我有一个包含两个字段用户名和密码的表格。输入用户名后,将启用下一个按钮,当我单击它时,它会显示一个密码字段,一旦输入该用户名,它将再次启用下一个按钮。如何等待表单更新之间启用按钮?
我尝试了以下方法,一种被评论,而另一种则没有。两者都不适合我。
(async () => {
const browser = await puppeteer.launch({headless: false});
const page = await browser.newPage();
await page.goto('http://localhost:9000/start#!');
await page.type('#login-form-un-field', 'xxxx')
// await page.waitForTarget('#default-next-btn:not([disabled])')
await page.$eval('#default-next-btn:not([disabled])', elem => elem.click());
// const btnNext = await page.$('#default-next-btn');
// btnNext.click();
await page.type('login-form-passcode', '1234');
await page.click('#default-next-btn');
await browser.close();
})();```
Thanks for the help in advance.
Edit: the button is always present on the page. It is just disabled while form entries are being validated.
你可以使用await page.waitForSelector(selector)
=> docs
你的代码变为:
(async () => {
const browser = await puppeteer.launch({headless: false});
const page = await browser.newPage();
await page.goto('http://localhost:9000/start#!');
await page.type('#login-form-un-field', 'xxxx')
await page.waitForSelector('YOUR_SELECTOR_1')
await page.click('YOUR_SELECTOR_1')
await page.type('login-form-passcode', '1234');
await page.waitForSelector('YOUR_SELECTOR_2')
await page.click('YOUR_SELECTOR_2')
await browser.close();
})();
谢谢@ettakhi