如何使用 Python 在 Firestore 中下载大型集合而不会出现 503 错误？

Question

尝试使用 python 计算 firestore 集合中的文档数量。当我使用时db.collection('xxxx").stream()，出现以下错误：

 503 The datastore operation timed out, or the data was temporarily unavailable.

大约进行到一半。它工作正常。这是代码：

    docs = db.collection(u'theDatabase').stream()
    count = 0
    for doc in docs:
        count += 1
    print (count)

每次我在大约 73,000 条记录中收到 503 错误时。有谁知道如何克服 20 秒超时？

Answer 1

尽管 Juan 的答案适用于基本计数，但如果您需要来自 Firebase 的更多数据，而不仅仅是id（其常见用例是不通过 GCP 的数据完全迁移），递归算法会占用您的内存。

所以我拿了 Juan 的代码，把它转换成标准的迭代算法。希望这可以帮助某人。

limit = 1000  # Reduce this if it uses too much of your RAM
def stream_collection_loop(collection, count, cursor=None):
    while True:
        docs = []  # Very important. This frees the memory incurred in the recursion algorithm.

        if cursor:
            docs = [snapshot for snapshot in
                    collection.limit(limit).order_by('__name__').start_after(cursor).stream()]
        else:
            docs = [snapshot for snapshot in collection.limit(limit).order_by('__name__').stream()]

        for doc in docs:
            print(doc.id)
            print(count)
            # The `doc` here is already a `DocumentSnapshot` so you can already call `to_dict` on it to get the whole document.
            process_data_and_log_errors_if_any(doc)
            count = count + 1

        if len(docs) == limit:
            cursor = docs[limit-1]
            continue

        break


stream_collection_loop(db_v3.collection('collection'), 0)

Answer 2

尝试使用递归函数来批量检索文档并将其保持在超时范围内。这是一个基于delete_collections 片段的示例：

from google.cloud import firestore

# Project ID is determined by the GCLOUD_PROJECT environment variable
db = firestore.Client()


def count_collection(coll_ref, count, cursor=None):

    if cursor is not None:
        docs = [snapshot.reference for snapshot
                in coll_ref.limit(1000).order_by("__name__").start_after(cursor).stream()]
    else:
        docs = [snapshot.reference for snapshot
                in coll_ref.limit(1000).order_by("__name__").stream()]

    count = count + len(docs)

    if len(docs) == 1000:
        return count_collection(coll_ref, count, docs[999].get())
    else:
        print(count)


count_collection(db.collection('users'), 0)