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            MongoDB語法與現有關系型數據庫SQL語法比較

            MongoDB語法??????????????????????????????????MySql語法

            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

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