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ViewsCómo agregar URL en el resultado de mi análisis junto con otros datos que se descartan
Quiero raspar muchos sitios web a la vez. Por lo tanto, preferiría tener la URL escrita en el resultado junto con los demás datos que se desechan. Pero no sé cómo. 
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() })()
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Juan Pablo Isaza
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Simplemente puede agregar la URL que usa a su 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(); })();
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Juan Pablo Isaza
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