Question

我想在木偶操纵者遇到任何错误时关闭页面，有时会翻页我试图加载崩溃的页面并且它不会调用.close（）;

(async () => {
const page = await browser.newPage();
await page.setViewport({width: resWidth, height: resHeight});
await page.goto(d["entities"]["urls"][0]["expanded_url"], {timeout :90000});
await page.screenshot({path: './resimdata/'+d['id']+'.png' ,fullPage: true});
await page.close();
})();

Answer 1

关于这个问题，木偶剧回购中存在一个问题/公关，这在类似情况下会有所帮助。

相关问题链接：https://github.com/GoogleChrome/puppeteer/issues/952

同时，你可以试试这个小小的黑客，如果PR版本在0.12+以上，我们不必担心以下代码。

(async() => {
  const browser = await puppeteer.launch({headless: false});
  const page = await browser.newPage();

    function handleClose(msg){
        console.log(msg);
        page.close();
        browser.close();
        process.exit(1);
    }

    process.on("uncaughtException", () => {
        handleClose(`I crashed`);
    });

    process.on("unhandledRejection", () => {
        handleClose(`I was rejected`);
    });

    await page.goto("chrome://crash");

})();

将输出如下内容，

▶ node app/app.js
I was rejected

Chrome puppeteer错误事件关闭页面

1 个答案: