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<html lang="zh"><head><meta charset="utf-8"><meta name="viewport" content="width=device-width, initial-scale=1, maximum-scale=1"><meta><title>吾途 - Studeyang.tech</title><link rel="manifest" href="/manifest.json"><meta name="application-name" content="吾途 - Studeyang.tech"><meta name="msapplication-TileImage" content="https://technotes.oss-cn-shenzhen.aliyuncs.com/2023/202306152245106.gif"><meta name="apple-mobile-web-app-capable" content="yes"><meta name="apple-mobile-web-app-title" content="吾途 - Studeyang.tech"><meta name="apple-mobile-web-app-status-bar-style" content="default"><meta name="description" content="愿你出走半生，归来仍是少年"><meta property="og:type" content="blog"><meta property="og:title" content="吾途 - Studeyang.tech"><meta property="og:url" content="https://studeyang.tech/"><meta property="og:site_name" content="吾途 - Studeyang.tech"><meta property="og:description" content="愿你出走半生，归来仍是少年"><meta property="og:locale" content="zh_CN"><meta property="og:image" content="https://studeyang.tech/img/og_image.png"><meta property="article:author" content="Studeyang"><meta property="article:tag" content="程序员"><meta property="article:tag" content="编程"><meta property="article:tag" content="架构"><meta property="article:tag" content="技术"><meta property="twitter:card" content="summary"><meta property="twitter:image:src" content="https://studeyang.tech/img/og_image.png"><meta property="twitter:creator" content="@studeyang"><meta property="twitter:site" content="studeyang"><script type="application/ld+json">{"@context":"https://schema.org","@type":"BlogPosting","mainEntityOfPage":{"@type":"WebPage","@id":"https://studeyang.tech"},"headline":"吾途 - Studeyang.tech","image":["https://studeyang.tech/img/og_image.png"],"author":{"@type":"Person","name":"Studeyang"},"publisher":{"@type":"Organization","name":"吾途 - Studeyang.tech","logo":{"@type":"ImageObject","url":"https://technotes.oss-cn-shenzhen.aliyuncs.com/2023/202306152245106.gif"}},"description":"愿你出走半生，归来仍是少年"}</script><link rel="alternate" href="/rss2.xml" title="吾途 - Studeyang.tech" type="application/atom+xml"><link rel="icon" href="https://technotes.oss-cn-shenzhen.aliyuncs.com/2023/202306152245106.gif"><link rel="stylesheet" href="https://use.fontawesome.com/releases/v6.0.0/css/all.css"><link rel="stylesheet" href="https://cdn.jsdelivr.net/npm/[email protected]/styles/atom-one-light.css"><link rel="stylesheet" href="https://fonts.googleapis.com/css2?family=Ubuntu:wght@400;600&family=Source+Code+Pro"><link rel="stylesheet" href="/css/default.css"><style>body>.footer,body>.navbar,body>.section{opacity:0}</style><!--!--><script>var _hmt = _hmt || [];
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</script><!-- hexo injector head_end end --><meta name="generator" content="Hexo 6.3.0"></head><body class="is-3-column"><nav class="navbar navbar-main"><div class="container navbar-container"><div class="navbar-brand justify-content-center"><a class="navbar-item navbar-logo" href="/"><img src="https://technotes.oss-cn-shenzhen.aliyuncs.com/2023/202306152245106.gif" alt="吾途 - Studeyang.tech" height="28"></a></div><div class="navbar-menu"><div class="navbar-start"><a class="navbar-item is-active" href="/">主页</a><a class="navbar-item" href="/technotes">学习笔记</a><a class="navbar-item" href="/about">关于我</a></div><div class="navbar-end"><a class="navbar-item" target="_blank" rel="noopener" title="Download on GitHub" href="https://github.com/studeyang/studeyang.tech"><i class="fab fa-github"></i></a><a class="navbar-item search" title="搜索" href="javascript:;"><i class="fas fa-search"></i></a></div></div></div></nav><section class="section"><div class="container"><div class="columns"><div class="column order-2 column-main is-8-tablet is-8-desktop is-6-widescreen"><div class="card"><div class="card-image"><a class="image is-7by3" href="/2024/0803.html"><img class="fill" src="https://technotes.oss-cn-shenzhen.aliyuncs.com/2024/202408022243347.jpeg" alt="【译】Apache Shiro介绍"></a></div><article class="card-content article" role="article"><div class="article-meta is-size-7 is-uppercase level is-mobile"><div class="level-left"><span class="level-item"><time dateTime="2024-08-03T01:00:00.000Z" title="2024/8/3 上午9:00:00">2024-08-03</time>发表</span><span class="level-item"><time dateTime="2024-08-02T14:46:23.598Z" title="2024/8/2 下午10:46:23">2024-08-02</time>更新</span><span class="level-item"><a class="link-muted" href="/categories/technotes/">technotes</a></span><span class="level-item">29 分钟读完 (大约4400个字)</span></div></div><p class="title is-3 is-size-4-mobile"><a class="link-muted" href="/2024/0803.html">【译】Apache Shiro介绍</a></p><div class="content"><h2 id="一、什么是-Apache-Shiro？"><a href="#一、什么是-Apache-Shiro？" class="headerlink" title="一、什么是 Apache Shiro？"></a>一、什么是 Apache Shiro？</h2><p>Apache Shiro 是一个功能强大且易于使用的 Java 安全框架，表现在身份认证、授权、加密和会话管理，可用于保护任何应用程序，小到命令行应用程序、移动应用程序，大到 Web 和企业应用程序。</p>
<p>应用程序安全性的 4 个基石：</p>
<ul>
<li>身份验证 - 证明用户身份，通常称为用户“登录”</li>
<li>授权 - 访问控制</li>
<li>加密 - 保护或隐藏数据免遭窥探</li>
<li>会话管理 - 每个用户的时间敏感状态</li>
</ul></div><a class="article-more button is-small is-size-7" href="/2024/0803.html#more">阅读更多</a></article></div><div class="card"><div class="card-image"><a class="image is-7by3" href="/2024/0725.html"><img class="fill" src="https://technotes.oss-cn-shenzhen.aliyuncs.com/2024/202407252241818.jpg" alt="MySQL是如何给表加字段的？"></a></div><article class="card-content article" role="article"><div class="article-meta is-size-7 is-uppercase level is-mobile"><div class="level-left"><span class="level-item"><time dateTime="2024-07-25T14:00:00.000Z" title="2024/7/25 下午10:00:00">2024-07-25</time>发表</span><span class="level-item"><time dateTime="2024-08-02T14:37:50.167Z" title="2024/8/2 下午10:37:50">2024-08-02</time>更新</span><span class="level-item"><a class="link-muted" href="/categories/technotes/">technotes</a></span><span class="level-item">15 分钟读完 (大约2266个字)</span></div></div><p class="title is-3 is-size-4-mobile"><a class="link-muted" href="/2024/0725.html">MySQL是如何给表加字段的？</a></p><div class="content"><p>在我最近的项目中，经常会有给大表加字段的需求，这个过程非常耗时。</p>
<p><img src="https://technotes.oss-cn-shenzhen.aliyuncs.com/2024/202407251951545.png"></p>
<p>可以看到，900 万数据量的一张表，加一个字段就需要 3 个小时左右。</p>
<p>我们知道，给一个表加字段，或者修改字段，或者加索引，需要扫描全表的数据。假设表数据量比较大，加字段的过程将会非常耗时。</p>
<p>不过我最关心的是，在加字段的过程中，会不会对业务的增删改查造成影响？在询问 DBA 后，他给出的答复是不会造成影响。这不禁让我思考这背后的实现原理。</p>
<p>下面，我们就来一探究竟。</p></div><a class="article-more button is-small is-size-7" href="/2024/0725.html#more">阅读更多</a></article></div><div class="card"><div class="card-image"><a class="image is-7by3" href="/2024/0705.html"><img class="fill" src="https://technotes.oss-cn-shenzhen.aliyuncs.com/2024/202407052309418.png" alt="MyBatis拦截器在实际项目中的应用"></a></div><article class="card-content article" role="article"><div class="article-meta is-size-7 is-uppercase level is-mobile"><div class="level-left"><span class="level-item"><time dateTime="2024-07-05T01:00:00.000Z" title="2024/7/5 上午9:00:00">2024-07-05</time>发表</span><span class="level-item"><time dateTime="2024-07-05T15:20:00.655Z" title="2024/7/5 下午11:20:00">2024-07-05</time>更新</span><span class="level-item"><a class="link-muted" href="/categories/technotes/">technotes</a></span><span class="level-item">13 分钟读完 (大约1881个字)</span></div></div><p class="title is-3 is-size-4-mobile"><a class="link-muted" href="/2024/0705.html">MyBatis拦截器在实际项目中的应用</a></p><div class="content"><p>MyBatis 是一个流行的 Java 持久层框架，它简化了数据库访问的复杂性，为开发者提供了强大的功能。其中，MyBatis 拦截器是一个非常有用的特性，可以帮助开发者灵活地解决各种问题。</p>
<p>本文将探讨 MyBatis 拦截器在实际项目中的应用场景和具体实现方法。</p>
<blockquote>
<p>文中代码：<a target="_blank" rel="noopener" href="https://github.com/studeyang/mybatis-interceptor-demo">https://github.com/studeyang/mybatis-interceptor-demo</a></p>
</blockquote></div><a class="article-more button is-small is-size-7" href="/2024/0705.html#more">阅读更多</a></article></div><div class="card"><div class="card-image"><a class="image is-7by3" href="/2023/1124.html"><img class="fill" src="https://technotes.oss-cn-shenzhen.aliyuncs.com/2023/4881406_2724_4.jpg" alt="Kafka 位移提交的正确姿势"></a></div><article class="card-content article" role="article"><div class="article-meta is-size-7 is-uppercase level is-mobile"><div class="level-left"><span class="level-item"><time dateTime="2023-11-24T01:00:00.000Z" title="2023/11/24 上午9:00:00">2023-11-24</time>发表</span><span class="level-item"><time dateTime="2023-11-24T14:10:37.947Z" title="2023/11/24 下午10:10:37">2023-11-24</time>更新</span><span class="level-item"><a class="link-muted" href="/categories/technotes/">technotes</a></span><span class="level-item">8 分钟读完 (大约1146个字)</span></div></div><p class="title is-3 is-size-4-mobile"><a class="link-muted" href="/2023/1124.html">Kafka 位移提交的正确姿势</a></p><div class="content"><p>你说你 Kafka 用了很多年了，但是位移提交的这些细节你未必清楚。</p>
<p>不久前有个同事出去面试了，他说在 Kafka 问题上被面试官藐视了。</p></div><a class="article-more button is-small is-size-7" href="/2023/1124.html#more">阅读更多</a></article></div><div class="card"><div class="card-image"><a class="image is-7by3" href="/2023/0912.html"><img class="fill" src="https://technotes.oss-cn-shenzhen.aliyuncs.com/2023/202309112237318.jpg" alt="Spring Security入门：保护Web应用"></a></div><article class="card-content article" role="article"><div class="article-meta is-size-7 is-uppercase level is-mobile"><div class="level-left"><span class="level-item"><time dateTime="2023-09-12T01:00:00.000Z" title="2023/9/12 上午9:00:00">2023-09-12</time>发表</span><span class="level-item"><time dateTime="2023-09-19T16:02:57.815Z" title="2023/9/20 上午12:02:57">2023-09-20</time>更新</span><span class="level-item"><a class="link-muted" href="/categories/technotes/">technotes</a></span><span class="level-item">14 分钟读完 (大约2110个字)</span></div></div><p class="title is-3 is-size-4-mobile"><a class="link-muted" href="/2023/0912.html">Spring Security入门：保护Web应用</a></p><div class="content"><p>本文我们将构建一个简单但完整的小型 Web 应用程序，以演示 Spring Security 的入门教程。大致逻辑是：当合法用户成功登录系统之后，浏览器会跳转到一个系统主页，并展示一些个人健康档案（HealthRecord）数据。</p>
<p>让我们开始吧！</p></div><a class="article-more button is-small is-size-7" href="/2023/0912.html#more">阅读更多</a></article></div><div class="card"><div class="card-image"><a class="image is-7by3" href="/2023/0911.html"><img class="fill" src="https://technotes.oss-cn-shenzhen.aliyuncs.com/2023/202309102143620.png" alt="今日算法13-剪绳子"></a></div><article class="card-content article" role="article"><div class="article-meta is-size-7 is-uppercase level is-mobile"><div class="level-left"><span class="level-item"><time dateTime="2023-09-11T01:00:00.000Z" title="2023/9/11 上午9:00:00">2023-09-11</time>发表</span><span class="level-item"><time dateTime="2023-09-10T13:49:16.952Z" title="2023/9/10 下午9:49:16">2023-09-10</time>更新</span><span class="level-item"><a class="link-muted" href="/categories/%E4%BB%8A%E6%97%A5%E7%AE%97%E6%B3%95/">今日算法</a></span><span class="level-item">5 分钟读完 (大约719个字)</span></div></div><p class="title is-3 is-size-4-mobile"><a class="link-muted" href="/2023/0911.html">今日算法13-剪绳子</a></p><div class="content"><h2 id="一、题目描述"><a href="#一、题目描述" class="headerlink" title="一、题目描述"></a>一、题目描述</h2><blockquote>
<p>题目链接：<a target="_blank" rel="noopener" href="https://leetcode.cn/problems/jian-sheng-zi-lcof/">https://leetcode.cn/problems/jian-sheng-zi-lcof/</a></p>
<p>难易程度：中等</p>
</blockquote>
<p>给你一根长度为 n 的绳子，请把绳子剪成整数长度的 m 段（m、n 都是整数，n > 1 并且 m > 1），每段绳子的长度记为 <code>k[0],k[1]...k[m-1]</code> 。请问 <code>k[0]*k[1]*...*k[m-1]</code> 可能的最大乘积是多少？例如，当绳子的长度是 8 时，我们把它剪成长度分别为 2、3、3 的三段，此时得到的最大乘积是 18。</p>
<figure class="highlight html"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">输入: 10</span><br><span class="line">输出: 36</span><br><span class="line">解释: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36</span><br></pre></td></tr></table></figure></div><a class="article-more button is-small is-size-7" href="/2023/0911.html#more">阅读更多</a></article></div><div class="card"><div class="card-image"><a class="image is-7by3" href="/2023/0803.html"><img class="fill" src="https://technotes.oss-cn-shenzhen.aliyuncs.com/2023/202308022332512.jpeg" alt="程序员的修炼-知识与体系"></a></div><article class="card-content article" role="article"><div class="article-meta is-size-7 is-uppercase level is-mobile"><div class="level-left"><span class="level-item"><time dateTime="2023-08-03T01:00:00.000Z" title="2023/8/3 上午9:00:00">2023-08-03</time>发表</span><span class="level-item"><time dateTime="2023-08-02T15:40:10.087Z" title="2023/8/2 下午11:40:10">2023-08-02</time>更新</span><span class="level-item"><a class="link-muted" href="/categories/technotes/">technotes</a></span><span class="level-item">18 分钟读完 (大约2722个字)</span></div></div><p class="title is-3 is-size-4-mobile"><a class="link-muted" href="/2023/0803.html">程序员的修炼-知识与体系</a></p><div class="content"><p>本文来自我最近正在学习的课程，极客时间胡峰的专栏文章《程序员进阶攻略》，内容如下。</p>
<p>今年年初，我学习了梁宁的《产品思维》课，其中有一篇叫《点线面体的战略选择》，我觉得特别有感触。虽然是讲产品，但假如把个人的成长当成产品演进一样来发展，会有一种异曲同工、殊途同归之感。</p>
<p>在我工作的经历中就曾碰到过这么一个人，他一开始做了几年开发，从前端到后端，后来又转做测试，接触的“点”倒是不少，但却没能连接起来形成自己的体系，那他个人最大的价值就局限在最后所在的“点”上了。</p>
<p>其实个人的成长有很多方面，但对于程序员的成长最重要的就是知识体系的构建，这其实就是一个 “点线面体” 的演进过程。</p>
<p>下面我会结合自己的成长路线来梳理下这个体系的建立过程。</p></div><a class="article-more button is-small is-size-7" href="/2023/0803.html#more">阅读更多</a></article></div><div class="card"><div class="card-image"><a class="image is-7by3" href="/2023/0802.html"><img class="fill" src="https://technotes.oss-cn-shenzhen.aliyuncs.com/2023/202307302248361.png" alt="今日算法12-矩阵中的路径"></a></div><article class="card-content article" role="article"><div class="article-meta is-size-7 is-uppercase level is-mobile"><div class="level-left"><span class="level-item"><time dateTime="2023-08-02T01:00:00.000Z" title="2023/8/2 上午9:00:00">2023-08-02</time>发表</span><span class="level-item"><time dateTime="2023-07-30T14:52:14.878Z" title="2023/7/30 下午10:52:14">2023-07-30</time>更新</span><span class="level-item"><a class="link-muted" href="/categories/%E4%BB%8A%E6%97%A5%E7%AE%97%E6%B3%95/">今日算法</a></span><span class="level-item">5 分钟读完 (大约687个字)</span></div></div><p class="title is-3 is-size-4-mobile"><a class="link-muted" href="/2023/0802.html">今日算法12-矩阵中的路径</a></p><div class="content"><h2 id="一、题目描述"><a href="#一、题目描述" class="headerlink" title="一、题目描述"></a>一、题目描述</h2><blockquote>
<p>题目链接：<a target="_blank" rel="noopener" href="https://leetcode.cn/problems/ju-zhen-zhong-de-lu-jing-lcof/">https://leetcode.cn/problems/ju-zhen-zhong-de-lu-jing-lcof/</a></p>
<p>难易程度：中等</p>
</blockquote>
<p>给定一个 <code>m x n</code> 二维字符网格 <code>board</code> 和一个字符串单词 <code>word</code> 。如果 <code>word</code> 存在于网格中，返回 <code>true</code> ；否则，返回 <code>false</code> 。</p>
<p>例如，在下面的 3×4 的矩阵中包含单词 “ABCCED”（单词中的字母已标出）。</p>
<p><img src="https://technotes.oss-cn-shenzhen.aliyuncs.com/2023/202307302242023.png"></p>
<p>示例：</p>
<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"</span><br><span class="line">输出：true</span><br></pre></td></tr></table></figure></div><a class="article-more button is-small is-size-7" href="/2023/0802.html#more">阅读更多</a></article></div><div class="card"><div class="card-image"><a class="image is-7by3" href="/2023/0801.html"><img class="fill" src="https://technotes.oss-cn-shenzhen.aliyuncs.com/2023/202307302209840.png" alt="今日算法11-旋转数组的最小数字"></a></div><article class="card-content article" role="article"><div class="article-meta is-size-7 is-uppercase level is-mobile"><div class="level-left"><span class="level-item"><time dateTime="2023-08-01T01:00:00.000Z" title="2023/8/1 上午9:00:00">2023-08-01</time>发表</span><span class="level-item"><time dateTime="2023-07-30T14:35:03.996Z" title="2023/7/30 下午10:35:03">2023-07-30</time>更新</span><span class="level-item"><a class="link-muted" href="/categories/%E4%BB%8A%E6%97%A5%E7%AE%97%E6%B3%95/">今日算法</a></span><span class="level-item">5 分钟读完 (大约698个字)</span></div></div><p class="title is-3 is-size-4-mobile"><a class="link-muted" href="/2023/0801.html">今日算法11-旋转数组的最小数字</a></p><div class="content"><h2 id="一、题目描述"><a href="#一、题目描述" class="headerlink" title="一、题目描述"></a>一、题目描述</h2><blockquote>
<p>题目链接：<a target="_blank" rel="noopener" href="https://leetcode.cn/problems/xuan-zhuan-shu-zu-de-zui-xiao-shu-zi-lcof">https://leetcode.cn/problems/xuan-zhuan-shu-zu-de-zui-xiao-shu-zi-lcof</a></p>
<p>难易程度：简单</p>
</blockquote>
<p>把一个数组最开始的若干个元素搬到数组的末尾，我们称之为数组的旋转。</p>
<p>给你一个可能存在 重复 元素值的数组 numbers ，它原来是一个升序排列的数组，并按上述情形进行了一次旋转。请返回旋转数组的最小元素。例如，数组 [3,4,5,1,2] 为 [1,2,3,4,5] 的一次旋转，该数组的最小值为 1。</p>
<p>注意，数组 [a[0], a[1], a[2], …, a[n-1]] 旋转一次 的结果为数组 [a[n-1], a[0], a[1], a[2], …, a[n-2]] 。</p>
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/0038204c-4b8a-42a5-921d-080f6674f989.png" width="210px"> </div>
<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br></pre></td><td class="code"><pre><span class="line">示例1：</span><br><span class="line">输入：numbers = [3,4,5,1,2]</span><br><span class="line">输出：1</span><br><span class="line"></span><br><span class="line">示例2：</span><br><span class="line">输入：numbers = [2,2,2,0,1]</span><br><span class="line">输出：0</span><br></pre></td></tr></table></figure></div><a class="article-more button is-small is-size-7" href="/2023/0801.html#more">阅读更多</a></article></div><div class="card"><div class="card-image"><a class="image is-7by3" href="/2023/0731.html"><img class="fill" src="https://technotes.oss-cn-shenzhen.aliyuncs.com/2023/202307302149261.png" alt="今日算法10-变态跳台阶"></a></div><article class="card-content article" role="article"><div class="article-meta is-size-7 is-uppercase level is-mobile"><div class="level-left"><span class="level-item"><time dateTime="2023-07-31T01:00:00.000Z" title="2023/7/31 上午9:00:00">2023-07-31</time>发表</span><span class="level-item"><time dateTime="2023-07-30T14:39:56.631Z" title="2023/7/30 下午10:39:56">2023-07-30</time>更新</span><span class="level-item"><a class="link-muted" href="/categories/%E4%BB%8A%E6%97%A5%E7%AE%97%E6%B3%95/">今日算法</a></span><span class="level-item">4 分钟读完 (大约621个字)</span></div></div><p class="title is-3 is-size-4-mobile"><a class="link-muted" href="/2023/0731.html">今日算法10-变态跳台阶</a></p><div class="content"><h2 id="一、题目描述"><a href="#一、题目描述" class="headerlink" title="一、题目描述"></a>一、题目描述</h2><blockquote>
<p>题目链接：<a target="_blank" rel="noopener" href="https://www.nowcoder.com/practice/22243d016f6b47f2a6928b4313c85387?tpId=13&tqId=11162&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github">牛客网</a></p>
<p>难易程度：简单</p>
</blockquote>
<p>一只青蛙一次可以跳上 1 级台阶，也可以跳上 2 级… 它也可以跳上 n 级。求该青蛙跳上一个 n 级的台阶总共有多少种跳法。</p>
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