("SELECT * FROM animals ORDER BY COLOR , HEIGHT , AGRESSIVE; ");
            animalList= query.list();
        }
...

结果集的示例如下：

NAME    COLOR   HEIGHT  AGRESSIVE
---------------------------------
JIMMY   BLACK   SHORT   NO
RIPPER  BLACK   SHORT   YES
GOOFY   BLACK   TALL    NO
MURPHY  BLACK   TALL    YES
PAUL    WHITE   SHORT   NO
ROB     WHITE   SHORT   YES
BOBBY   WHITE   TALL    NO
JACK    WHITE   TALL    YES

我的实际结果列表至少有100,000条记录，因此解析列表并为每个记录执行list.add(..)并不是很明智; 我想要做的是将结果集拆分为8个较小的列表，其中包含COLOR，HEIGHT和AGGRESSIVE的每个组合。之后，列表将作为参数提供给方法，以便执行一系列操作。

我只能使用Java 1.6或更低版本。我的主要目标是逃避我采取的IndexOutOfBounds措施以及IF's的疯狂行为，如下所示：

    for (int i = 0; i < animalList.size(); i++) {
previousAnimal = animals.get(i-1);
currentAnimal = animals.get(i);
nextAnimal = animals.get(i+1);
    ...//extra code
    IF ( previousAnimal.getColor() != currentAnimal.getColor() || previousAnimal.getHeight() != currentAnimal.getColor() || previousAnimal.getAgressive() != currentAnimal.getAgressive() )
    ...//extra code
    IF ( currentAnimal.getColor() != nextAnimal.getColor() || currentAnimal.getHeight() != nextAnimal.getColor() || currentAnimal.getAgressive() != nextAnimal.getAgressive() )
    ...//extra code

Answer 1

也许您可以使用DENSE_RANK分析函数来识别具有相似特征的动物：

...
List<Animals> animalList= null;
        try {
            Query query = session.createQuery("SELECT a.*, DENSE_RANK() OVER (ORDER BY COLOR, HEIGHT, AGRESSIVE) GRP FROM animals a ORDER BY COLOR, HEIGHT, AGRESSIVE; ");
            animalList= query.list();
        }
...

结果集的示例如下：

NAME    COLOR   HEIGHT  AGRESSIVE GRP
-------------------------------------
JIMMY   BLACK   SHORT   NO          1
RIPPER  BLACK   SHORT   YES         2
GOOFY   BLACK   TALL    NO          3
MURPHY  BLACK   TALL    YES         4
PAUL    WHITE   SHORT   NO          5
ROB     WHITE   SHORT   YES         6
BOBBY   WHITE   TALL    NO          7
JACK    WHITE   TALL    YES         8

Color，Height和Aggressive的每个不同组合会为GRP列产生不同的值。然后你可以简化你的其他代码：

for (int i = 0; i < animalList.size(); i++) {
previousAnimal = animals.get(i-1);
currentAnimal = animals.get(i);
nextAnimal = animals.get(i+1);
    ...//extra code
    IF ( previousAnimal.getGrp() != currentAnimal.getGrp() )
    ...//extra code
    IF ( currentAnimal.getGrp() != nextAnimal.getGrp() )
    ...//extra code

按多个条件拆分列表

1 个答案: