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Node.js将变量从事件处理程序返回到父函数

[英]Node.js Return variable from event handler to parent function

I need the function scrape to return the value obtained in the page.on("request") event handler. 我需要函数scrape返回在page.on("request")事件处理程序中获得的值。

async function scrape(url) {
        const page = await browser.newPage();
        await page.setRequestInterception(true);

        page.on("request", async(request) => {
            return "fish"
        }
        await page.goto(url)
    }

Currently: 目前:

const ans = await scrape(url)
console.log(ans)
"undefined'

Expected: 预期:

const ans = await scrape(url)
console.log(ans)
"fish"

you'll need to return promise that is resolved when you see event that you are waiting 您需要返回看到正在等待的事件时已解决的承诺

const matchRequest = request => request.method() === 'GET'; // your filter
async function scrape(url) {
  return new Promise(resolve => {
    const page = await browser.newPage();
    // not sure what your logic is, but if you don't need to cancel or modify requests/resposes you probably don't need interception
    // await page.setRequestInterception(true);
    page.on("response", async(response) => {
        if (matchRequest(response.request())) {
           resolve(response.buffer());
        }
    }
    await page.goto(url);
  })
}

const body = await scrape('https://example.com');

Try follow like: 尝试按照以下步骤操作:

async function scrape(url) {
    let sendRequest = []
    const browser = await puppeteer.launch()
    const page = await browser.newPage()
    await page.setRequestInterception(true)
    page.on('request', request => {
      request.continue()
      sendRequest.push('fish')
    })
    await page.goto(url)
    return sendRequest
}

Using request.continue() for continues request with optional request overrides. 使用request.continue()通过可选的请求覆盖继续请求。 To use this, request interception should be enabled with page.setRequestInterception . 要使用此功能,应使用page.setRequestInterception启用请求拦截。 Exception is immediately thrown if the request interception is not enabled. 如果未启用请求拦截,则会立即引发异常。

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