内容简介: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。
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
1234567public class Tree {private Integer id;private String name;private Integer parentId;private List<Integer> childIds;} - 往Redis中添加测试数据:
1234567891011121314151617181920[@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 = 100000tree.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。
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