java - 将部门层次结构打印到表中

Question

我有一个部门关系表如下:

+---------------+----------------+
|     Dept.     | superior Dept. |
+---------------+----------------+
| "00-01"       | "00"           |
| "00-02"       | "00"           |
| "00-01-01"    | "00-01"        |
| "00-01-02"    | "00-01"        |
| "00-01-03"    | "00-01"        |
| "00-02-01"    | "00-02"        |
| "00-02-03"    | "00-02"        |
| "00-02-03-01" | "00-02-03"     |
+---------------+----------------+

现在我想按照这样的等级列出它们:

+-----------+--------------+--------------+--------------+
| Top Dept. | 2-tier Dept. | 3-tire Dept. | 4-tier Dept. |
+-----------+--------------+--------------+--------------+
|        00 |              |              |              |
|           | 00-01        |              |              |
|           |              | 00-01-01     |              |
|           |              | 00-01-02     |              |
|           | 00-02        |              |              |
|           |              | 00-02-01     |              |
|           |              | 00-02-03     |              |
|           |              |              | 00-02-03-01  |
+-----------+--------------+--------------+--------------+

我可以使用下面的代码构建关系树:
树节点.java

import java.util.LinkedList;
import java.util.List;

public class TreeNode {
  public String value;
  public List children = new LinkedList();

  public TreeNode(String rootValue) {
    value = rootValue;
  }

}

PairsToTree.java

import java.util.*;

public class PairsToTree {

  public static void main(String[] args) throws Exception {
    // Create the child to parent hash map
    Map <String, String> childParentMap = new HashMap<String, String>(8);
    childParentMap.put("00-01", "00");
    childParentMap.put("00-02", "00");
    childParentMap.put("00-01-01", "00-01");
    childParentMap.put("00-01-02", "00-01");
    childParentMap.put("00-01-03", "00-01");
    childParentMap.put("00-02-01", "00-02");
    childParentMap.put("00-02-03", "00-02");
    childParentMap.put("00-02-03-01", "00-02-03");

    // All children in the tree
    Collection<String> children = childParentMap.keySet();

    // All parents in the tree
    Collection<String> values = childParentMap.values();

    // Using extra space here as any changes made to values will
    // directly affect the map
    Collection<String> clonedValues = new HashSet();
    for (String value : values) {
      clonedValues.add(value);
    }

    // Find parent which is not a child to any node. It is the
    // root node
    clonedValues.removeAll(children);

    // Some error handling
    if (clonedValues.size() != 1) {
      throw new Exception("More than one root found or no roots found");
    }

    String rootValue = clonedValues.iterator().next();
    TreeNode root = new TreeNode(rootValue);

    HashMap<String, TreeNode> valueNodeMap = new HashMap();
    // Put the root node into value map as it will not be present
    // in the list of children
    valueNodeMap.put(root.value, root);

    // Populate all children into valueNode map
    for (String child : children) {
      TreeNode valueNode = new TreeNode(child);
      valueNodeMap.put(child, valueNode);
    }

    // Now the map contains all nodes in the tree. Iterate through
    // all the children and
    // associate them with their parents
    for (String child : children) {
      TreeNode childNode = valueNodeMap.get(child);
      String parent = childParentMap.get(child);
      TreeNode parentNode = valueNodeMap.get(parent);
      parentNode.children.add(childNode);
    }

    // Traverse tree in level order to see the output. Pretty
    // printing the tree would be very
    // long to fit here.
    Queue q1 = new ArrayDeque();
    Queue q2 = new ArrayDeque();
    q1.add(root);
    Queue<TreeNode> toEmpty = null;
    Queue toFill = null;
    while (true) {
      if (false == q1.isEmpty()) {
        toEmpty = q1;
        toFill = q2;
      } else if (false == q2.isEmpty()) {
        toEmpty = q2;
        toFill = q1;
      } else {
        break;
      }
      while (false == toEmpty.isEmpty()) {
        TreeNode node = toEmpty.poll();
        System.out.print(node.value + ", ");
        toFill.addAll(node.children);
      }
      System.out.println("");
    }
  }

}

但是想不通 了解如何格式化输出以类似于表格。或者是否有 sql 语句/存储过程(如 this question )来执行此操作？

编辑:为方便起见，部门名称只是一个示例，它可以是任意字符串。

Answer 1

您没有注意到您使用的是什么 RDBMS，但如果它恰好是 MS SQL Server 或某些版本的 Oracle(请参阅下面的编辑)，并且仅作为我们这些使用 T-SQL 并感兴趣的人的服务在这个问题中，你可以使用 recursive CTE在 SQL Server 中完成此操作。

编辑:Oracle(我对它的经验非常非常少)似乎也支持递归，包括自 11g 第 2 版以来的递归 CTE。See this question for more information .

鉴于此设置:

CREATE TABLE hierarchy
    ([Dept] varchar(13), [superiorDept] varchar(12))
;

INSERT INTO hierarchy
    ([Dept], [superiorDept])
VALUES
    ('00',NULL),
    ('00-01', '00'),
    ('00-02', '00'),
    ('00-01-01', '00-01'),
    ('00-01-02', '00-01'),
    ('00-01-03', '00-01'),
    ('00-02-01', '00-02'),
    ('00-02-03', '00-02'),
    ('00-02-03-01', '00-02-03')
;

这个递归 CTE:

WITH cte AS (
SELECT 
  Dept,
  superiorDept,
  0 AS depth,
  CAST(Dept AS NVARCHAR(MAX)) AS sort
FROM hierarchy h
WHERE superiorDept IS NULL

UNION ALL

SELECT 
  h2.Dept,
  h2.superiorDept,
  cte.depth + 1 AS depth,
  CAST(cte.sort + h2.Dept AS NVARCHAR(MAX)) AS sort
FROM hierarchy h2
INNER JOIN cte ON h2.superiorDept = cte.Dept
)

SELECT 
  CASE WHEN depth = 0 THEN Dept END AS TopTier,
  CASE WHEN depth = 1 THEN Dept END AS Tier2,
  CASE WHEN depth = 2 THEN Dept END AS Tier3,
  CASE WHEN depth = 3 THEN Dept END AS Tier4
FROM cte
ORDER BY sort

将呈现所需的输出:

 TopTier    Tier2   Tier3     Tier4
 00 
            00-01   
                    00-01-01
                    00-01-02
                    00-01-03
            00-02
                    00-02-01
                    00-02-03
                              00-02-03-01

Here is a SQLFiddle to demonstrate

无论邻接表中涉及的两列的内容或类型如何，这都应该有效。只要父值与 id 值匹配，CTE 生成深度和排序信息应该没有问题。

完全在 SQL 中执行此操作的唯一警告是您必须手动定义一定数量的列，但在处理 SQL 时几乎总是如此。您可能希望生成 CTE，然后将已排序的结果传递给您的程序并使用深度信息来简化列操作。