【发布时间】:2018-06-11 22:03:33
【问题描述】:
database name is test1 and collection name is feedbacks
this is app.js
我无法在我的 html 表单字段中获取 mongodb 数据正在插入数据,但我想在我的表单中获取初始值帮助我解决这个问题。因为我是新手,所以我想有一个简单的显式解决方案。
var express = require('express');
var path = require('path');
var bodyParser = require('body-parser');
var mongodb = require('mongodb');
var dbConn = mongodb.MongoClient.connect('mongodb://localhost:27017/test1');
var app = express();
app.use(bodyParser.urlencoded({ extended: false }));
app.use(express.static(path.resolve(__dirname, 'public')));
app.get('/retrieve', function(req, res){
post.find({}, function(err, feedbacks){
if(err) res.json(err);
else res.render('retrieve', {posts: feedbacks});
});
});
app.post('/post-feedback', function (req, res) {
dbConn.then(function(db) {
delete req.body._id; // for safety reasons
var dd = db.db("test1");
dd.collection('feedbacks').insertOne(req.body);
});
res.send('Data received:\n' + JSON.stringify(req.body));
});
app.get('/view-feedbacks', function(req, res) {
dbConn.then(function(db) {
var dd=db.db("test1");
dd.collection('feedbacks').find({"Name":"shubham"}).toArray().then(function(feedbacks) {
res.status(200).json(feedbacks);
});
});
});
app.get('/view-sahil', function(req, res) {
dbConn.then(function(db) {
var dd=db.db("test1");
dd.collection('feedbacks').insertOne({"name":"sahil","E-mail":"sahil@vibhuti.guru","comment":"jhdsfji"},function(feedbacks) {
res.status(200).json("Data inserted");
});
});
});
app.listen(process.env.PORT || 3000, process.env.IP || '0.0.0.0' );
这是我的index.html
<!doctype html>
<html lang="en">
<head>
<script src="..app.js"></script>
<meta charset="UTF-8">
<title>mongodb</title>
</head>
<body>
<h1>Please fill data in the form below:</h1>
<form method="POST" action="/">
<label>Name:<input type="text" name="" value= feedbacks.name required></label>
<br>
<label>Email:<input type="text" name="Email" value="" required></label>
<br>
<label>Comment:<br><textarea name="comment"></textarea></label>
<br>
<input type="submit" value="Submit">
</form>
<a href="/view-feedbacks">View Records</a>
<a href="/view-sahil">insert dummy record </a>
</body>
</html>
【问题讨论】:
-
这是错误的方式
<a href="/view-feedbacks">View Records</a>这里<a href="/view-feedbacks">View Records</a>是RESTapi 所以在AJAX的帮助下调用这个api -
@ManjeetThakur 先生,您能详细说明一下这个答案吗?我是新手,对此我没有太多了解。如果可能的话,请给我写完整的代码,我将不胜感激
-
你使用的是哪个框架?
-
@manjeet thakur nodejs 和 angularjs
标签: javascript html node.js mongodb