Jest.js 错误:“已接收:序列化为相同的字符串”
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【中文标题】Jest.js 错误:“已接收:序列化为相同的字符串”【英文标题】:Jest.js error: "Received: serializes to the same string" 【发布时间】:2019-11-12 08:29:53 【问题描述】:我在这个测试中遇到了一个奇怪的问题:
deal.test.js
import Deal from "../src/models/Deal";
import apiProducts from "../__mocks__/api/products";
describe("Deal", () =>
describe("Deal.fromApi", () =>
it("takes an api product and returns a Deal", () =>
const apiDeal = apiProducts[0];
const newDeal = Deal.fromApi(apiDeal);
const expected = expectedDeal();
expect(newDeal).toEqual(expected);
);
);
);
Deal.js
export default class Deal
// no constructor since we only ever create a deal from Deal.fromApi
static fromApi(obj: Object): Deal
const deal = new Deal();
deal.id = obj.id;
deal.name = obj.name;
deal.slug = obj.slug;
deal.permalink = obj.permalink;
deal.dateCreated = obj.date_created;
deal.dateModified = obj.date_modified;
deal.status = obj.status;
deal.featured = obj.featured;
deal.catalogVisibility = obj.catalog_visibility;
deal.descriptionhtml = obj.description;
deal.shortDescriptionHTML = obj.short_description;
deal.price = Number(obj.price);
deal.regularPrice = Number(obj.regular_price);
deal.salePrice = Number(obj.sale_price);
deal.dateOnSaleFrom = obj.date_on_sale_from;
deal.dateOnSaleTo = obj.date_on_sale_to;
deal.onSale = obj.on_sale;
deal.purchasable = obj.purchasable;
deal.relatedIds = obj.related_ids;
deal.upsellIds = obj.upsell_ids;
deal.crossSellIds = obj.cross_sell_ids;
deal.categories = obj.categories;
deal.tags = obj.tags;
deal.images = obj.images;
return deal;
descriptionWithTextSize(size: number): string
return this.descriptionWithStyle(`font-size:$size`);
descriptionWithStyle(style: string): string
return `<div style="$style">$this.description</div>`;
distanceFromLocation = (
location: Location,
unit: unitOfDistance = "mi"
): number =>
return distanceBetween(this.location, location);
;
distanceFrom = (otherDeal: Deal, unit: unitOfDistance = "mi"): number =>
return distanceBetween(this.location, otherDeal.location);
;
static toApi(deal: Deal): Object
return ...deal ;
测试失败并出现以下错误:
● Deal › Deal.fromApi › takes an api product and returns a Deal
expect(received).toEqual(expected) // deep equality
Expected: "catalogVisibility": "visible", "categories": ["id": 15, "name": "New York", "slug": "new-york"], "crossSellIds": [34, 31], "dateCreated": "2019-05-18T17:36:14", "dateModified": "2019-05-18T17:39:02", "dateOnSaleFrom": null, "dateOnSaleTo": null, "descriptionHTML": "<p>Pete's Tavern<br />
129 E 18th St<br />
New York, NY 10003</p>
<p>Weekdays from 4 p.m. to 7 p.m.<br />
$5 wines and beers</p>
", "distanceFromLocation": [Function anonymous], "featured": false, "id": 566, "images": ["alt": "", "date_created": "2019-05-18T17:38:52", "date_created_gmt": "2019-05-18T17:38:52", "date_modified": "2019-05-18T17:38:52", "date_modified_gmt": "2019-05-18T17:38:52", "id": 567, "name": "wine and beers2", "src": "https://tragodeals.com/wp-content/uploads/2019/05/wine-and-beers2.jpg"], "name": "Wines and beers", "onSale": true, "permalink": "https://tragodeals.com/product/wines-and-beers/", "price": 5, "purchasable": true, "regularPrice": 11, "relatedIds": [552, 564, 390, 37, 543], "salePrice": 5, "shortDescriptionHTML": "<p>$5 wines and beers</p>
", "slug": "wines-and-beers", "status": "publish", "tags": ["id": 58, "name": "beers", "slug": "beers", "id": 54, "name": "Cocktails", "slug": "cocktails", "id": 45, "name": "drink", "slug": "drink", "id": 57, "name": "wine", "slug": "wine"], "upsellIds": [53]
Received: serializes to the same string
> 15 | expect(newDeal).toEqual(expected);
| ^
16 | );
17 | );
18 | );
at Object.toEqual (__tests__/deal.test.js:15:23)
我插入了这个循环来调查:
for (let key in expected)
expect(expected[key]).toEqual(newDeal[key]);
我发现问题出在函数上。所以我把整个测试改成这样:
for (let key in expected)
if (typeof expected[key] === "function") continue;
expect(expected[key]).toEqual(newDeal[key]);
// expect(newDeal).toEqual(expected);
它通过了,并且在它应该失败的时候也失败了。 (如果您阅读了这个问题的旧版本,我在其中通过了我不理解的测试,那是因为我在循环中是 return
ing 而我应该是 continue
ing)。
但我希望能够使用标准断言expect(newDeal).toEqual(expected)
来做到这一点。关于检查类对象 (Deal
) 是否与函数相等,我似乎有些不理解。
PS。您可能会建议使用toMatchObject
。但是,很遗憾:
● Deal › Deal.fromApi › takes an api product and returns a Deal
expect(received).toMatchObject(expected)
- Expected
+ Received
@@ -1,6 +1,6 @@
- Deal
+ Object
"address": "129 E 18th St New York, NY 10003",
"catalogVisibility": "visible",
"categories": Array [
Object
"id": 15,
13 | expect(expected[key]).toEqual(newDeal[key]);
14 |
> 15 | expect(newDeal).toMatchObject(expected);
| ^
16 | );
17 | );
18 | );
【问题讨论】:
嗨,乔纳森,您是否可以传递一个 apiProducts 样本来重现此错误?您能否提供您在控制台中遇到的确切错误? 我注意到你的构造函数 Object.assign(this, obj: Object) 【参考方案1】:与其他同事类似,我在数组比较中遇到了这个问题,我基本上是在测试一个函数,它可以获取数组中最大的字符串,此外,如果超过 1 个字符串与可能的最大长度匹配,它应该返回一个数组。
当我开始测试时,我收到以下消息:
所以我替换了toBe
方法
expect(function(array1)).toBe('one result')
与toStrictEqual
进行深度相等比较
expect(function(array2)).toStrictEqual(['more than one', 'more than one']);
【讨论】:
【参考方案2】:当尝试比较一个数组中存在一个索引为 -1 的元素的数组时遇到了这个问题(想象一下要设置的任何其他键,除了从 0 到 N 的数字)。 所以你可能会在以下场景中出现这个错误:
const arr = [1, 2]
arr[-1] = 'foo'
expect(arr).toEqual([1, 2])
它们都序列化为相同的字符串,但它们不相等。
【讨论】:
我的问题是我们在数组上放置了一个静态属性,类似于这个【参考方案3】:我有一个类似的情况,对象有一个 base64 编码的字符串,我管理测试以使用 JSON.stringify
比较对象的序列化:
expect(JSON.stringify(newDeal)).toMatchObject(JSON.stringify(expected));
【讨论】:
expect(newDeal).toMatchObject(expected)
可以。无需转换为字符串。
@AVC 你确定这是正确的吗?开玩笑说这是关于expect(newDeal).toMatchObject(expected)
:It will match received objects with properties that are not in the expected object.
。当我尝试它时,它对我不起作用。为了使toEqual()
比较成功,我必须在等式两边都做JSON.stringify()
。【参考方案4】:
在我的情况下,我在检查代理对象和常规对象时非常平等。
const proxyObject = new Proxy(,
get: function(target, prop)
return Object.hasOwnProperty.call(target, prop) ? target[prop] : '';
)
expect(proxyObject).toEqual()
// Expected:
// Received: serializes to the same string
应该是:
expect(proxyObject).toMatchObject()
【讨论】:
【参考方案5】:当我尝试比较一个数组从 Graphql 的解析器返回而另一个数组来自我的测试输入的数组时,我遇到了这个问题。
所有字段都是相同的,除了来自 Graphql 的数组中的条目没有任何 __proto__
而来自我的测试输入的条目有 __proto__: Object
并且导致 toStrictEqual 失败,因为它检查结构以外的类型。所以,在我的情况下,类型导致失败。将其更改为 toEqual
即可解决问题。
expect(res.data.myMutation.values).toEqual(
testInput.values,
);
【讨论】:
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