Create Table
Create table tablename(sNo int , name nvarchar(250))
Set Primarykey
Create table tablename(sNo int identity Primary Key, name nvarchar(250))
Set ForeignKey
Create table tablename(sNo int identity Primary Key, name nvarchar(250) foreign key references secondTableName(columnName))
Insert query
Insert into tableName(colName1,colName1)values(value1,value2);
Update query
Update tableName set ColName1=value1,ColName1=value1 where Condition;
Delete query
Delete from tableName where condition; //condition -->columnName=value like this
Select query
Select * from tableName; //get all column value
Select tableName.columnName1,tableName.columnName1 from tableName; //get specified column only
Select * from tableName where Condition;// get all column value based on the condition
Select tableName.columnName1 from tableName; //get specified column value based on the condition
Inner join
Select tableName1.columnName1,tableName1.columnName2,tableName2.columnName from tableName1 inner join tableName1 on tableName1.columnName1=tableName2.columnName1
Left Outer join
Use this when you only want to return rows that have matching data in the left table, even if there's no matching rows in the right table.
SELECT * FROM TableName1 LEFT JOIN TableName2 AS ON TableName1.Id = TableName1.Id
Right Outer join
Use this when you only want to return rows that have matching data in the right table, even if there's no matching rows in the left table.
SELECT * FROM TableName1 RIGHT JOIN TableName2 ON TableName1.Id = TableName1.Id
Full Outer join
Use this when you want to all rows, even if there's no matching rows in the right table.
Count(*)
If we only want to see how many records are in a table (but not actually view those records), we could use COUNT(*). COUNT(*) returns everything - including null values and duplicates.
Sql Ordey By
Select * from Tablename Order By columnName
Between for date
select columnName from Tablename where DatecolumnName between '4/6/2010' and '4/10/2010'
Create table tablename(sNo int , name nvarchar(250))
Set Primarykey
Create table tablename(sNo int identity Primary Key, name nvarchar(250))
Set ForeignKey
Create table tablename(sNo int identity Primary Key, name nvarchar(250) foreign key references secondTableName(columnName))
Insert query
Insert into tableName(colName1,colName1)values(value1,value2);
Update query
Update tableName set ColName1=value1,ColName1=value1 where Condition;
Delete query
Delete from tableName where condition; //condition -->columnName=value like this
Select query
Select * from tableName; //get all column value
Select tableName.columnName1,tableName.columnName1 from tableName; //get specified column only
Select * from tableName where Condition;// get all column value based on the condition
Select tableName.columnName1 from tableName; //get specified column value based on the condition
Inner join
Select tableName1.columnName1,tableName1.columnName2,tableName2.columnName from tableName1 inner join tableName1 on tableName1.columnName1=tableName2.columnName1
Left Outer join
Use this when you only want to return rows that have matching data in the left table, even if there's no matching rows in the right table.
SELECT * FROM TableName1 LEFT JOIN TableName2 AS ON TableName1.Id = TableName1.Id
Right Outer join
Use this when you only want to return rows that have matching data in the right table, even if there's no matching rows in the left table.
SELECT * FROM TableName1 RIGHT JOIN TableName2 ON TableName1.Id = TableName1.Id
Full Outer join
Use this when you want to all rows, even if there's no matching rows in the right table.
SELECT * FROM TableName1 FULL JOIN
TableName2
ON
TableName1
.Id=
TableName2
.Id
Count(*)
If we only want to see how many records are in a table (but not actually view those records), we could use COUNT(*). COUNT(*) returns everything - including null values and duplicates.
SELECT COUNT(*) FROM Individual
Sql Ordey By
Select * from Tablename Order By columnName
Between for date
select columnName from Tablename where DatecolumnName between '4/6/2010' and '4/10/2010'
No comments:
Post a Comment