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javascript-如何等待按钮被启用并单击puppeteer?

(javascript - How to wait for a button to be enabled and click with puppeteer?)

发布于 2020-01-23 15:58:00

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.

Questioner
DJR
Viewed
0
3,693 2020-11-11 00:43:07

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();
})();