从树构建JPA规范

时间：2018-08-27 21:59:01

标签： java algorithm spring-data-jpa jpa-criteria

我创建了一个API，该API允许用户使用树来构建查询。该树是根据SearchOperationRequest类构建的。

@Data
@ApiModel(value = "SearchOperationRequest", description = "Condition for the query")
public class SearchOperationRequest {

    @ApiModelProperty(value = "Conditional statement for the where clause", allowableValues = "EQUALS, NOT_EQUALS, GREATER_THAN, LESS_THAN, LIKE, STARTS_WITH, ENDS_WITH, CONTAINS")
    private SearchConditionOperation condition;

    @ApiModelProperty(value = "Column name to be searched on")
    private String column;

    @ApiModelProperty(value = "Value of column")
    private Object value;

    @ApiModelProperty(value = "Value of OR / AND")
    private SearchComparator comparator;

    @ApiModelProperty(value = "Left node of comparator condition")
    private SearchOperationRequest left;

    @ApiModelProperty(value = "Right node of comparator condition")
    private SearchOperationRequest right;

    public boolean isTreeLeaf() {
        return left == null && right == null;
    }

    public boolean isComparator() {
        return comparator != null;
    }
}

因此，从这个示例中，我可以创建一个SearchOperationRequest，要求所有 Where隐藏=假AND X = 88 

"searchOperation": {
    "left": {
        "column": "Hidden",
        "condition": "EQUALS",
        "value": false
    },
    "comparator": "AND",
    "right": {
        "left": {
            "column": "X",
            "condition": "EQUALS",
            "value": 88
        },
        "comparator": "AND"
    }
}

此请求是使用通用规范构建器内置到规范中的。

public class GenericSpecificationsBuilder<U> {

    public Specification<U> buildSpecificationFromSearchOperationRequest(SearchOperationRequest root, Function<SpecificationSearchCriteria, Specification<U>> converter) {

        Stack<SearchOperationRequest> stack = new Stack<>();

        Stack<SearchOperationRequest> comparatorStack = new Stack<>();
        Deque<Specification<U>> specStack = new LinkedList<>();

        SearchOperationRequest pointer = root;

        while (pointer != null || !stack.empty()) {

            if (pointer.isTreeLeaf()) {
                specStack.push(converter.apply(new SpecificationSearchCriteria(pointer.getColumn(), pointer.getCondition(), pointer.getValue())));
            }

            if (pointer.isComparator()) {
                comparatorStack.push(pointer);
            }

            if (pointer.getRight() != null) {
                stack.push(pointer.getRight());
            }

            if (pointer.getLeft() != null) {
                pointer.setRight(pointer.getLeft());
                pointer.setLeft(null);
            } else if (!stack.empty()) {
                SearchOperationRequest temp = stack.pop();
                pointer.setRight(temp);
            }

            pointer = pointer.getRight();
        }


        while (specStack.size() <= comparatorStack.size()) {
            comparatorStack.pop();
        }

        while (!comparatorStack.empty()) {

            SearchOperationRequest searchOperationRequest = comparatorStack.pop();

            Specification<U> operand1 = specStack.pop();
            Specification<U> operand2 = specStack.pop();
            if (searchOperationRequest.getComparator().equals(SearchComparator.AND)) {
                specStack.push(Specification.where(operand1)
                        .and(operand2));
            } else if (searchOperationRequest.getComparator().equals(SearchComparator.OR)) {
                specStack.push(Specification.where(operand1)
                        .or(operand2));
            }
        }


        return specStack.pop();
    }
}

我当前的作品非常适合RIGHT重树。含义查询，例如：

WHERE X = 6 AND X = 9
WHERE Z = 5 OR T=9
WHERE Z = 5 OR T=9 OR H=6

但是，这不适用于构建更复杂的树的情况，在这些树中，花括号中的条件应优先考虑并首先执行。

WHERE (X = 6 OR Z = 9) AND (T=6 OR H=8)

更复杂的SearchOperationRequest的模型为：

"searchOperation": {
    "left": {
        "left": {
            "column": "X",
            "condition": "EQUALS",
            "value": 6
        },
        "comparator": "AND",
        "right": {
            "column": "Z",
            "condition": "EQUALS",
            "value": 9
        }
    },
    "comparator": "AND",
    "right": {
        "left": {
            "column": "T",
            "condition": "EQUALS",
            "value": 6
        },
        "comparator": "AND",
        "right": {
            "column": "H",
            "condition": "EQUALS",
            "value": 8
        }
    }
}

如何修改GenericSpecificationsBuilder以便能够处理更复杂的SearchOperationRequest树？

1 个答案:

答案 0 :(得分：6)

让我们使用示例树来遵循执行流程。

         AND
      /       \
  leftOR     rightOR
  /    \     /    \ 
X=6   Z=9  T=6   H=8

当我们退出第一个while循环时，我们的堆栈如下所示：

stack = {}
comparatorStack = { AND, leftOR, rightOR }
specStack = { X=6, Z=9, T=6, H=8 }

同一状态进入最后的while循环。

while (!comparatorStack.empty()) {

    SearchOperationRequest searchOperationRequest = comparatorStack.pop();

    Specification<U> operand1 = specStack.pop();
    Specification<U> operand2 = specStack.pop();
    if (searchOperationRequest.getComparator().equals(SearchComparator.AND)) {
        specStack.push(Specification.where(operand1)
                .and(operand2));
    } else if (searchOperationRequest.getComparator().equals(SearchComparator.OR)) {
        specStack.push(Specification.where(operand1)
                .or(operand2));
    }
}

这里的问题是您将结果推回到specStack上。因此，在第二次迭代中，您将弹出第一次迭代的结果（rightOR）和Z=9，并对其应用leftOr逻辑。

分解树

让我们退后一步，看看如何分解树，更具体地说：

if (pointer.getLeft() != null) {
    pointer.setRight(pointer.getLeft());
    pointer.setLeft(null);
} else if (!stack.empty()) {
    SearchOperationRequest temp = stack.pop();
    pointer.setRight(temp);
}

此代码的问题是您正在更改树中的节点。在第一个示例中，有时我们的指针指向该节点：

    Z=9      
  /     \ 
null   rightOR

那看起来不对。不用使用堆栈（Depth First Search）分解树，您可以使用队列（Breadth First Search）并免费获得所需的顺序。

BFS

是否解决了将每个逻辑运算符（comparator）应用于正确的操作数的问题？不完全是，为了能够解决下面的两种布局，我们可以分解不同工作流中的运算符和操作数，而不是将它们全部分解。

         AND                    |                    rootAND                 
      /       \                 |                  /         \    
  leftOR     rightOR            |              leftOR       rightOR  
  /    \     /    \             |              /    \       /    \   
X=6   Z=9  T=6   H=8            |            X=6    AND    Z=9   H=8   
                                |                 /    \      
                                |               T=6   Y=3

解决方案

您的帖子中的第一个类似于json的表示形式具有不合逻辑的布局，因为逻辑运算符应同时在左侧和右侧操作数上进行操作。相反，您有：

"right": {
    "left": {
        "column": "X",
        "condition": "EQUALS",
        "value": 88
    },
    "comparator": "AND"
}

让我们考虑一种对称表示的解决方案，在该解决方案中，每个逻辑运算符都同时具有左右操作数。

首先，我们逐级处理树 Breadth First 。同时，我们将每个comparator放在堆栈上，因此我们在第二个while循环中首先取出最后一个。

然后在第二个循环中，当我们回到根目录时，使用新的Queue存储“中间结果”。

Queue<SearchOperationRequest> queue = new LinkedList<>();
Deque<SearchOperationRequest> comparatorStack = new LinkedList<>();

if (root == null || !root.isComparator()) return;
queue.add(root);
while(!queue.isEmpty()){
    SearchOperationRequest node = queue.poll();
    comparatorStack.push(node);
    if(node.left != null && node.left.isComparator()) queue.add(node.left);
    if(node.right != null && node.right.isComparator()) queue.add(node.right);
}

Queue<Specification> specQueue = new LinkedList<>();
while(!comparatorStack.isEmpty()){
    SearchOperationRequest comparator = comparatorStack.pop();
    // reverse operand order so already computed values are polled correctly
    Specification operand2; 
    SearchOperationRequest pointer = comparator.getRight();
    if(pointer.isTreeLeaf()) {
        operand2 = converter.apply(new SpecificationSearchCriteria(pointer.getColumn(), pointer.getCondition(), pointer.getValue()));
    } else {
        operand2 = specQueue.poll();
    }
    Specification operand1; 
    pointer = comparator.getLeft();
    if(pointer.isTreeLeaf()) {
        operand1 = converter.apply(new SpecificationSearchCriteria(pointer.getColumn(), pointer.getCondition(), pointer.getValue()));
    } else {
        operand1 = specQueue.poll();
    }
    if (comparator.equals(SearchComparator.AND)) {
        specQueue.add(Specification.where(operand1).and(operand2));
    } else if (comparator.equals(SearchComparator.OR)) {
        specQueue.add(Specification.where(operand1).or(operand2));
    }
} 

return specQueue.poll();

我尚未测试代码，但是您应该能够提取（并重构）工作流程。

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