Question

我尝试从失败的请求和js错误中收集数据。

我正在使用以下网站：https://nitzani1.wixsite.com/marketing-automation/3rd-page

该网站向https://api.fixer.io/1latest发出请求，该请求返回状态代码404，

该页面还包含以下js错误：

"Uncaught (in promise) Fetch did not succeed"

我试图用下面的代码来捕获404和js错误，但是不能。不知道我做错了什么，有关如何解决它的想法？

const puppeteer = require('puppeteer');

function wait (ms) {
    return new Promise(resolve => setTimeout(() => resolve(), ms));
}

var  run = async () => {
    const browser = await puppeteer.launch({
        headless: false,
        args: ['--start-fullscreen']
    });

    page = await browser.newPage();

    page.on('error', err=> {
        console.log('err: '+err);
    });

    page.on('pageerror', pageerr=> {
        console.log('pageerr: '+pageerr);
    });

    page.on('requestfailed', err => console.log('requestfailed: '+err));

    collectResponse = [];

    await page.on('requestfailed', rf => {
        console.log('rf: '+rf);
    });

    await page.on('response', response => {
        const url = response.url();
        response.buffer().then(
            b => {
                // console.log(url+' : '+response.status())
            },
            e => {
                console.log('response err');
            }
        );
    });
    await wait(500);

    await page.setViewport({ width: 1920, height: 1080 });
    await page.goto('https://nitzani1.wixsite.com/marketing-automation/3rd-page', {
    });
};

run();

Answer 1

完整的有效答案是：

const puppeteer = require('puppeteer');

const run = async () => {
    const browser = await puppeteer.launch({
        headless: true
    });

    const page = await browser.newPage();
    // Catch all failed requests like 4xx..5xx status codes
    page.on('requestfailed', request => {
        console.log(`url: ${request.url()}, errText: ${request.failure().errorText}, method: ${request.method()}`)
    });
    // Catch console log errors
    page.on("pageerror", err => {
        console.log(`Page error: ${err.toString()}`);
    });
    // Catch all console messages
    page.on('console', msg => {
        console.log('Logger:', msg.type());
        console.log('Logger:', msg.text());
        console.log('Logger:', msg.location());

    });

    await page.setViewport({ width: 1920, height: 1080 });
    await page.goto('https://nitzani1.wixsite.com/marketing-automation/3rd-page', { waitUntil: 'domcontentloaded' });
    await page.waitFor(10000); // To be sure all exceptions logged and handled
    await browser.close();
};

run();

保存为.js文件并轻松运行。

Answer 2

当前的 puppeteer 8.0.0^ 在 message.text() 中的信息量非常少。所以我们需要从 JSHandle 获取错误的描述。请使用来自 JSHandle 对象

的完整描述性控制台错误检查此评论

检查这里的链接https://stackoverflow.com/a/66801550/9026103

Puppeteer无法捕捉失败的请求＆amp;错误

2 个答案: