Adr*_*ien 1 javascript unit-testing mocking jestjs
我尝试覆盖以下代码：
\n\n// @flow strict\n\nimport { bind, randomNumber } from 'Utils'\nimport { AbstractOperator } from './AbstractOperator'\n\nexport class Randomize extends AbstractOperator {\n // ...\n\n randomPick (dataset: Array<string>, weights: ?Array<number>): number {\n if (!weights) { return randomNumber(0, (dataset.length - 1)) }\n\n const sumOfWeights: number = weights.reduce((a, b) => a + b)\n let randomWeight = randomNumber(1, sumOfWeights)\n let position: number = -1\n\n for (let i = 0; i < dataset.length; i++) {\n randomWeight = randomWeight - weights[i]\n if (randomWeight <= 0) {\n position = i\n break\n }\n }\n\n return position\n }\n}\n这是测试覆盖率：
\n\nimport { Randomize } from './Randomize'\n\nconst dataset = [\n 'nok',\n 'nok',\n 'nok',\n 'ok',\n 'nok'\n]\n\nconst weights = [\n 0,\n 0,\n 0,\n 1,\n 0\n]\n\nconst randomNumber = jest.fn()\n\ndescribe('operator Randomize#randomPick', () => {\n test('without weights, it calls `randomNumber`', () => {\n const randomizeOperator = new Randomize({}, [dataset], {})\n randomizeOperator.randomPick(dataset)\n\n expect(randomNumber).toBeCalledWith(0, dataset.length - 1)\n })\n})\n我试图确保它randomNumber被调用，但我得到的只是：
\xe2\x97\x8f operator Randomize#randomPick \xe2\x80\xba without weights, it calls `randomNumber`\n\n expect(jest.fn()).toBeCalledWith(...expected)\n\n Expected: 0, 4\n\n Number of calls: 0\n\n 33 | randomizeOperator.randomPick(dataset)\n 34 |\n > 35 | expect(randomNumber).toBeCalledWith(0, dataset.length - 1)\n | ^\n 36 | })\n 37 | })\n 38 |\n\n at Object.toBeCalledWith (node_modules/jest-chain/dist/chain.js:15:11)\n at Object.toBeCalledWith (src/app/Services/Providers/Result/Resolvers/Operators/Randomize.test.js:35:26)\n\n我的两分钱是，模拟randomNumber依赖关系并不是测试此功能的正确方法。
不过，我将在这里回答主要问题，看看我们如何才能通过测试。然后将在未来的更新中得到我关于更好的测试方法的额外想法。
\n\nrandomNumber调用代码的实际问题是randomNumber模拟函数悬而未决。正如错误所示，它没有被调用。
缺少的部分是拦截模块导入并使其外部调用能够Utils.randomNumber触发模拟函数；这样我们就可以反对它。以下是如何拦截Utils导入并模拟它：
// Signature is: \n// jest.mock(pathToModule: string, mockModuleFactory: Function)\njest.mock(\'Utils\', () => ({\n randomNumber: jest.fn()\n}))\n现在，测试期间的每次调用Utils.randomNumber都会触发模拟函数，并且它不再悬而未决。
如果您想了解它在幕后是如何工作的，请研究一下如何在babel-plugin-jest-hoist编译为 CommonJS Jest 劫持调用的jest.mock之上进行提升调用。importrequire
根据具体情况，模拟整个模块可能会出现问题。如果测试依赖于Utils模块的其他导出怎么办？例如bind？
有一些方法可以部分模拟一个模块，只是一个函数，一个或两个类。然而，为了让您的测试通过，还有一种更简单的方法。
\n\n解决方案就是简单地监听通话randomNumber。这是一个完整的例子：
import { Randomize } from \'./Randomize\'\nimport * as Utils from \'Utils\'\n\n// Sidenote: This values should probably be moved to a beforeEach()\n// hook. The module-level assignment does not happen before each test.\nconst weights = [0, 0, 0, 1, 0]\nconst dataset = [\'nok\', \'nok\', \'nok\', \'ok\', \'nok\']\n\ndescribe(\'operator Randomize#randomPick\', () => {\n test(\'without weights, it calls `randomNumber`\', () => {\n const randomizeOperator = new Randomize({}, [dataset], {})\n const randomNumberSpy = jest.spyOn(Utils, \'randomNumber\')\n\n randomizeOperator.randomPick(dataset)\n\n expect(randomNumberSpy).toBeCalledWith(0, dataset.length - 1)\n })\n})\n希望这是一项通过测试，但非常脆弱。
\n\n总结一下，以下是关于笑话背景下该主题的非常好的读物：
\n\n\n\n主要是因为测试与代码紧密耦合。如果您比较测试和 SUT，就会发现重复的代码。
\n\n更好的方法是根本不模拟/监视任何东西（研究Classist 与 Mockist TDD学校）并使用一组动态生成的数据和权重来练习 SUT，这反过来又断言它“足够好” 。
\n\n我将在更新中详细说明这一点。
\n\nrandomPick出于另一个原因，测试其实现细节也不是一个好主意。此类测试无法验证算法的正确性，因为它仅验证其发出的调用。如果存在边缘情况错误，则它的覆盖范围不足以解决该错误。
当我们想要断言对象的通信时，模拟/间谍通常是有益的；在通信实际上断言正确性的情况下，例如“断言它到达了数据库”；但这里不是。
\n\n更好的测试用例的一个想法可能是“大力”运用 SUT 并断言它对于它正在做的事情来说“足够好”；选择一个随机元素。这个想法得到了大数定律的支持：
\n\n\n\n\n“在概率论中，大数定律（LLN）是描述多次进行相同实验的结果的定理。根据该定律，从大量试验中获得的结果的平均值应为接近预期值，并且随着进行更多试验，将趋于更接近预期值。” \xe2\x80\x94维基百科
\n
为 SUT 提供相对较大的、动态生成的随机输入集，并断言它每次都会通过：
\n\nimport { Randomize } from \'./Randomize\'\n\nconst exercise = (() => {\n // Dynamically generate a relatively large random set of input & expectations:\n // [ datasetArray, probabilityWeightsArray, expectedPositionsArray ]\n //\n // A sample manual set: \n return [\n [[\'nok\', \'nok\', \'nok\', \'ok\', \'nok\'], [0, 0, 0, 1, 0], [3]],\n [[\'ok\', \'ok\', \'nok\', \'ok\', \'nok\'], [50, 50, 0, 0, 0], [0, 1]],\n [[\'nok\', \'nok\', \'nok\', \'ok\', \'ok\'], [0, 0, 10, 60, 30], [2, 3, 4]]\n ]\n})()\n\ndescribe(\'whatever\', () => {\n test.each(exercise)(\'look into positional each() params for unique names\', (dataset, weights, expected) => {\n const randomizeOperator = new Randomize({}, [dataset, weights], {})\n\n const position = randomizeOperator.randomPick(dataset, weights)\n\n expect(position).toBeOneOf(expected)\n })\n})\n这是基于相同想法的另一个观点，不一定需要生成动态数据：
\n\nimport { Randomize } from \'./Randomize\'\n\nconst exercise = (() => {\n return [\n [\n [\'moreok\'], // expect "moreok" to have been picked more during the exercise.\n [\'lessok\', \'moreok\'], // the dataset.\n [0.1, 99.90] // weights, preferring the second element over the first.\n ],\n [[\'moreok\'], [\'moreok\', \'lessok\'], [99, 1]],\n [[\'moreok\'], [\'lessok\', \'moreok\'], [1, 99]],\n [[\'e\'], [\'a\', \'b\', \'c\', \'d\', \'e\'], [0, 10, 10, 0, 80]],\n [[\'d\'], [\'a\', \'b\', \'c\', \'d\'], [5, 20, 0, 75]],\n [[\'d\'], [\'a\', \'d\', \'c\', \'b\'], [5, 75, 0, 20]],\n [[\'b\'], [\'a\', \'b\', \'c\', \'d\'], [0, 80, 0, 20]],\n [[\'a\', \'b\'], [\'a\', \'b\', \'c\', \'d\'], [50, 50]],\n [[\'b\'], [\'a\', \'b\', \'c\'], [10, 60, 30]],\n [[\'b\'], [\'a\', \'b\', \'c\'], [0.1, 0.6, 0.3]] // This one pinpoints a bug.\n ]\n})()\n\nconst mostPicked = results => {\n return Object.keys(results).reduce((a, b) => results[a] > results[b] ? a : b )\n}\n\ndescribe(\'randompick\', () => {\n test.each(exercise)(\'picks the most probable: %p from %p with weights: %p\', (mostProbables, dataset, weights) => {\n const operator = new Randomize({}, [dataset, weights], {})\n const results = dataset.reduce((carry, el) => Object.assign(carry, { [el]: 0 }), {})\n // e.g. { lessok: 0, moreok: 0 }\n\n for (let i = 0; i <= 2000; i++) {\n // count how many times a dataset element has win the lottery!\n results[dataset[operator.randomPick(dataset, weights)]]++\n }\n\n // console.debug(results, mostPicked(results))\n\n expect(mostPicked(results)).toBeOneOf(mostProbables)\n })\n})\n当测试被如上所述的“功能噪音”污染时，它们就会变得难以阅读；它们不再充当文档。
\n\n在这种情况下，开发自定义匹配器或测试替身有助于提高可读性。
\n\ntest.each([\n // ...\n])(\'picks the most probable: %p from %p with weights: %p\', mostProbables, dataset, weights) => {\n const results = []\n const operator = new Randomize(...whatever)\n\n ;[...Array(420).keys()].forEach(() => results.push(\n operator.randomPick(...whatever)\n )\n\n expect(results).toHaveMostFrequentElements(mostProbables)\n}\n这个自定义toHaveMostFrequentElements断言匹配器有助于消除测试中的“噪音”。
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