使用 Redis 解决“树”形数据的复杂查询-Linuxeden开源社区

原文出处： qiujiayu

最近因业务需要，研究了一下树数据结果的存储及查询解决方案。最初的想法是使用neo4j,可是在网上看了一下开源的不支持集群，感觉用的人不多。

网上也查了一些树形结构数据存储方案但每种实现方案都有它的一定局限性。

想了一短时间后，想出了下面的方案：

一、因为复杂的查询都由Redis来处理，所以数据库表的设计就变得非常简单：tree 表

| 字段名称 | 数据类型 | 备注说明 | | —- | —- | —- | | id | int | 主键 | | parent_id | int | 上级节点ID |

二、Redis的数据存储方案：

把表的数据存储到一个Hash表中，使用表中的id值做为此hash表的key, value值为：

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{
   id: 10,
   parentId: 9,
   childIds: [11]
}

代码实现

为了简化测试，这里只演示Redis相关的操作

Tree 类定义

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public class Tree {

     private Integer id;

     private String name;

     private Integer parentId;

     private List<Integer> childIds;

}

往Redis中添加测试数据：

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[@Test](https://my.oschina.net/azibug)
public void addTestData() throws Exception {
     String key = “tree-test-key”;
     Tree tree = new Tree();
     List<Integer> childIds = new ArrayList<>();
     int max = 100000
     tree.setChildIds(childIds);
     for (int i = 0; i < max; i++) {
         tree.setId(i);
         tree.setName(“tree” + i);
         if (i > 0) {
             tree.setParentId(i – 1);
         }
         childIds.clear();
         if(i < (max – 1)){
             childIds.add(i + 1);
         }
         redis.setHash(key, “” + i, JsonUtil.toJson(tree));
     }
}

Lua 代码的实现

在Lua中使用递归时，需要使用“尾调用”来优化代码。关于尾调用的知识，大家可以上网去搜索。

获取所有子节点 get-tree-childs.lua

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local treeKey = KEYS[1]
local fnodeId  = ARGV[1]
local function getTreeChild(currentnode, t, res)
  if currentnode == nil or t == nil  then
    return res
  end
  local nextNode = nil
  local nextType = nil
  if t == “id” and (type(currentnode) == “number” or type(currentnode) == “string”) then
    local treeNode = redis.call(“HGET”, treeKey, currentnode)
    if treeNode then
      local node = cjson.decode(treeNode)
      table.insert(res, treeNode)
      if node and node.childIds then
        nextNode = node.childIds
        nextType = “childIds”
      end
    end
  elseif t == “childIds” then
    nextNode = {}
    nextType = “childIds”
    local treeNode  = nil
    local node = nil
    local cnt = 0
    for _, val in ipairs(currentnode) do
      treeNode = redis.call(“HGET”, treeKey, tostring(val))
      if treeNode then
        node = cjson.decode(treeNode)
        table.insert(res, treeNode)
        if node and node.childIds then
          for _, val2 in ipairs(node.childIds) do
            table.insert(nextNode, val2)
            cnt = cnt + 1
          end
        end
      end
    end
    if cnt == 0 then
      nextNode = nil
      nextType = nil
    end
  end
  return getTreeChild(nextNode, nextType, res)
end
if treeKey and fnodeId then
  return getTreeChild(fnodeId, “id”, {})
end
return {}

获取所有子节点数目 get-tree-childs-cnt.lua

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local treeKey = KEYS[1]
local fnodeId  = ARGV[1]
local function getTreeChildCnt(currentnode, t, res)
  if currentnode == nil or t == nil  then
    return res
  end
  local nextNode = nil
  local nextType = nil
  if t == “id” and (type(currentnode) == “number” or type(currentnode) == “string”) then
    local treeNode = redis.call(“HGET”, treeKey, currentnode)
    if treeNode then
      local node = cjson.decode(treeNode)
      res = res + 1
      if node and node.childIds then
        nextNode = node.childIds
        nextType = “childIds”
      end
    end
  elseif t == “childIds” then
    nextNode = {}
    nextType = “childIds”
    local treeNode  = nil
    local cnt = 0
    for _, val in ipairs(currentnode) do
      treeNode = redis.call(“HGET”, treeKey, tostring(val))
      if treeNode then
        local node = cjson.decode(treeNode)
        res = res + 1
        if node and node.childIds then
          for _, val2 in ipairs(node.childIds) do
            table.insert(nextNode, val2)
            cnt = cnt + 1
          end
        end
      end
    end
    if cnt == 0 then
      nextNode = nil
      nextType = nil
    end
  end
  return getTreeChildCnt(nextNode, nextType, res)
end
if treeKey and fnodeId then
  return getTreeChildCnt(fnodeId, “id”, 0)
end
return 0

获取所有子节点数目 get-tree-parent.lua

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local treeKey = KEYS[1]
local nodeId  = ARGV[1]
local function getTreeParent(treeKey, res, nodeId)
  if nodeId == nil or not (type(nodeId) == “number” or type(nodeId) == “string”) then
    return res
  end
  local treeNode = redis.call(“HGET”, treeKey, nodeId)
  local nextNodeId = nil
  if treeNode then
    local node = cjson.decode(treeNode)
    table.insert(res, treeNode)
    if node then
      nextNodeId = node.parentId
    end
  end
  return getTreeParent(treeKey, res, nextNodeId)
end
if treeKey and nodeId then
  return getTreeParent(treeKey, {}, nodeId)
end
return {}

获取所有子节点数目 get-tree-parent-cnt.lua

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local treeKey = KEYS[1]
local nodeId  = ARGV[1]
local function getTreeParentCnt(treeKey, nodeId, res)
  if nodeId == nil or not (type(nodeId) == “number” or type(nodeId) == “string”) then
    return res
  end
  local treeNode = redis.call(“HGET”, treeKey, nodeId)
  local nextNodeId = nil
  if treeNode then
    local node = cjson.decode(treeNode)
    res = res + 1
    if node then
      nextNodeId = node.parentId
    end
  end
  return getTreeParentCnt(treeKey, nextNodeId, res)
end
if treeKey and nodeId then
  return getTreeParentCnt(treeKey, nodeId, 0)
end
return 0

以上代码因为使用了“尾调用”，所以变得相对比较复杂

总结

此方案相对比较灵活，能支持相对比较大量的数据。

缺点：过于依赖Redis。数据同步会麻烦些，好在操作不是很复杂。

转自 http://blog.jobbole.com/112358/

代码实现

获取所有子节点 get-tree-childs.lua

获取所有子节点数目 get-tree-childs-cnt.lua

获取所有子节点数目 get-tree-parent.lua

获取所有子节点数目 get-tree-parent-cnt.lua

总结

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