在proxyquired对象中存根函数

mon*_*ist 10 unit-testing stubs sinon proxyquire

我想对以下简化模块进行单元测试:

const Logger = require('logplease');
const logger = Logger.create('utils');

const tester = {

    one: () => {
        logger.log('called real one()');
        tester.two();
    },
    two: () => {
        logger.log('called real two()');
    },
};

module.exports = {
    one: tester.one,
    two: tester.two
};
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我正在logplease使用Proxyquire替换外部依赖,这非常有效.但是我需要存根,two()因为我想进行单元测试,one()同时消除two()在实际代码中运行时产生的副作用.

it.only('stubbing functions on the "proxyquired" object under test', function(done) {

    const loggerStub = {
        create: () => {
            return { log: (msg) => { console.log('fake logger: ', msg); } };
        }
    };

    let tester = proxyquire('../tester', { 'logplease': loggerStub });

    let stub2 = sinon.stub(
        tester, 
        'two', 
        () => { 
            console.log('called fake stub of two()'); 
        }
    );

    tester.one();

    console.log('call count 2: ', stub2.callCount);
    done();
});
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输出我得到:

fake logger:  called real one() 
fake logger:  called real two() 
call count 2: 0
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输出我期望:

fake logger:  called real one() 
called fake stub of two()
call count 2: 1
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为什么我的存根函数不运行?

Naz*_*nko 1

简短回答:

const Logger = require('logplease');
const logger = Logger.create('utils');

const tester = {

    one: () => {
        logger.log('called real one()');
        tester.two();
    },
    two: () => {
        logger.log('called real two()');
    },
};

module.exports = tester;
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说明:范围

您将一和二导出为:

module.exports = {
    one: tester.one,
    two: tester.two
};
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在这种情况下tester.one只知道这个函数:

two: () => {
    logger.log('called real two()');
}
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并且不知道存根两个。所以你有两个版本的Two,只需尝试在 test 中调用tester.two()