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[英]How do I use puppeteer to take full screenshots of several websites?
[英]How can you take several hundred screenshots with Puppeteer?
我有一個包含數百個 html 文件的文件夾。 我需要為每一個截圖,我想我可以在我的 Gulp 任務中使用 Puppeteer。
function takeScreenshots() {
return gulp.src(path.join(paths.dest.stage, "**/*.html"))
.pipe(tap(async (file) => {
const browser = await puppeteer.launch({ headless: true });
const page = await browser.newPage();
await page.setViewport({
width: 1024,
height: 768,
deviceScaleFactor: 1,
});
await page.goto("file://" + file.path);
await page.screenshot({ path: path.join(path.dirname(file.path), path.basename(file.basename, ".html") + ".jpg"), quality: 10 });
await browser.close();
}));
}
當我運行它時,我的電腦風扇繼續運轉,事情被鎖定,我收到這條消息:
(node:85389) MaxListenersExceededWarning: Possible EventEmitter memory leak detected. 11 exit listeners added to [process]. Use emitter.setMaxListeners() to increase limit
(Use `node --trace-warnings ...` to show where the warning was created)
我認為它試圖同時做所有的截圖,但我不知道如何告訴它拍攝一張,完成,然后移動到下一張等等。
我進行了快速搜索,但沒有找到任何同步管道的方法。所以我寫了一個解決方法。
1-首先,您將使用 pipe 和 gulp-tap 將所有文件添加到“文件”數組
2-然后在“結束”事件訂閱
3-然后編寫依賴於文件數組的業務邏輯
function takeScreenshots() {
const files = [];
return gulp
.src(path.join(paths.dest.stage, "**/*.html"))
.pipe(
tap((file) => {
files.push(file);
})
)
.on("end", async () => {
const browser = await puppeteer.launch({ headless: true });
const page = await browser.newPage();
await page.setViewport({
width: 1024,
height: 768,
deviceScaleFactor: 1,
});
for (let i = 0, len = files.length; i < len; i++) {
const file = files[i];
await page.goto("file://" + file.path);
await page.screenshot({
path: path.join(
path.dirname(file.path),
path.basename(file.basename, ".html") + ".jpg"
),
quality: 10,
});
}
await browser.close();
});
}
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