我创建了一个 API，允许用户使用树构建查询。该树是从 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，根据这个示例，我可以创建一个 SearchOperationRequest，它要求所有 WHERE hidden = false AND X = 88
"searchOperation": {
"left": {
"column": "Hidden",
"condition": "EQUALS",
"value": false
},
"comparator": "AND",
"right": {
"left": {
"column": "X",
"condition": "EQUALS",
"value": 88
},
"comparator": "AND"
}
}
此请求使用通用规范构建器构建到规范中。
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();
}
}
我目前的工作非常适合右重树。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
但它不适用于构建更复杂的树，其中大括号中的条件应优先并首先执行。
WHERE (X = 6 OR Z = 9) AND (T=6 OR H=8)
这个更复杂的 SearchOperationRequest 的模型是:
"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
}
}
}
如何修改我的 GenericSpecificationsBuilder 以便能够处理更复杂的 SearchOperationRequest 树？
最佳答案
让我们使用您的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，我们的堆栈看起来像:
stack = {}
comparatorStack = { AND, leftOR, rightOR }
specStack = { X=6, Z=9, T=6, H=8 }
同样的状态进入最后的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 specStack。因此，在第二次迭代中，您将弹出第一次迭代的结果(rightOR)以及 Z=9，并应用 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);
}
此代码的问题在于您正在更改树中的节点。在第一个示例中，有一次我们的指针指向节点:
Z=9
/ \
null rightOR
这看起来不对。您可以使用队列 (Depth First Search) 而不是使用堆栈 (Breadth First Search) 分解树，并免费获得您想要的顺序。
这是否解决了将每个逻辑运算符(比较器)应用于正确操作数的问题？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在一起。
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
您帖子中的第一个类似 json 的表示具有不合逻辑的布局，因为逻辑运算符应该同时对左右操作数进行操作。相反，你有:
"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，其中每个逻辑运算符都存在左右操作数。
首先我们处理树广度优先，逐层处理。同时，我们将每个 比较器 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。
在第二个循环中，我们使用新的 Queue 来存储我们返回根的“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();
我没有测试代码，但您应该能够提取(和重构)工作流。
关于java - 从树构建 JPA 规范，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/52047509/