mongodb
robomongo
结果排序
db.products.find({},
{'brand_name':1, 'series':1, 'product_name':1,'volume':1,"review_summary.count":1})
.sort({'brand_name':-1,'product_name':-1})
pymongo
基本用法
from pymongo import MongoClient
db = MongoClient()['databas']
reviews = db.products.find({'review_summary':{'$exists':1}})
for rev in reviews:
print(rev['product_name'])
ObjectId 可以多,不可以少。。
from pymongo import MongoClient
from bson.objectid import ObjectId
db = MongoClient()['test']
result = db.product.find_one()
print(ObjectId(result['_id']))
myid = "53799b0595ca313f2aad412a"
print(ObjectId(myid))
#所以ObjectId嵌套没有问题的哟
mongo aggregate!!!
mongo export import dump restore
mongodump -h xxx.xxx.xx.x -d jdmilk -o jd
mongorestore -d jdmilk -directoryperdb jd/jdmilk
mongoexport -h xxx.xxx.xx.x -d dev -c mycollection -o mcollection.json
mongoimport -d dev -c mycollection --file mcollection.json
mongo update
from bson.objectid imoprt ObjectId
db.collection.update({'_id': ObjectId("xxxxxxxx"), {"$set":{'target': 'value'}}})
如果在robomongo下对所有记录做set:
db.products.update({}, {$set: {'industry': 'beer'}}, {multi:true})
db.products.update({}, {"$unset": {"industry": 1}}, multi=True)
fucking things in robomongo
db.collection.insert({'value': '好-p'})
it will display 好R in robomongo’s GUI, but it is correct inside mongo..
mongo upsert
r = p_item.update({'source':'taobao'}, {'$set':{'price':21}}, upsert=True)
##PyMongo
data = []
for src_, ids in item['_productsIds'].items():
data.extend(list(env.MONGO.product_item_changes.find(
{'source': src_, 'product_id': {'$in': ids}, 'date': date})))