[NoSQL] basic 코드 구현

Select * from emplyees where empno=7369

db.employees.find({empno:7369})

{ "_id" : "E1", "empno" : 7369, "ename" : "SMITH", "job" : "CLERK", "hiredate" : "17-12-1980", "sal" : 800, "deptno" : 20 }

Select ename from emlpoyees where empno=7900

db.employees.find({empno:7900}, {ename:1})

{ "_id" : "E2", "ename" : "JAMES" }

Select empno, ename from employees where empno > 7500 and empno <=7600

db.employees.find({empno:{$gt:7500, $lte:7600}}, {empno:1, ename:1, _id:0})

{ "empno" : 7566, "ename" : "JONES" }

{ "empno" : 7521, "ename" : "WARD" }

Select empno from employees where empno = 7782 or empno=7844

db.employees.find($or:[{empno:7782}, {empno:7844}]}, {empno:1, _id:0})

{ "empno" : 7782 }

{ "empno" : 7844 }

Select count(*) from employees

db.employees.count()

14

Select count(*) from employees where empno > 7900

방법1

db.employees.find({empno:{$gt:7900}}).count()

2
방법2

db.employees.count({empno:{$gt:7900}})

2

Select distinct deptno from employees

db.employees.distinct("deptno")

[ 10, 20, 30 ]

Select ename, job from employees where deptno=10 prder by ename desc

db.employees.find({deptno:10}, {ename:1, job:1, _id:0}).sort({ename:-1})

{ "ename" : "PRESIDENT", "job" : "CEO" }

{ "ename" : "CLERK", "job" : "CLERK" }

{ "ename" : "CLARK", "job" : "MANAGER" }

Select sum(salary) from employees

db.employees.aggregate([ $group: {_id:null, total_salary: {$sum: "$sal"}}}])

{ "_id" : null, "total_salary" : 29025 }

Select deptno, avg(salary) from employees group by deptno order by deptno

db.employees.aggregate([{ $group: {_id: "$deptno", avg_sal: {$avg:"$sal"}}}, {$sort: {_id:1}}])

{ "_id" : 10, "avg_sal" : 2916.6666666666665 }

{ "_id" : 10, "avg_sal" : 2175 }

{ "_id" : 10, "avg_sal" : 1566.6666666666667 }