[英]puppeteer newly loaded page after the form submit
I am using puppeteer to measure the performance of the pages and their features (this is still in the initial phase, since I am just starting). 我正在使用puppeteer来衡量页面及其功能的性能(这仍处于初始阶段,因为我才刚刚开始)。 Here's an example I am using:
这是我正在使用的示例:
const puppeteer = require('puppeteer');
(async () => {
const browser = await puppeteer.launch({headless: false});
const page = await browser.newPage();
await page._client.send('Performance.enable');
await page.goto('https://xxxx.com');
await page.focus("#formulario input#username");
await page.type("test2222@gmail.com", {delay: 200});
await page.focus("#formulario input#inputPassword3");
await page.type("pass", {delay: 200});
const loginForm = await page.$("#formulario");
await loginForm.evaluate(loginForm => loginForm.submit());
loginForm._client.send('Performance.getMetrics');
//BELOW DOESN'T WORK BECAUSE THE PAGE IS GONE
await page.click("span.show-filters");
await browser.close();
})();
How to access the newly loaded page after a form is submitted? 提交表单后如何访问新加载的页面? I mean the user has logged in and how can we get a handle of that newly loaded page?
我的意思是用户已登录,我们如何获得该新加载页面的句柄?
I've got it working by waiting for the page to load first. 我通过等待页面首先加载来使其工作。 Then I was able to click the correct element.
然后,我能够单击正确的元素。
So I've added the following line: 因此,我添加了以下行:
await page.waitForNavigation();
before this one: 在此之前:
await page.click(".show-filters");
I must say that the documentation is rather scarce for the tool :( 我必须说,该工具的文档非常少:(
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