Question

有人能告诉我从node.js Express发送帖子请求的最简单方法，包括如何传递和检索一些数据？我期待在PHP中类似于cURL。

Answer 1

var express = require('express');
var bodyParser = require('body-parser');
var app = express();
app.use(bodyParser.urlencoded({ extended: false }));
app.use(bodyParser.json())

var port = 9000;

app.post('/sample/put/data', function(req, res) {
    console.log('receiving data ...');
    console.log('body is ',req.body);
    res.send(req.body);
});

// start the server
app.listen(port);
console.log('Server started! At http://localhost:' + port);

为此，您的服务器必须是：

{{1}}

Answer 2

正如here所述的帖子请求：

var http = require('http');

var options = {
  host: 'www.host.com',
  path: '/',
  port: '80',
  method: 'POST'
};

callback = function(response) {
  var str = ''
  response.on('data', function (chunk) {
    str += chunk;
  });

  response.on('end', function () {
    console.log(str);
  });
}

var req = http.request(options, callback);
//This is the data we are posting, it needs to be a string or a buffer
req.write("data");
req.end();

Answer 3

您可以尝试这样：

var request = require('request');
request.post({ headers: {'content-type' : 'application/json'}
               , url: <your URL>, body: <req_body in json> }
               , function(error, response, body){
   console.log(body); 
});

Answer 4

我使用superagent，这与jQuery相似。

以下是docs

演示就像：

var sa = require('superagent');
sa.post('url')
  .send({key: value})
  .end(function(err, res) {
    //TODO
  });

Answer 5

在服务器端，代码如下：

app.post('/add', function(req, res){
  console.log(req.body);
  request.post(
    {
    url:'http://localhost:6001/add',
    json: {
      unit_name:req.body.unit_name,
      unit_price:req.body.unit_price
        },
    headers: {
        'Content-Type': 'application/json'
    }
    },
  function(error, response, body){
    // console.log(error);
    // console.log(response);
    console.log(body);
    res.send(body);
  });
  // res.send("body");
});

接收最终服务器代码中的代码如下：

app.post('/add', function(req, res){
console.log('received request')
console.log(req.body);
let adunit = new AdUnit(req.body);
adunit.save()
.then(game => {
res.status(200).json({'adUnit':'AdUnit is added successfully'})
})
.catch(err => {
res.status(400).send('unable to save to database');
})
});

模式仅是两个属性unit_name和unit_price。

如何从node.js Express发送POST请求？

5 个答案: