进入，查看，使用
进入 mongodb show dbs use mydatabase show collections db["mycol"].find({},{_id:0,name:1}) # 第一个参数为条件，第二个参数为显示结果设置
db["mycol"].inseret({"key":"value",title:"tutorial",name:"jkmiao"})
MongoDB语法与现有关系型数据库SQL语法比较 MongoDB语法 MySql语法 db.test.find({'name':'foobar'})<==> select * from test where name='foobar' db.test.find() <==> select *from test db.test.find({'ID':10}).count()<==> select count(*) from test where ID=10 db.test.find().skip(10).limit(20)<==> select * from test limit 10,20 db.test.find({'ID':{$in:[25,35,45]}})<==> select * from test where ID in (25,35,45) db.test.find().sort({'ID':-1}) <==> select * from test order by IDdesc db.test.distinct('name',{'ID':{$lt:20}}) <==> select distinct(name) from testwhere ID<20 db.test.group({key:{'name':true},cond:{'name':'foo'},reduce:function(obj,prev){prev.msum+=obj.marks;},initial:{msum:0}}) <==> select name,sum(marks) from testgroup by name db.test.find('this.ID<20',{name:1}) <==> select name from test whereID<20 db.test.insert({'name':'foobar','age':25})<==>insertinto test ('name','age') values('foobar',25) db.test.remove({}) <==> delete * from test db.test.remove({'age':20}) <==> delete test where age=20 db.test.remove({'age':{$lt:20}}) <==> elete test where age<20 db.test.remove({'age':{$lte:20}}) <==> delete test where age<=20 db.test.remove({'age':{$gt:20}}) <==> delete test where age>20 db.test.remove({'age':{$gte:20}})<==> delete test where age>=20 db.test.remove({'age':{$ne:20}}) <==> delete test where age!=20 db.test.update({'name':'foobar'},{$set:{'age':36}})<==> update test set age=36 where name='foobar' db.test.update({'name':'foobar'},{$inc:{'age':3}})<==> update test set age=age+3 where name='foobar'
[转]MongoDB基本使用
http://www.cnblogs.com/TankMa/archive/2011/06/08/2074947.html