从Unirest块返回的空响应

Question

我正试图用Unirest和Cheerio搜索谷歌的搜索结果。所以在这里，我试图得到这个刮过的标题数组。但是，当我在外部安慰它时，它什么也不返回，而当我在Unirest块中安慰它时，它返回给我数据。

这是我的代码：

const unirest = require('unirest')
const cheerio = require('cheerio')


var titles = []
unirest
.get('https://www.google.com/search?q=oxylabs')
.headers({'Accept': 'application/json', 'Content-Type': 'application/json'})
.proxy(proxy)//hided
.then((response) =>
{
    const $ = cheerio.load(response.body)

      $('.uEierd').each((i,el) =>
      {
        titles[i] = $(el)
        .find('.ZINbbc div.v5yQqb a.cz3goc div.CCgQ5 span')
        .text()

        
      })
})

for (let i = 0; i < titles.length; i++)
{
    
    console.log(titles[i]);
    
} 

Answer 1

您的代码无法工作，因为unirest是异步的，并且运行在for循环之后。代码以及在线IDE中的完整示例

const unirest = require("unirest");
const cheerio = require("cheerio");

function getData() {
  return new Promise((resolve, reject) => {
    const titles = [];
    unirest
      .get("https://www.google.com/search?q=oxylabs")
      .headers({
        "Accept": "application/json",
        "Content-Type": "application/json",
        "User-Agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/101.0.4951.54 Safari/537.36"
      })
      // .proxy(proxy) //hided
      .then((response) => {
        const $ = cheerio.load(response.body);

        $(".uEierd").each((i, el) => {
          titles[i] = $(el).find(".v0nnCb span").text();
        });
        resolve(titles);
      })
      .catch((error) => {
        console.error(error);
        reject(error);
      });
  });
}

function logData() {
  getData().then((titles) => {
    for (let i = 0; i < titles.length; i++) {
      console.log(titles[i]);
    }
  });
}

logData();

输出：

Oxylabs Premium Proxies - Residential Proxy Network
72M+ Residential IPs Network - 7 Day-Free Trial - Join Now

或者，您可以使用来自SerpApi的SerpApi。如果您不想知道如何解决captchas、旋转代理、从头创建解析器并维护它，API方法就更容易了。有关更多信息，请查看游乐场。

用法：

const SerpApi = require("google-search-results-nodejs");
const mySecret = process.env['API_KEY']     //your API key from serpapi.com 
const search = new SerpApi.GoogleSearch(mySecret);

const params = {
  engine: "google",                         // search engine 
  q: "oxylabs",                             // search query
  location: "Austin, Texas, United States", // location parameter
  google_domain: "google.com",              // google domain of the search
  gl: "us",                                 // contry of the search
  hl: "en",                                 // language of the search  
};

const getAdTitles = function (data) {
  const titles = [];
  const adResults = data.ads;
  adResults?.forEach((result) => {
    const { title } = result;
    titles.push(title);
  });
  console.log(titles);
};

search.json(params, getAdTitles);

输出：

[
  'Oxylabs Premium Proxies - Residential Proxy Network',
  '72M+ Residential IPs Network - 7 Day-Free Trial - Join Now'
]

免责声明，我为SerpApi工作。