javascript - 如何通过调用 ajax 将 html 表单数据发送到 node.js 服务器？

Question

我想使用 Ajax 将表单数据发送到 node.js 服务器，我正在使用以下方法发送数据。

我不知道如何在 node.js 服务器程序中接收它我没有使用 node.js 的快速框架

客户端.HTML

<script>
 function myFunction() {
   var region = document.getElementById("region").value;
   var os = document.getElementById("os").value;
   var data = {};
   data.region = region;
   data.os = os;
   $.ajax({
     type: 'post',
     datatype: 'jsonp',
     data: JSON.stringify(data),
     contentType: 'application/json',
     url: 'http://127.0.0.1:8083/', //node.js server is running
     success: function(data) {
       alert("success");

     }
   });

</script>
<form>
<select id="region" name="region" class="region"></select>
<select id="os" name="os" class="os"></select>  
<input type="button" value="search" class="fil_search" onclick="myFunction()"/>
</form>

服务器.js

var http = require('http');
var fs = require('fs');
var url = require('url');
var MongoClient = require('mongodb').MongoClient;
var assert = require('assert')
var ObjectId = require('mongodb').ObjectID;
var express = require('express');
var bodyParser = require('body-parser');
var app     = express();
var result1=[];
var result2=[];
var result3=[];
var result4=[];
var result5=[];
var result6=[];
var result7=[];
var i=0;
var region;
var os;
app.use(bodyParser.urlencoded({ extended: true })); 

MongoClient.connect("mongodb://192.168.1.22:27017/test", function(err, db) {

if(err) { return console.dir(err); }
else {
app.get('/',function(req, res) {

var url_parts = url.parse(req.url, true);
var url_parsed = url.parse(req.url, true);

"i want data here from ajax to perform"

console.log("connected");


var instance = db.collection('instance');
var region = db.collection('region');

region.findOne(({$and: [{"region_name": region_name},{"os_type": os}]}), function(err, result){
if(err){    
throw(err);
}
else{
console.log(region);
var newr = result.inst_name;

instance.find({ "inst_id": { "$in": newr } }).toArray(function(err, resultn){
if(err){
throw(err);
}
else{       
    var len=resultn.length;

    console.log(resultn);
    console.log(len);

    for(var i=0;i<len;i++)
    {
        result1[i]=resultn[i].inst_type;
        result2[i]=resultn[i].vcpu;
        result3[i]=resultn[i].memory_gib;
        result4[i]=resultn[i].storage;
        result5[i]=resultn[i].phy_processor;
        result6[i]=resultn[i].clock_spd;
        result7[i]=resultn[i].netwk_pef;            
    }
    var wstream = fs.createWriteStream('myOutput.txt');
    wstream.write(result1.toString()+"~"+result2.toString());       
    //var str = "Hi man"
    res.writeHead(200, {'Content-Type':'text/html'});
    //res.end(url_parsed.query.callback+'("'+resultn.toString()+'")'); 
    res.end(url_parsed.query.callback+'("'+result1.toString()+"~"+result2.toString()+"~"
    +result3.toString()+"~"+result4.toString()+"~"+result5.toString()+"~"+result6.toString()
    +"~"+result7.toString()+'")');      
}
});
}
}); 
}).listen(process.env.PORT || 8083);

});

我不知道如何在 node.js 服务器程序中接收数据，在接收到数据后我想处理该数据并将处理后的数据发送回同一个 HTML 页面。

Answer 1

如果您的 server.js 文件中有 app.use(bodyParser.urlencoded({ extended: true }));，那么您可以使用 bodyParser 从 ajax 中检索数据，如下所示:

app.post('/', function(req, res) { //your ajax should also post to url: '/'
    var region = req.body.region,
    os = req.body.os;
    // ...
});