178 lines
7.4 KiB
JavaScript
178 lines
7.4 KiB
JavaScript
'use strict';
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/**
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* 入库绑定 + 限额清理：应用里唯一会主动删用户数据的两条路径。
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*
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* 起因（2026-07-31 体检）：insertText / insertImage 的调用把 `now` 绑到了
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* top_at 列（占位符第 3 个是 top_at），导致每条新记录都"被置顶"——
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* trimOldest / clearNonTop 只删 top_at IS NULL 的行，于是限额清理和
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* 「清空历史」在会话内全部静默失效，history.db 可无限膨胀。启动时会清空
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* 所有 top_at，所以重启后"看起来正常"，藏了很久。
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*
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* main.js 不能脱离 Electron 加载，这里两层守：
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* 1. 静态：插入调用的第 3 个实参必须是 null（top_at）；
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* 2. 行为：把 main.js 源码里的 trimOldest / clearNonTop / INSERT **原样
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* 提取出来**在真 SQLite 上跑 —— 测的是生产 SQL 本身，不是克隆副本。
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*/
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const assert = require('node:assert');
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const fs = require('node:fs');
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const os = require('node:os');
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const path = require('node:path');
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const Database = require('better-sqlite3');
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const mainSrc = fs.readFileSync(path.join(__dirname, '..', 'main.js'), 'utf8');
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let passed = 0;
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function check(name, fn) {
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try { fn(); console.log(' ok -', name); passed++; }
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catch (e) { console.error(' FAIL -', name, '\n', e.message); process.exitCode = 1; }
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}
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// ---------- 静态：插入调用必须给 top_at 绑 null ----------
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check('insertText 调用给 top_at 绑 null，不是 now', () => {
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assert.match(
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mainSrc,
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/stmt\.insertText\.run\(\s*text\s*,\s*makePreview\(text\)\s*,\s*null\s*,\s*now\s*\)/,
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'insertText.run 的第 3 个实参（top_at 占位符）必须是 null —— 绑 now 会让每条新记录被当成置顶，限额清理全废'
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);
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});
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check('insertImage 调用给 top_at 绑 null，不是 now', () => {
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assert.match(
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mainSrc,
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/stmt\.insertImage\.run\(\s*resized\s*,\s*preview\s*,\s*null\s*,\s*now\s*\)/,
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'insertImage.run 的第 3 个实参（top_at 占位符）必须是 null'
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);
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});
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// ---------- 提取生产 SQL ----------
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function extractSql(anchor) {
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const re = new RegExp(`stmt\\.${anchor}\\s*=\\s*db\\.prepare\\(\\s*\`([\\s\\S]*?)\`\\s*\\)`);
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const m = mainSrc.match(re);
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assert.ok(m, `main.js 里找不到 stmt.${anchor} 的 prepare`);
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return m[1];
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}
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const INSERT_TEXT = extractSql('insertText');
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const TRIM_OLDEST = extractSql('trimOldest');
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const CLEAR_NON_TOP = extractSql('clearNonTop');
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const MOVE_TO_TOP = extractSql('moveToTop');
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check('INSERT 语句的占位符顺序确实是 (text, preview, top_at, created_at)', () => {
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// 静态断言成立的前提：占位符顺序变了而调用没跟着变，null 会落到别的列
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assert.match(INSERT_TEXT, /INSERT INTO clips\(\s*type\s*,\s*text\s*,\s*preview\s*,\s*top_at\s*,\s*created_at\s*\)/);
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assert.match(INSERT_TEXT, /VALUES\(\s*'text'\s*,\s*\?\s*,\s*\?\s*,\s*\?\s*,\s*\?\s*\)/);
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});
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check('trimOldest 只删 top_at IS NULL、按 created_at ASC', () => {
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assert.match(TRIM_OLDEST, /top_at IS NULL/, '少了 top_at IS NULL 会删掉用户置顶项');
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assert.match(TRIM_OLDEST, /ORDER BY created_at ASC/, '删的顺序必须是最旧优先');
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});
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// ---------- 行为：拿提取出来的 SQL 在真库上跑 ----------
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function makeDb() {
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const tmp = fs.mkdtempSync(path.join(os.tmpdir(), 'clip-limit-'));
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const db = new Database(path.join(tmp, 'history.db'));
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db.exec(`
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CREATE TABLE clips (
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id INTEGER PRIMARY KEY AUTOINCREMENT,
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type TEXT NOT NULL,
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text TEXT,
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image BLOB,
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preview TEXT,
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top_at REAL,
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created_at REAL NOT NULL
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)
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`);
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return db;
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}
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// 与 main.js 现在一致的绑定方式：(text, preview, top_at=null, created_at)
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function seedNatural(db, n) {
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const ins = db.prepare(INSERT_TEXT);
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for (let i = 1; i <= n; i++) ins.run(`clip-${i}`, `clip-${i}`, null, i);
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}
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check('新插入的记录 top_at 为 NULL（自然位置）', () => {
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const db = makeDb();
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seedNatural(db, 3);
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const rows = db.prepare('SELECT id, top_at FROM clips').all();
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assert.strictEqual(rows.length, 3);
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for (const r of rows) assert.strictEqual(r.top_at, null, `id=${r.id} 不应被置顶`);
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db.close();
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});
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check('超限清理：只删最旧的自然项，置顶项不动', () => {
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const db = makeDb();
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seedNatural(db, 5); // created_at 1..5
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db.prepare('UPDATE clips SET top_at = 100 WHERE created_at IN (4, 5)').run();
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const trim = db.prepare(TRIM_OLDEST);
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const r = trim.run(2); // 5 条保留 3 条 → 删 2
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assert.strictEqual(r.changes, 2);
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const left = db.prepare('SELECT created_at FROM clips ORDER BY created_at').all().map(x => x.created_at);
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assert.deepStrictEqual(left, [3, 4, 5], '应删掉最旧的两条自然项（1、2），置顶的 4、5 保留');
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db.close();
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});
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check('过量删除也碰不到置顶项', () => {
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const db = makeDb();
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seedNatural(db, 5);
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db.prepare('UPDATE clips SET top_at = 100 WHERE created_at IN (4, 5)').run();
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const r = db.prepare(TRIM_OLDEST).run(99);
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assert.strictEqual(r.changes, 3, '只能删掉 3 条自然项');
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const left = db.prepare('SELECT created_at FROM clips ORDER BY created_at').all().map(x => x.created_at);
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assert.deepStrictEqual(left, [4, 5], '置顶项必须全部幸存');
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db.close();
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});
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check('clearNonTop（清空历史）同样只动自然项', () => {
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const db = makeDb();
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seedNatural(db, 4);
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db.prepare('UPDATE clips SET top_at = 100 WHERE created_at = 4').run();
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const r = db.prepare(CLEAR_NON_TOP).run();
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assert.strictEqual(r.changes, 3);
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const left = db.prepare('SELECT created_at FROM clips').all().map(x => x.created_at);
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assert.deepStrictEqual(left, [4], '只剩置顶的那条');
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db.close();
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});
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check('置顶是会话级的：启动清顶后这些行又能被正常清理', () => {
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const db = makeDb();
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seedNatural(db, 3);
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db.prepare('UPDATE clips SET top_at = 100').run();
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// 模拟 main.js openDb 里的启动清顶
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db.prepare('UPDATE clips SET top_at = NULL WHERE top_at IS NOT NULL').run();
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const r = db.prepare(TRIM_OLDEST).run(3);
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assert.strictEqual(r.changes, 3, '清顶后所有行都回到自然位置，可被删');
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db.close();
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});
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check('moveToTop 同时更新 top_at 和 created_at（让点过的项保持「最近活跃」）', () => {
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// 用户点过复制的项要排到顶 —— 单纯更新 top_at 不够，created_at 也得一起
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// bump 到 now，否则列表排序（ORDER BY created_at DESC）看不到这次活跃。
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assert.match(MOVE_TO_TOP, /UPDATE\s+clips\s+SET\s+top_at\s*=\s*\?\s*,\s*created_at\s*=\s*\?\s+WHERE\s+id\s*=\s*\?/,
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'moveToTop 必须同时更新 top_at 和 created_at');
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});
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check('moveToTop 把 created_at bump 到 now —— 已点项重新活跃，排到顶', () => {
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const db = makeDb();
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seedNatural(db, 3); // created_at 1..3
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const move = db.prepare(MOVE_TO_TOP);
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// 点 id=1 的「复制」：把它的 top_at / created_at 都置到 1000
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move.run(1000, 1000, 1);
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// 用新排序取 → 刚被点过的 id=1 必须在最顶
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const rows = db.prepare(`SELECT id FROM clips ORDER BY created_at DESC, id DESC`).all();
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assert.deepStrictEqual(rows.map(r => r.id), [1, 3, 2],
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'点过的项（created_at=1000）必须排在原始 created_at=3、2 的前面');
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// 而且 top_at 也被置上了，trim / 清空历史 仍会跳过它
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const r1 = db.prepare('SELECT top_at, created_at FROM clips WHERE id = 1').get();
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assert.strictEqual(r1.top_at, 1000);
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assert.strictEqual(r1.created_at, 1000);
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db.close();
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});
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console.log(`\n${passed} checks passed`);
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