使用Express / Node编写的嵌套Google DataStore查询无法获取预期数据

Question

我想在datastore中实现嵌套的Node/express查询。父查询是get上的简单kind请求，子查询将根据上面提取的父kind结果的特定列值聚合其他kind的特定列。请查看以下代码以获得更好的洞察力。

app.get("/api/venues/", (req, res, next) => {
  const query = datastore
  .createQuery('venue');

  // parent query run here
  query.run().then(([venuesList]) => {
  venuesList.forEach(
    venue => {
      startDate = moment(new Date()).format('L');
      endDate = moment(new Date()).startOf('week').format('L');
      const queryVenueInvoice = datastore
      .createQuery('invoices')
      .filter('targetacntkey', '=', venue.userid);

      // child query run here
      queryVenueInvoice.run().then(([invoicesList]) => {
        const filteredInvoiceList = invoicesList.filter( invoice =>
          (new Date(invoice.timestamp).toISOString().split('T')[0])
          <= startDate && (new Date(i.timestamp).toISOString().split('T')[0]) >= (endDate));

          venue['weeklySummary'] = filteredInvoiceList.reduce((sum, invoice) => {
            return sum + invoice.totalamount; }, 0);
        })

      venue['venueKey'] = venue[datastore.KEY]
    }
    );
  // venuesList.forEach(venue => console.log(venue));
  res.status(200).json(
    {
      message: "Request was processed successfully!",
      venues: venuesList
    }
  );
})
})

我能够以venuesList填充的venueKey数组取回响应。但是我无法在响应中看到聚合属性weeklySummary。我在这里失踪了什么？任何人都指导我，所以看到预期的结果？

Answer 1

看起来你正试图在你的场地中做异步操作List.forEach（）

因此，响应将在完成之前发送。你需要使用类似的东西

async function asyncForEach(array, callback) {
  for (let index = 0; index < array.length; index++) {
    await callback(array[index], index, array);
  }
}

app.get("/api/venues/", (req, res, next) => {
  const query = datastore
  .createQuery('venue');

    query.run().then(async ([venuesList]) => {
      await asyncForEach(venuesList, async(venue) => {
        startDate = moment(new Date()).format('L');
        endDate = moment(new Date()).startOf('week').format('L');
        const queryVenueInvoice = datastore
          .createQuery('invoices')
          .filter('targetacntkey', '=', venue.userid);

        // child query run here
        await queryVenueInvoice.run().then(([invoicesList]) => {...

另外我会将res.status（200）...放在query.run（）中。然后（...和asyncForEach之外...（我希望这会有帮助）