如何在我的分析结果以及其他被报废的数据中添加URL

Question

我想一次刮擦许多网站。因此，我希望将URL与被废弃的其他数据一起写入结果。但是我不知道如何。

const puppeteer = require('puppeteer');

(async () => {
const browser = await puppeteer.launch()
const page = await browser.newPage()
await page.goto('https://www.amazon.com/')


 await page.waitForTimeout( 10000 );
const localStorageData = await page.evaluate(() => {
let json = {};
for (let i = 0; i < localStorage.length; i++) {
  const key = localStorage.key(i);
  json[key] = localStorage.getItem(key);
}
return json;
});
const data = {};
for (let entry of Object.entries(data)) {
data[entry.key] = entry.value;
}
console.log(localStorageData)

await browser.close()
})()

Answer 1

您可以将使用的URL添加到JSON：

const puppeteer = require("puppeteer");

(async () => {
  const browser = await puppeteer.launch();
  const page = await browser.newPage();

  const url = "https://www.amazon.com/";
  await page.goto(url);

  await page.waitForTimeout(10000);
  const localStorageData = await page.evaluate((url) => {
    let json = {};
    for (let i = 0; i < localStorage.length; i++) {
      const key = localStorage.key(i);
      json[key] = localStorage.getItem(key);
    }
    return json;
  });
  localStorageData["url"] = url;
  const data = {};
  for (let entry of Object.entries(data)) {
    data[entry.key] = entry.value;
  }
  console.log(localStorageData);

  await browser.close();
})();