具有 sinon 和 proxyquire 的存根相关函数不打印存根结果

Question

我为 _getuser() 编写了单元测试，但我没有看到 console.log 打印存根 result.Also 测试覆盖率显示行 'let user = result.user; console.log('User'+JSON.stringify(结果));' uncovered.Why 在 LogInCommand class 中的 getUser() 函数的控制台日志中未打印存根结果。我在单元测试中看到结果显示未定义。

// userApi.js

    'use strict';
    const config = require('../../config/config');
    const api = require('./apiService');

    class UserApi {
        constructor() {
        }
        getUser(userId) {
            return api.get({
                url: config.url,
                qs: {
                    includeInactive: true,
                    id: userId,
                    startIndex: 0,
                    maxResults: 1
                },
                headers: {
                    Accept: 'application/json;',
                    'Connection': 'Keep-Alive'
                }
            });
        }

    }
    module.exports = UserApi;

    // LoginCommand.js

    'use restrict';

    const userApi = require('../../api/userApi');
    class LogInCommand {
        constructor() {
        }
        async _getUser(userId) {
            let result = await new userApi().getUser(userId);
            let user = result.user;
            console.log('User'+JSON.stringify(result));
            return user;
        }

    }

    module.exports = LogInCommand;

    //LoginCommand.test.js
    describe('Test LogInCommand Class',function(){
      it.only('_getUser function should return user',    async ()=> { 
            let apiData= {
                user:'abc'
            };
          let logincmd = proxyquire('../LogInCommand.js', {
                userApi : { getUser : Promise.resolve(apiData) },'@noCallThru':true});
            let obj= new logincmd();
            let result= await obj._getUser(client);
        });

    });

Answer 1

您当前设置中的 proxyquire 配置不正确。 Proxyquire 将 require 调用中传递的字符串值映射到所需的 mock/stub 值。请尝试以下操作：

let logincmd = proxyquire('../LogInCommand.js', {
  '../../api/userApi' : { getUser : Promise.resolve(apiData) },
  '@noCallThru':true
});

Answer 2

下面的代码对我有用 // userApi.js

'use strict';
const config = require('../../config/config');
const api = require('./apiService');

class UserApi {
    constructor() {
    }
    getUser(userId) {
        return api.get({
            url: config.url,
            qs: {
                includeInactive: true,
                id: userId,
                startIndex: 0,
                maxResults: 1
            },
            headers: {
                Accept: 'application/json;',
                'Connection': 'Keep-Alive'
            }
        });
    }

}
module.exports = UserApi;

// LoginCommand.js

'use restrict';

const userApi = require('../../api/userApi');
class LogInCommand {
    constructor() {
    }
    async _getUser(userId) {
        let result = await new userApi().getUser(userId);
        let user = result.user;
        console.log('User'+JSON.stringify(result));
        return user;
    }

}

module.exports = LogInCommand;

//LoginCommand.test.js
describe('Test LogInCommand Class',function(){
  it.only('_getUser function should return user',    async ()=> { 
        class userApiStub{
        constructor(){}
        getUser() {
            return  Promise.resolve({
                user:4
            });
        }
    }
      let logincmd = proxyquire('../LogInCommand.js', {
            '../../api/userApi' :userApiStub },'@noCallThru':true});
        let obj= new logincmd();
        let result= await obj._getUser(client);
    });

});

具有 sinon 和 proxyquire 的存根相关函数不打印存根结果

Stub dependent function with sinon and proxyquire does not print stub result

javascript

unit-testing

mocha.js

node.js

sinon-chai