【发布时间】:2018-11-27 05:16:06
【问题描述】:
我在运行单元测试时出错。我正在使用 NodeJS 的快速应用程序。我想测试返回消息(200)和(最终)函数的输出。
const {Router} = require('express');
const {query} = require('../dbConfig');
module.exports = (router = new Router()) => {
//get only one user from the id
router.post('/getUser', async (req,res)=>{
id = req.body.id;
sql = "SELECT lastname, firstname, email, title_id, id, state FROM users WHERE id=" + id;
let result = await query(sql);
if(result.length >= 1){
res.send(result[0]);
}
})
return router;
};
但是当我尝试运行以下测试时:
const app = require('../../app')
const request = require('supertest')
jest.mock('../../dbConfig');;
describe('POST /getUser', () => {
let data = {
id:1
}
it('Response 200 + response type',(done)=>{
request.agent(app)
.post('/getUser')
.send(data)
.expect(200, done)
})
});
它返回以下错误
Timeout - Async callback was not invoked within the 5000ms timeout specified by jest.setTimeout.
我尝试了不同的变体但没有成功...我还尝试了在线文档,但对这个错误并没有真正的帮助...我用npx jest 开玩笑,模拟文件看起来像这样:
const rep={
"lastname": "admin",
"firstname": "admin",
"email": "admin@etsmtl.ca",
"title_id": 1,
"id": 1,
"state": 1
};
const query = (sqli) =>{
return new Promise((resolve, reject) =>{
console.log("SQL request : " + sqli)
if(sqli === "SELECT lastname, firstname, email, title_id, id, state FROM users WHERE id=1"){
console.log("Passed")
resolve(rep);
}
});
}
module.exports = {query};
【问题讨论】:
标签: node.js unit-testing express jestjs supertest