Question

我正在尝试改进firestore get函数，我有类似的东西：

    return admin.firestore().collection("submissions").get().then(
        async (x) => {
            var toRet: any = [];
            for (var i = 0; i < 10; i++) {
                try {
                    var hasMedia = x.docs[i].data()['mediaRef'];
                    if (hasMedia != null) {
                        var docData = (await x.docs[i].data()) as MediaSubmission;

                        let submission: MediaSubmission = new MediaSubmission();
                        submission.author = x.docs[i].data()['author'];
                        submission.description = x.docs[i].data()['description'];

                        var mediaRef = await admin.firestore().doc(docData.mediaRef).get();
                        submission.media = mediaRef.data() as MediaData;
                        toRet.push(submission);
                    }
                }
                catch (e) {
                    console.log("ERROR GETTIGN MEDIA: " + e);
                }
            }
            return res.status(200).send(toRet);
        });

第一次获得很好，但性能最差：

var mediaRef = await admin.firestore().doc(docData.mediaRef).get();

我认为这是因为电话没有批量处理。

是否可以批量访问mediaRefs数组以提高性能？

基本上我有一个文档集合，这些文档的外部引用由一个指向单独集合中的路径的字符串存储，并且已经证明获取这些引用很慢。

Answer 1

这个怎么样？我做了一些重构以使用更多的等待/异步代码，希望我的评论很有帮助。

主要思想是使用Promise.all并等待所有mediaRefs检索

async function test(req, res) {
  // get all docs
  const { docs } = await admin
    .firestore()
    .collection('submissions')
    .get();

  // get data property only of docs with mediaRef
  const datas = await Promise.all(
    docs.map(doc => doc.data()).filter(data => data.mediaRef),
  );

  // get all media in one batch - this is the important change
  const mediaRefs = await Promise.all(
    datas.map(({ mediaRef }) =>
      admin
        .firestore()
        .doc(mediaRef)
        .get(),
    ),
  );

  // create return object
  const toRet = datas.map((data: MediaSubmission, i) => {
    const submission = new MediaSubmission();
    submission.author = data.author;
    submission.description = data.description;

    submission.media = mediaRefs[i].data() as MediaData;

    return submission;
  });
  return res.status(200).send(toRet);
}

批量获取DocumentReferences？

1 个答案: