Question

嗨，我正在尝试为足球比赛建立一个奇怪的展示网站
我试图从betfair获取数据，然后正常工作，然后将其插入我的数据库问题是该功能仅插入第一项，我该怎么办？

const request = require('request');
const cheerio = require('cheerio');
const fs = require('fs');
const util = require('util');
const appendFile = util.promisify(fs.appendFile);
const async = require("async");

var mysql      = require('mysql');
var connection = mysql.createConnection({
  host     : 'localhost',
  user     : 'root',
  password : '',
  database : 'betfair'
});

connection.connect(function(err) {
  if (err) {
    console.error('error connecting: ' + err.stack);
    return;
  }

  console.log('connected as id ' + connection.threadId);
});

var url = 'https://www.betfair.com/sport/football';

var customHeaderRequest = request.defaults({
    headers: {'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/57.0.2987.133 Safari/537.36'}
})

customHeaderRequest.get(url, async function(err, resp, body){
  try {
    let $ = cheerio.load(body);
    let links = $('.section-list .section:nth-child(2) .event-list li').toArray();
    for (let link of links) {

        const id = $(link).find('.event-information .avb-col-markets a:nth-child(1)').attr('href');
        const idx = id.replace('/sport/football/event?eventId=', '');

        const home = $(link).find('.event-information div:nth-child(3) div a div span.team-name:nth-child(1)').text().trim();
        const away = $(link).find('.event-information div:nth-child(3) div a div span.team-name:nth-child(2)').text().trim();

        const x1 = $(link).find('.event-information .avb-col-markets .market-3-runners .runner-list li:nth-child(1) a .ui-runner-price').text().trim();
        const x = $(link).find('.event-information .avb-col-markets .market-3-runners .runner-list li:nth-child(2) a .ui-runner-price').text().trim();
        const x2 = $(link).find('.event-information .avb-col-markets .market-3-runners .runner-list li:nth-child(3) a .ui-runner-price').text().trim();

        const over = $(link).find('.event-information .avb-col-markets .market-2-runners .runner-list li:nth-child(1) a .ui-runner-price').text().trim();
        const under = $(link).find('.event-information .avb-col-markets .market-2-runners .runner-list li:nth-child(2) a .ui-runner-price').text().trim();
        if (home && away) {

          connection.query("INSERT INTO bf (id, home, away, x1, x, x2, over, under) VALUES ('"+idx+"', '"+home+"', '"+away+"', '"+x1+"', '"+x+"', '"+x2+"', '"+over+"', '"+under+"')", function (err, result) {
            if (err) throw err;
          });
        }
    }
  } catch(e) {
  }
});

我已经尝试了很多想法，但是在nodejs中不是很好，我正在寻找您的帮助
谢谢

Answer 1

您没有提到错误是什么，但是我想您真的不需要ID，因此将其删除就可以了。

P.S。您的代码还有很多问题，最重要的是sql查询。

您应该使用knex或至少将其更改为：

INSERT INTO bf (home, away, x1, x, x2, over, under) VALUES (?, ?, ?, ?, ?, ?, ?)", [home, away, x1, x, x2, over, under], function (err, result) {

Source

Nodejs Mysql仅插入一项

1 个答案: