如何使用 Puppeteer 从 XHR 请求中获取 body/json 响应
问题描述：
我想从我用 Puppeteer 抓取的网站获取 JSON 数据，但我不知道如何取回请求的正文.这是我尝试过的:
I want to get the JSON data from a website I'm scraping with Puppeteer, but I can't figure how to get the body of the request back. Here's what I've tried:
const puppeteer = require('puppeteer')
const results = [];
(async () => {
const browser = await puppeteer.launch({
headless: false
})
const page = await browser.newPage()
await page.goto("https://capuk.org/i-want-help/courses/cap-money-course/introduction", {
waitUntil: 'networkidle2'
});
await page.type('#search-form > input[type="text"]', 'bd14ew')
await page.click('#search-form > input[type="submit"]')
await page.on('response', response => {
if (response.url() == "https://capuk.org/ajax_search/capmoneycourses"){
console.log('XHR response received');
console.log(response.json());
}
});
})()
这只是返回一个承诺挂起的函数.任何帮助都会很棒.
This just returns a promise pending function. Any help would be great.
答
由于 response.json 返回一个 promise，我们需要等待它.
As response.json returns a promise we need to await it.
page.on('response', async (response) => {
if (response.url() == "https://capuk.org/ajax_search/capmoneycourses"){
console.log('XHR response received');
console.log(await response.json());
}
});