如何同步向mongo插入数据(Nodejs,Express)

时间:2016-10-22 11:34:31

标签: node.js mongodb express asynchronous synchronous

使用node(express)

将数据插入mongo db时遇到问题

我的代码如下所示:

router.get('/data/:section/:sort', function(req, res, next) {
//Deleting old data always before writing new
//Image.remove().exec();
var section = req.params.section;
var sort = req.params.sort;
//Link to Igmur API
var url = 'https://api.imgur.com/3/gallery/'+section+'/'+sort+'/1'; //1 at the end is used to get more than 60 images(gives only 60 without it)
    request.get({
    url: url,
    method: 'GET',
    headers: {
        'Authorization': 'Client-Id XXXXXXXXXXXXX'
    }}, function(e, r, body){
var metadata = JSON.parse(body);
for(var i = 0; i<100; i++){
        var image = new Image(metadata.data[i])
        image.save(function(err, result){
        });
};res.render('index', { title: 'SearchAPI' });});});

问题是我只能在循环中看到大约20个对象而不是100个。 所有因为节点只是在save方法完成之前向前跳转。 我怎样才能解决这个问题?提前谢谢

2 个答案:

答案 0 :(得分:2)

使用承诺库,例如Q

您基本上需要做的是等待所有save方法完成。使用等待执行所有内容的Q.all方法。

var Q = require('q');

var promiseArr = [];

for(var i = 0; i<100; i++){
        var imgDefer = Q.defer();
        var image = new Image(metadata.data[i])
        image.save(function(err, result){
             if(err)imgDefer.reject(err);
             else imgDefer.resolve()
        });
       promiseArr.push(imgDefer);
}
Q.all(promiseArr).then (function (){
    res.render('index', { title: 'SearchAPI' });
})

答案 1 :(得分:0)

一次又一次地调用数据库是一种错误的做法,最好使用create并一次传递所有内容。看看下面的代码片段。希望对您有所帮助。

router.get('/data/:section/:sort', function(req, res, next) {
  //Deleting old data always before writing new
  //Image.remove().exec();
  var section = req.params.section;
  var sort = req.params.sort;
  //Link to Igmur API
  var url = 'https://api.imgur.com/3/gallery/'+section+'/'+sort+'/1'; //1 at the end is used to get more than 60 images(gives only 60 without it)
  request.get({
    url: url,
    method: 'GET',
    headers: {
    'Authorization': 'Client-Id XXXXXXXXXXXXX'
  }}, function(e, r, body){
    var metadata = JSON.parse(body);

    Image.create(metadata.data,(err, result)=>{
      //do operations here
      res.render('index', { title: 'SearchAPI' });
    })

    /* WRONG PRACTICE */
    // for(var i = 0; i<100; i++){
    //   var image = new Image(metadata.data[i])
    //   image.save(function(err, result){

    //   });
    // };

  });
});