使用 sinon 存根 redis 发布方法?
stubbing redis publish method using sinon?
如何存根 redis 发布方法?
// module ipc
const redis = require('redis');
module.exports = class IPC {
constructor() {
this.pub = redis.createClient();
}
publish(data) {
this.pub.publish('hello', JSON.stringify(data));
}
}
和另一个模块
// module service
module.exports = class Service {
constructor(ipc) {
this.ipc = ipc;
}
sendData() {
this.ipc.publish({ data: 'hello' })
}
}
如何在 IPC
class 中存入私有变量 pub
?
我可以使用 proxyquire
存根 redis.createClient
,如果我这样做,它会抱怨 publish
undefined
我目前的测试代码
let ipcStub;
before(() => {
ipcStub = proxyquire('../ipc', {
redis: {
createClient: sinon.stub(redis, 'createClient'),
}
})
});
it('should return true', () => {
const ipc = new ipcStub();
const ipcPublishSpy = sinon.spy(ipc, 'publish')
const service = new Service(ipc);
service.sendData();
assert.strictEqual(true, ipcPublishSpy.calledOnce);
})
你只需要在 publish 方法上设置间谍,不需要 proxyquire
.
例如
import {expect} from 'chai';
import sinon from 'sinon';
class IPC {
constructor() {
this.pub = {
publish:() => {} //here your redis requirement
};
}
publish(data) {
this.pub.publish('hello', JSON.stringify(data));
}
}
class Service {
constructor(ipc) {
this.ipc = ipc;
}
sendData() {
this.ipc.publish({ data: 'hello' })
}
}
describe('Test Service', () => {
it('should call publish ', () => {
const ipc = new IPC;
sinon.spy(ipc.pub,'publish');
const service = new Service(ipc);
service.sendData();
expect(ipc.pub.publish.calledOnce).to.be.true;
});
});
我找到了一种方法,只需使用 sinon
只需要使用sinon.createStubInstance
创建存根实例,
那么这个存根将具有 sinon
中的所有功能,而无需实现对象(仅 class 方法名称)
let ipcStub;
before(() => {
ipcStub = sinon.createStubInstance(IPC)
});
it('should return true', () => {
const ipc = new ipcStub();
const service = new Service(ipc);
service.sendData();
assert.strictEqual(true, ipc.publishSystem.calledOnce);
})
如何存根 redis 发布方法?
// module ipc
const redis = require('redis');
module.exports = class IPC {
constructor() {
this.pub = redis.createClient();
}
publish(data) {
this.pub.publish('hello', JSON.stringify(data));
}
}
和另一个模块
// module service
module.exports = class Service {
constructor(ipc) {
this.ipc = ipc;
}
sendData() {
this.ipc.publish({ data: 'hello' })
}
}
如何在 IPC
class 中存入私有变量 pub
?
我可以使用 proxyquire
存根 redis.createClient
,如果我这样做,它会抱怨 publish
undefined
我目前的测试代码
let ipcStub;
before(() => {
ipcStub = proxyquire('../ipc', {
redis: {
createClient: sinon.stub(redis, 'createClient'),
}
})
});
it('should return true', () => {
const ipc = new ipcStub();
const ipcPublishSpy = sinon.spy(ipc, 'publish')
const service = new Service(ipc);
service.sendData();
assert.strictEqual(true, ipcPublishSpy.calledOnce);
})
你只需要在 publish 方法上设置间谍,不需要 proxyquire
.
例如
import {expect} from 'chai';
import sinon from 'sinon';
class IPC {
constructor() {
this.pub = {
publish:() => {} //here your redis requirement
};
}
publish(data) {
this.pub.publish('hello', JSON.stringify(data));
}
}
class Service {
constructor(ipc) {
this.ipc = ipc;
}
sendData() {
this.ipc.publish({ data: 'hello' })
}
}
describe('Test Service', () => {
it('should call publish ', () => {
const ipc = new IPC;
sinon.spy(ipc.pub,'publish');
const service = new Service(ipc);
service.sendData();
expect(ipc.pub.publish.calledOnce).to.be.true;
});
});
我找到了一种方法,只需使用 sinon
只需要使用sinon.createStubInstance
创建存根实例,
那么这个存根将具有 sinon
中的所有功能,而无需实现对象(仅 class 方法名称)
let ipcStub;
before(() => {
ipcStub = sinon.createStubInstance(IPC)
});
it('should return true', () => {
const ipc = new ipcStub();
const service = new Service(ipc);
service.sendData();
assert.strictEqual(true, ipc.publishSystem.calledOnce);
})