我的返回结果并没有让它一直回到API端点。 你看到我做错了吗?
app.js
const express = require('express');
const app = express();
app.use(express.static('client'));
var GetContracts = require('./contractsService');
app.get('/contracts', async (req, res) => {
var results = await GetContracts.get();
console.log(results);
res.send(results);
});
module.exports = app;
contractsService.js
var mysql = require('mysql');
const config = require('./config')
var con = mysql.createConnection({
host: config.HOST,
user: config.USER,
password: config.PASSWORD,
database: config.DATABASE
});
exports.get = function () {
con.connect(function (err) {
if (err) {
throw new Error('Error by Rodney')
};
con.query("SELECT * FROM " + config.DATABASE + ".Contracts", function (err, result, fields) {
if (err) {
throw new Error('Error by Rodney')
};
return result;
//console.log(result); //works
});
});
}
答案 0 :(得分:0)
query
方法接受不受返回值影响的错误优先回调。 GetContracts.get
不会返回承诺,await
将不会做任何事情。
为了在承诺控制流程中使用它应该被宣传:
exports.get = function () {
return new Promise((resolve, reject) => {
con.connect(function (err) {
if (err) {
reject(new Error('Error by Rodney'))
};
con.query("SELECT * FROM " + config.DATABASE + ".Contracts", function (err, result, fields) {
if (err) {
reject(new Error('Error by Rodney'));
} else
resolve(result);
});
});
});
}
或者最好使用现有的基于promise的MySQL库,如promise-mysql
,类似于:
var mysql = require('promise-mysql');
const conPromise = mysql.createConnection({
host: config.HOST,
user: config.USER,
password: config.PASSWORD,
database: config.DATABASE
});
exports.get = async () => {
const con = await conPromise;
const result = await con.query("SELECT * FROM " + config.DATABASE + ".Contracts");
return result;
};