我一直在阅读 mongodocs 并阅读教程。
出于测试目的,我使用 python/pymongo 制作了这个小脚本。
基本上,3个db,每个有2个集合,“订单”和“产品”
在products集合中插入相同的商品,每个order集合中下一个订单,区别是 产品与订单“相关”的方式,一种是链接的,另一种是嵌入的,最后一种是引用的。 最后,更新产品中的一个值并打印它。
我的问题是: 1 - “关系”处理得好吗?重新表述:这是引用/嵌入/链接的完成方式吗?
2-“嵌入”订单是否应该反射(reflect)商品价格的变化、反射(reflect)产品的变化,还是必须由脚本完成?
# -*- coding: utf-8 *-*
from pymongo import *
from bson import *
import sys
def connect():
try:
auxcon = Connection('localhost', 27017)
print "Connection: %s database_names: %s" % (
auxcon, auxcon.database_names())
return auxcon
except error.ConnectionFailure as cf:
print "Conection error: %s" % cf
sys.exit(0)
def newDB(db_name, conx):
try:
ldb = database.Database(conx, db_name + "_linked")
edb = database.Database(conx, db_name + "_embedded")
rdb = database.Database(conx, db_name + "_referenced")
return ldb, edb, rdb
except (error.TypeError, error.InvalidName) as err:
print "Error: %s" % err
sys.exit(0)
def newCollections(db_name):
try:
colprod = db_name.create_collection("products")
colord = db_name.create_collection("orders")
return colprod, colord
except errors.CollectionInvalid as err:
print "Collection alrready exists %s" % err
return db_name["products"], db_name["orders"]
def insertProducts(colname):
product = {"name": "Tablet", "price": 200, "desc": "Android tablet"}
product2 = {"name": "Phone", "price": 100, "desc": "Samsung Phone"}
try:
p1 = colname.insert(product, safe=True)
p2 = colname.insert(product2, safe=True)
return p1, p2
except errors.OperationFailure as err:
print "Error inserting %s" % err
return None, None
def updateProducts(colname):
for product in colname.find({}):
product["price"] = product["price"] * 110 / 100
colname.save(product)
def printProducts(colname):
print "DATABASE: %s COLLECTION: %s" % (
colname.database.name, colname.name)
for product in colname.find({}):
print product
def findOrders(colname):
print "DATABASE: %s COLLECTION: %s" % (
colname.database.name, colname.name)
for order in colname.find({}):
for key, value in order.items():
print "%s : %s" % (key, value)
if __name__ == "__main__":
cx = connect()
try:
cx.drop_database("carritodb_linked")
cx.drop_database("carritodb_embedded")
cx.drop_database("carritodb_referenced")
except errors.TypeError as err:
print "Error %s" % err
dbl, dbe, dbr = newDB("carritodb", cx)
licoll = newCollections(dbl)
emcoll = newCollections(dbe)
recoll = newCollections(dbr)
lp1, lp2 = insertProducts(licoll[0])
ep1, ep2 = insertProducts(emcoll[0])
rp1, rp2 = insertProducts(recoll[0])
linkedOrder = {"userInfo": "Alex Martinavarro", "items_chart": [lp1, lp2]}
linkedOrder = licoll[1].insert(linkedOrder, safe=True)
embeddedOrder = {"userInfo": "Alex Martinavarro", "items_chart": []}
embeddedOrder = emcoll[1].insert(embeddedOrder, safe=True)
embeddedOrder = emcoll[1].find_one(embeddedOrder)
for product in emcoll[0].find({}):
embeddedOrder["items_chart"].append(product)
emcoll[1].save(embeddedOrder)
p1ref = dbref.DBRef(recoll[0].name, rp1)
p2ref = dbref.DBRef(recoll[0].name, rp2)
referencedOrder = {"userInfo": "Alex Martinavarro", "items": [p1ref, p2ref]}
referencedOrder = recoll[1].insert(referencedOrder, safe=True)
print "INSERTED PRODUCTS"
printProducts(licoll[0])
printProducts(emcoll[0])
printProducts(recoll[0])
print "ORDERS"
findOrders(licoll[1])
findOrders(emcoll[1])
findOrders(recoll[1])
"""UPDATING"""
updateProducts(licoll[0])
updateProducts(emcoll[0])
updateProducts(recoll[0])
print "UPDATED PRODUCTS"
printProducts(licoll[0])
printProducts(emcoll[0])
printProducts(recoll[0])
print "ORDERS AFTER UPDATE"
findOrders(licoll[1])
findOrders(emcoll[1])
findOrders(recoll[1])
最佳答案
我是从 Google 群组中回答的,所以如果答案与上面发布的答案不一致,请原谅我。
“2- 如果“嵌入”订单反射(reflect)了商品价格的变化,则反射(reflect)该变化。”
永远不要想象用户下了订单,然后在发货之前突然意识到价格无明显原因发生了变化。您应该在订购时复制价格并将其放入嵌入式产品列表中。这是大多数(如果不是全部)电子商务网站所做的。
"在products集合中插入相同的商品,每个order集合中下一个订单,区别是 产品与订单“相关”的方式,一种是链接的,另一种是嵌入的,最后一种是引用的。”
这三个中的任何一个都有效,具体取决于嵌入的完成方式。如果您将订购的产品嵌入到订单中,那就没问题,将订单嵌入到产品中并不是一个好的解决方案。由于我怀疑是否有人会订购足够的产品来超过 16Mb 限制,因此您应该是安全的。
希望这有帮助,
关于python - MongoDB 链接/嵌入/引用,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/9143934/