我的代码是:
app.get('/main', function(req, res) {
Posts.findAll().then(function(posts){
res.render(__dirname + "/home.pug", {posts:posts});
})
div
each val in posts
li= val
对不起我的英语，如果问题令人困惑
最佳答案
因为您正在尝试渲染 Posts.findAll() 方法返回的 sequelize 的 model instances。
您可能想要呈现 post 的属性，而不是整个模型实例对象。
例如。app.ts :
import express from 'express';
import path from 'path';
import { DataTypes, Model } from 'sequelize';
import { sequelize } from '../../db';
class Posts extends Model {}
Posts.init(
{
name: DataTypes.STRING,
},
{ sequelize, tableName: 'posts' },
);
const app = express();
app.set('view engine', 'pug');
app.set('views', path.resolve(__dirname, './views'));
app.get('/', (req, res) => {
Posts.findAll().then(function(posts) {
res.render('home.pug', { posts: posts });
});
});
(async function main() {
await sequelize.sync({ force: true });
//seed
await Posts.bulkCreate([{ name: 'teresa' }, { name: 'teng' }]);
app.listen(3000, () => console.log('Server started at http://localhost:3000'));
})();
views/home.pug :<!DOCTYPE html>
html(lang="en")
head
meta(charset="UTF-8")
meta(name="viewport", content="width=device-width, initial-scale=1.0")
title Document
body
div
each val in posts
li= val.name
源代码:https://github.com/mrdulin/node-sequelize-examples/tree/master/src/examples/stackoverflow/65376890
关于mysql - findAll() 返回 [对象 SequelizeInstance :table name] (mysql)，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/65376890/