Javascript异步函数永远不会返回

时间:2018-04-27 01:43:39

标签: javascript asynchronous redis sequelize.js

我正在尝试运行一个从Redis中提取并插入数据库的简单程序。这个程序似乎永远不会退出,有人能指出我正确的方向吗?

require('newrelic')

const client = require('./config/redis')
const models = require('./models')
const sequelize = require('sequelize')

async function flushTweets() {
  try {
    var tweets = await client.lrangeAsync('writer', 0, 100)
    client.ltrim('writer', 0, 100)
    if (tweets && tweets.length > 0) {
      tweets = tweets.map(tweet => JSON.parse(tweet))

      models.Tweet.bulkCreate(tweets, { returning: true });

      for (var i = 0; i < tweets.length; ++i) {
        models.User.update(
            { numTweets: sequelize.literal(`"Users"."numTweets" + 1`) },
            { where: { id: tweets[i].user.id }
        });
      }

    }
  } catch (e) {

  }
}

flushTweets();

0 个答案:

没有答案