javascript - 如何在自调用的异步函数中插入参数？

Question

我有一个带有发布请求的 Express 设置。我尝试将 req 添加到 (async (req, res):

router.post('/search', (req, res) => {
  ;(async (req, res) => {
    const browser = await puppeteer.launch()
    const page = await browser.newPage()
    await page.goto(`https://www.google.com/search?tbm=bks&q=%22this+is%22`)
    const result = await page.evaluate(() => {
      console.log('CLAUSESS:', req.body.clauses)
      const clauses = req.body.clauses
      return clauses.map(clause => clause.textContent)
    })
    result.join('\n')
    await browser.close()
    res.send(result)
  })()
})

但是，我仍然收到此错误:

(node:5757) UnhandledPromiseRejectionWarning: Unhandled promise rejection (rejection id: 1): Error: Evaluation failed: ReferenceError: req is not defined at :2:32

这样做的正确方法是什么？

Answer 1

你的 linter(如果有的话)应该提示说你在那个 IIFE 上隐藏了变量。我将其重写为

router.post('/search', (req, res) => {
  ;(async () => {
    const browser = await puppeteer.launch()
    const page = await browser.newPage()
    await page.goto(`https://www.google.com/search?tbm=bks&q=%22this+is%22`)
    const result = await page.evaluate(() => {
      console.log('CLAUSESS:', req.body.clauses)
      const clauses = req.body.clauses
      return clauses.map(clause => clause.textContent)
    })
    result.join('\n')
    await browser.close()
    res.send(result)
  })()
})

让它使用来自路由器的 req 和 res 而不是它自己的阴影空变量。