[英]How to get body / json response from XHR request with Puppeteer
我想從我用 Puppeteer 抓取的網站獲取 JSON 數據,但我不知道如何取回請求的正文。 這是我嘗試過的:
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());
}
});
})()
這只是返回一個承諾掛起函數。 任何幫助都會很棒。
當response.json
返回一個 promise 時,我們需要等待它。
page.on('response', async (response) => {
if (response.url() == "https://capuk.org/ajax_search/capmoneycourses"){
console.log('XHR response received');
console.log(await response.json());
}
});
你能幫我提建議嗎? 如何提取所有對數組的json響應並從此函數中提取出來?
await page.on('response', async (response) => {
if (response.url()){
console.log('XHR response received');
let rest = await response.json();
}
});
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.