puppeteer在公开函数被调用后拍摄屏幕截图

问题描述:

我正在尝试使用puppeteer导航到页面,等待webapp达到特定状态,截屏并退出。当SPA处于我想截图的状态时,它会调用一个函数。我很难绕过异步JS代码。puppeteer在公开函数被调用后拍摄屏幕截图

我的代码如下:

const puppeteer = require('puppeteer'); 

const url = 'http://example.com/'; 
const width = 300; 
const height = 250; 
const path = '/vagrant/tmp/screenshot.jpg'; 

async function takeScreenshoot(url, width, height, path) { 
    "use strict"; 

    const browser = await puppeteer.launch(); 
    const page = await browser.newPage(); 
    await page.setViewport({width: width, height: height}); 

    await page.exposeFunction('interestingFunction', (async data => { 
     console.log('Interesting function has been called. Taking a screenshot now.'); 
     await page.screenshot({path: path}); 
     await browser.close(); 
    })); 

    await page.goto(url); 
} 

(async() => { 
    "use strict"; 
    await takeScreenshoot(url, width, height, path); 
})(); 

但是当我打电话screenshot.js,我得到一个未处理的承诺,说警告“会议闭幕最有可能的页面已被关闭。”

node --trace-warnings screenshot.js 

Interesting function has been called. Taking a screenshot now. 
(node:2572) Error: Protocol error (Runtime.evaluate): Session closed. Most likely the page has been closed. 
    at Session.send (/vagrant/node_modules/puppeteer/lib/Connection.js:167:29) 
    at Page._onConsoleAPI (/vagrant/node_modules/puppeteer/lib/Page.js:296:20) 
    at <anonymous> 
    at process._tickCallback (internal/process/next_tick.js:188:7) 
(node:2572) [DEP0018] DeprecationWarning: Unhandled promise rejections are deprecated. In the future, promise rejections that are not handled will terminate the Node.js process with a non-zero exit code. 
    at emitWarning (internal/process/promises.js:78:15) 
    at emitPendingUnhandledRejections (internal/process/promises.js:95:11) 
    at process._tickCallback (internal/process/next_tick.js:189:7) 

如果我删除线18 await browser.close(),没有警告,但是脚本永远不会完成。

现在,interstingFunction()做了一些更多,但它是安全的暴露给web应用程序的窗口。我只是试图举一个上面的小脚本的例子,但仍然失败。

我正在使用节点v8.5.0。

我接近这个错误吗?

+0

你'page.goto(URL)'调用似乎是错误的。 – nilobarp

+0

我按照https://github.com/GoogleChrome/puppeteer/blob/master/examples/custom-event.js中的示例首先公开该函数,然后转到该页面。另请注意,'await browser.close();'发生在应用程序加载后几秒钟内调用的回调函数中,所以它无关紧要。 移动'await page.goto(url);'并没有使警告消失。 – hjm27

我假设有一些函数的调用者,你不需要自己用评估来调用它。

Imo,browser.close应该被移动到最后并且去暴露函数之前。至于在文档中阅读exposeFunction如果返回一个承诺它将被等待。到底异步/等待是糖承诺

await page.exposeFunction('interestingFunction', (async data => { 
    console.log('Interesting function has been called. Taking a screenshot now.'); 
    await page.screenshot({path: path}); 
    await browser.close(); 
})); 

await page.goto(url); 

这就是我的全部例子

const url = 'http://example.com/'; 
const width = 300; 
const height = 250; 
const path = './screenshot.jpg'; 

async function takeScreenshoot(url, width, height, path) { 
    const browser = await puppeteer.launch(); 
    const page = await browser.newPage(); 

    await page.goto(url, {waitUntil: 'load'}); 

    await page.exposeFunction('jjj', (async() => { 
     console.log('Interesting function has been called. Taking a screenshot now.'); 
     return await page.screenshot({path: path}); 
    })); 

    await page.evaluate(async() => { 
     const myHash = await window.jjj(); 
    }); 

    await browser.close(); 
    } 

(async() => { 
    await takeScreenshoot(url, width, height, path); 
})(); 
+0

非常感谢您的时间!是的,我自己并没有调用“interestingFunction()”。不幸的是,您提出的解决方案会返回而无需等待公开的函数至少被调用一次,因此不会进行任何截图。这就是在原始示例中试图关闭浏览器内的浏览器的原因。 – hjm27