如何同步打开页面列表

Question

let downloadPageLinks = [];
fetchStreamingLinks.forEach(async (item) => {
    page = await browser.newPage();
    await page.goto(item, { waitUntil: "networkidle0" });
    const fetchDownloadPageLinks = await page.evaluate(() => {
        return loc4;
    });

    console.log(fetchDownloadPageLinks);
});

I have an array of links( fetchStreamingLinks ). Above function opens all the links simultaneously present in fetchDownloadPageLinks . Suppose the array contains 100 links then it opens all the 100 links simultaneously.

Now what I want to do is, open all the links one by one present in fetchStreamingLinks , perform some logic in page context's and close it then open next link.

Answer 1

.forEach() is not promise-aware so when you pass it an async callback, it doesn't pay any attention to the promise that it returns. Thus, it runs all your operations in parallel. .forEach() should be essentially considered obsolete these days, especially for asynchronous operations because a plain for loop gives you so much more control and is promise-aware (eg the loop will pause for an await ).

let downloadPageLinks = [];
for (let item of fetchStreamingLinks) {
    let page = await browser.newPage();
    await page.goto(item, { waitUntil: "networkidle0" });
    const fetchDownloadPageLinks = await page.evaluate(() => {
        return loc4;
    });
    await page.close();
    console.log(fetchDownloadPageLinks);
}

FYI, I don't know the puppeteer API really well, but you probably should close the page (as I show) when you're done with it to avoid pages stacking up as you process.