背景
异步树展开如果要实现展开回调比较困难,因为展开的过程是异步的。
前端:js引擎虽然是单线程执行,但是操作ui的线程是单独的,树的展开过程,就经历了js引擎线程+ui线程的过程,展开代码和展开回调的代码在不同时机执行的,本质上就是异步的。
展开回调的实现
展开回调的实现困难点在于判断展开结束的时机。树的展开是一个递归过程,本质上就是判断递归的结束时机。递归的过程,其实就是进栈出栈的过程。如果我们能够统计进栈出栈的次数,就能够判断出结束的时机。下面是验证代码(Java)
1.树
package com.hdwang.test.recursize; import java.util.ArrayList; import java.util.List; /** * @author wanghuidong * 时间: 2022/8/13 22:36 */ public class TreeNode { private int id; private int level; List<TreeNode> children = new ArrayList<>(); public int getId() { return id; } public void setId(int id) { this.id = id; } public int getLevel() { return level; } public void setLevel(int level) { this.level = level; } public List<TreeNode> getChildren() { return children; } public void setChildren(List<TreeNode> children) { this.children = children; } @Override public String toString() { return "TreeNode{" + "id=" + id + ", level=" + level + '}'; } }
2.回调接口
package com.hdwang.test.recursize; /** * @author wanghuidong * 时间: 2022/8/13 22:54 */ @FunctionalInterface public interface Callback { void callback(); }
3.树的展开回调实现
package com.hdwang.test.recursize; import java.util.Timer; import java.util.TimerTask; import java.util.concurrent.atomic.AtomicInteger; /** * 递归测试 * * @author wanghuidong * 时间: 2022/8/13 22:34 */ public class RecursiveTest { static AtomicInteger treeRecursiveEnterCount = new AtomicInteger(0); static AtomicInteger treeRecursiveExitCount = new AtomicInteger(0); static AtomicInteger treeRecursiveRemainCount = new AtomicInteger(0); static AtomicInteger treeRecursiveCount = new AtomicInteger(0); static Timer timer = new Timer(); public static void main(String[] args) { //构造一颗树 TreeNode rootNode = buildTree(); //打印树 System.out.println("==================printTree=================="); printTree(rootNode); //展开回调(若异步退出需写在异步函数里) System.out.println("==================expandTree=================="); expandTree(rootNode, new Callback() { @Override public void callback() { System.out.println("callback execute"); //结束时间器 timer.cancel(); } }); } private static TreeNode buildTree() { TreeNode rootNode = new TreeNode(); rootNode.setId(1); rootNode.setLevel(0); TreeNode treeNode21 = new TreeNode(); treeNode21.setId(2); treeNode21.setLevel(1); rootNode.getChildren().add(treeNode21); TreeNode treeNode211 = new TreeNode(); treeNode211.setId(4); treeNode211.setLevel(2); treeNode21.getChildren().add(treeNode211); TreeNode treeNode22 = new TreeNode(); treeNode22.setId(3); treeNode22.setLevel(1); rootNode.getChildren().add(treeNode22); return rootNode; } /** * 打印树 * * @param node */ static void printTree(TreeNode node) { System.out.println(node); if (!node.getChildren().isEmpty()) { for (TreeNode childNode : node.getChildren()) { printTree(childNode); } } } static void expandTree(TreeNode node, Callback callback) { //进入递归函数,统计递归进入次数 treeRecursiveCount.incrementAndGet(); treeRecursiveEnterCount.incrementAndGet(); treeRecursiveRemainCount.incrementAndGet(); printRecursiveCounts(); //展开结点(模拟异步) timer.schedule(new TimerTask() { @Override public void run() { System.out.println("展开结点:" + node); //递归孩子 if (!node.getChildren().isEmpty()) { for (TreeNode childNode : node.getChildren()) { expandTree(childNode, callback); } treeRecursiveExitCount.incrementAndGet(); treeRecursiveRemainCount.decrementAndGet(); printRecursiveCounts(); if (treeRecursiveRemainCount.get() == 0) { callback.callback(); } } else { treeRecursiveExitCount.incrementAndGet(); treeRecursiveRemainCount.decrementAndGet(); printRecursiveCounts(); if (treeRecursiveRemainCount.get() == 0) { callback.callback(); } } } }, 1000); } private static void printRecursiveCounts() { System.out.printf("递归次数:%d,进入次数:%d,退出次数:%d,剩余次数:%d/n", treeRecursiveCount.get(), treeRecursiveEnterCount.get(), treeRecursiveExitCount.get(), treeRecursiveRemainCount.get()); } }
4.测试结果
==================printTree================== TreeNode{id=1, level=0} TreeNode{id=2, level=1} TreeNode{id=4, level=2} TreeNode{id=3, level=1} ==================expandTree================== 递归次数:1,进入次数:1,退出次数:0,剩余次数:1 展开结点:TreeNode{id=1, level=0} 递归次数:2,进入次数:2,退出次数:0,剩余次数:2 递归次数:3,进入次数:3,退出次数:0,剩余次数:3 递归次数:3,进入次数:3,退出次数:1,剩余次数:2 展开结点:TreeNode{id=2, level=1} 递归次数:4,进入次数:4,退出次数:1,剩余次数:3 递归次数:4,进入次数:4,退出次数:2,剩余次数:2 展开结点:TreeNode{id=3, level=1} 递归次数:4,进入次数:4,退出次数:3,剩余次数:1 展开结点:TreeNode{id=4, level=2} 递归次数:4,进入次数:4,退出次数:4,剩余次数:0 callback execute
javascript实现版
下面是对 vue框架element-ui库中的tree组件 实现的展开回调,实现功能:展开异步树指定结点然后勾选结点
//展开匹配的结点,根结点默认展开 expandNodes(nodeDataIds){ let that = this; let tree = this.$refs.tree; let rootNode = tree.root; this.expandNode(rootNode, nodeDataIds, function(){ that.checkNodes(['node001','node002']); }); }, //展开指定结点下匹配的结点 expandNode(node, nodeDataIds, callback){ let that = this; //递归进入 that.recursiveEnter(); //当前结点需要展开未展开,则展开(根结点默认展开) if(node.level==0 || nodeDataIds.indexOf(node.data.id) != -1){ //展开孩子结点 let expandChildren = function(){ let childNodes = node.childNodes; if(childNodes.length > 0){ for (let i = 0; i < childNodes.length; i++) { let childNode = childNodes[i]; //递归展开孩子结点 that.expandNode(childNode, nodeDataIds, callback); } } } if(!node.expanded){ //当前结点未展开则先展开,展开后再展开孩子结点 node.expand(function(){ //展开孩子结点 expandChildren(); //递归退出 that.recursiveExit(callback); }); }else{ //当前结点已展开,直接展开孩子结点 expandChildren(); //递归退出 that.recursiveExit(callback); } }else{ //递归退出 that.recursiveExit(callback); } }, //递归计入计数剩余递归次数 recursiveEnter(){ this.recursiveRemainCount++; console.log('enter recursiveRemainCount', this.recursiveRemainCount) }, //递归退出计数剩余递归次数 recursiveExit(callback){ this.recursiveRemainCount--; console.log('exit recursiveRemainCount', this.recursiveRemainCount) if(this.recursiveRemainCount==0){ if(callback){ callback(); } } }, checkNodes(nodeDataIds){ let tree = this.$refs.tree; tree.setCheckedKeys(nodeDataIds, false) }
总结
递归回调的实现,可以用一个变量统计剩余递归次数,当剩余递归次数为0时,则表示递归结束,可以执行回调函数。前端js是单线程,无需考虑线程安全问题,后端则需要考虑线程安全问题,java可以用原子类AtomicInteger统计次数。
原创文章,作者:ItWorker,如若转载,请注明出处:https://blog.ytso.com/280405.html