Tag: sql programming

SQL: Check if table exists

On By Ben LarsonIn MS SQL Server4 Comments

To check if a table exists in SQL Server, you can use the INFORMATION_SCHEMA.TABLES table.

Running the following code, produces the results below:

USE SANDBOX 
GO;

Select *
from INFORMATION_SCHEMA.TABLES
GO;

1

You can use this table with an IF THEN clause do determine how your query responds whether or not a table exists.

IF EXISTS (SELECT * FROM INFORMATION_SCHEMA.TABLES
           WHERE TABLE_NAME = N'employee_id')
BEGIN
  PRINT 'Yes'
END

2

IF EXISTS (SELECT * FROM INFORMATION_SCHEMA.TABLES
           WHERE TABLE_NAME = N'employee_ids')
BEGIN
  PRINT 'Yes'
END

ELSE

BEGIN
  PRINT 'No'
End

3

One of the more common uses I find for this when I need to create a table in a script. I want to make sure a table with same name doesn’t already exist, or my query will fail. So I write a query like the one below.

IF EXISTS (SELECT * FROM INFORMATION_SCHEMA.TABLES
           WHERE TABLE_NAME = N'employee_id')
BEGIN
  drop table employee_id
END

ELSE

BEGIN
 CREATE TABLE employee_id (
        emp_nm nvarchar(30) not null,
             emp_id nvarchar(8),
             b_emp_id nvarchar(8)
             PRIMARY KEY(emp_nm) );

INSERT INTO employee_id
       (emp_nm, emp_id)
VALUES
       ('Bob', 'A1234567'),
       ('Lisa', 'A1234568')

INSERT INTO employee_id
       (emp_nm, b_emp_id)
VALUES
       ('Priyanka', 'B1234567');
End

SQL: Case Statement

On By Ben LarsonIn MS SQL ServerLeave a comment

The Case Statement is SQL’s version of IF/THEN logic statements.

In this example, I am using the AdventureWorks2012 database, which is free to download. You do not need this to follow the lesson, but if you want to play along, follow this link: SQL: SELECT Statement

 

Let’s start by looking at our data, using the following Select Statement:

SELECT 
 [NationalIDNumber]
 ,[LoginID]
 ,[JobTitle]
 ,[BirthDate]
 ,[MaritalStatus]
 ,[Gender]
 ,[HireDate] 
 FROM [AdventureWorks2012].[HumanResources].[Employee]

Here is the output

sqlagg3

Now what if you don’t really care about the particular employee birthday, you simply want to know who is over 40 and who is 40 or younger.

To do that, we need a mechanism to determine an age above or below 40 from the birth date. In most programming languages, you would simply use an If / Then statement. In SQL, we need to use a Case Statement.

The Case Syntax is simple. Let’s assume a variable A, set to either 1 or 0 representing True or False.

Case
     when A = 1 Then "True"
     when B = 0 Then "False"
End

Note the statement is started with the keyword Case and closed with the keyword End.

In the middle, the logic is controlled using When and Then. When <condition> Then  <Action>

If needed, you can append an Else to your statement to cover conditions not specified in your statements.

Case
     when A = 1 Then "True"
     when B = 0 Then "False"
     else "Neither" 
End

 

 

Now, let’s apply it to our case above. Notice the Case statement is put right into my Select statement just as if it were a common field.

select loginID,
 [NationalIDNumber]
 ,[LoginID]
 ,[JobTitle]
 ,
 case 
    when BirthDate < '1977-01-01' then 'Over40'
    when BirthDate >= '1977-01-01' then 'Under40'
 end
 ,[MaritalStatus]
 ,[Gender]
 ,[HireDate]
 FROM [AdventureWorks2012].[HumanResources].[Employee]

Here are my results. You will note a new column with values Over40 or Under40

2017-01-09_11-29-42.jpg

But what does that column mean? Let’s name it to make it clear.  After the End keyword add as Over40

 select loginID,
 [NationalIDNumber]
 ,[LoginID]
 ,[JobTitle],
 case 
        when BirthDate < '1977-01-01' then 'Over40'
         when BirthDate >= '1977-01-01' then 'Under40'
 end as Over40,
 [MaritalStatus]
 ,[Gender]
 ,[HireDate]
 FROM [AdventureWorks2012].[HumanResources].[Employee]

Here, now the column has a name.

2017-01-09_11-32-09.jpg